John Bookstaver St. Charles Community College Cottleville, MO Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Gases John Bookstaver St. Charles Community College Cottleville, MO  2009, Prentice-Hall, Inc.

Characteristics of Gases Unlike liquids and solids, gases expand to fill their containers; are highly compressible; have extremely low densities.  2009, Prentice-Hall, Inc.

Force = mass x acceleration Units of Force Force = mass x acceleration  2009, Prentice-Hall, Inc.

Force = mass x acceleration Units of Force Force = mass x acceleration F = m x a  2009, Prentice-Hall, Inc.

Force = mass x acceleration Units of Force Force = mass x acceleration F = m x a m = mass  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say: A body’s mass = 68 kg  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say: A body’s mass = 68 kg a = 9.8 m/s2  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say: A body’s mass = 68 kg a = 9.8 m/s2 F = 68 kg x 9.8 m/s2 = 670 kg m/s2  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say: A body’s mass = 68 kg a = 9.8 m/s2 F = 68 kg x 9.8 m/s2 = 670 kg m/s2 1 N = kg m/s2  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say: A body’s mass = 68 kg a = 9.8 m/s2 F = 68 kg x 9.8 m/s2 = 670 kg m/s2 1 N = kg m/s2 F = 670 N  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say Your on the Moon: A body’s mass = 68 kg  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say Your on the Moon: A body’s mass = 68 kg A = 1.6 m/s2  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say Your on the Moon: A body’s mass = 68 kg a = 1.6 m/s2 F = m x a = 68 kg x 1.6 m/s2 = 108.8 kg m/s2 = 110 N  2009, Prentice-Hall, Inc.

Units of Force Force = mass x acceleration F = m x a m = mass a = acceleration due to gravity Let’s Say Your on the Moon: A body’s mass = 68 kg a = 1.6 m/s2 F = m x a = 68 kg x 1.6 m/s2 = 108.8 kg m/s2 = 110 N Weight = 108.8 N x 1 lb/4.47 N = 24 lbs.  2009, Prentice-Hall, Inc.

Pressure F P = A Pressure is the amount of force applied to an area. Atmospheric pressure is the weight of air per unit of area.  2009, Prentice-Hall, Inc.

Units of Pressure Pressure = Force/Area  2009, Prentice-Hall, Inc.

Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/ss /( 0.5 m x 0.5 m)  2009, Prentice-Hall, Inc.

Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/s2 /( 0.25 m2)  2009, Prentice-Hall, Inc.

P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2)  2009, Prentice-Hall, Inc.

P = F/A = 670 kg m/s2 /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/s2 /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2  2009, Prentice-Hall, Inc.

P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 The SI unit of Pressure is the Pascal (Pa) …  2009, Prentice-Hall, Inc.

P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 The SI unit of Pressure is the Pascal (Pa) … 1 Pa = 1kg / m/s2 = 1 N/m2  2009, Prentice-Hall, Inc.

P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 Units of Pressure Pressure = Force/Area Let’s Say: P = F/A = 670 kg m/ss /( 0.25 m2) = 2700 kg/(m/s2) = 2700 N/m2 The SI unit of Pressure is the Pascal (Pa) … 1 Pa = 1kg / m/s2 = 1 N/m2 So … the pressure exerted is … 2700 Pa  2009, Prentice-Hall, Inc.

P = F/A = 670 kg m/ss /(1 x 10-4 m2) = 6.7 x 106 N/m2 = 6.7 x 106 Pa Units of Pressure Let’s Say you are wearing high heels … P = F/A = 670 kg m/ss /(1 x 10-4 m2) = 6.7 x 106 N/m2 = 6.7 x 106 Pa  2009, Prentice-Hall, Inc.

The “standard atmosphere” is = 101,325 Pa Units of Pressure The “standard atmosphere” is = 101,325 Pa  2009, Prentice-Hall, Inc.

Units of Pressure The “standard atmosphere” is = 101,325 Pa Pressure is commonly expressed in units of atmospheres (atm)  2009, Prentice-Hall, Inc.

Units of Pressure The “standard atmosphere” is = 101,325 Pa Pressure is commonly expressed in units of atmospheres (atm), torr,  2009, Prentice-Hall, Inc.

The “standard atmosphere” is = 101,325 Pa Units of Pressure The “standard atmosphere” is = 101,325 Pa Pressure is commonly expressed in units of atmospheres (atm), torr, mmHg  2009, Prentice-Hall, Inc.

The “standard atmosphere” is = 101,325 Pa Units of Pressure The “standard atmosphere” is = 101,325 Pa Pressure is commonly expressed in units of atmospheres (atm), torr, mmHg 1 atm = 760 torr = 760 mmHg  2009, Prentice-Hall, Inc.

The “standard atmosphere” is = 101,325 Pa Units of Pressure The “standard atmosphere” is = 101,325 Pa Pressure is commonly expressed in units of atmospheres (atm), torr, mmHg 1 atm = 760 torr = 760 mmHg  2009, Prentice-Hall, Inc.

Interconnecting Units of Pressure Let’s Say the Pressure exerted by a gas is 0.985 atm …  2009, Prentice-Hall, Inc.

Interconnecting Units of Pressure Let’s Say the Pressure exerted by a gas is 0.985 atm … Calculate the Pressure in torr …  2009, Prentice-Hall, Inc.

Interconnecting Units of Pressure Let’s Say the Pressure exerted by a gas is 0.985 atm … Calculate the Pressure in torr … 0.985 atm x 760 torr/1 atm = 749 torr (749 mmHg)  2009, Prentice-Hall, Inc.

Interconnecting Units of Pressure Let’s Say the Pressure exerted by a gas is 0.985 atm … Calculate the Pressure in torr … 0.985 atm x 760 torr/1 atm = 749 torr (749 mmHg) Calculate the Pressure in pascals …  2009, Prentice-Hall, Inc.

Interconnecting Units of Pressure Let’s Say the Pressure exerted by a gas is 0.985 atm … Calculate the Pressure in torr … 0.985 atm x 760 torr/1 atm = 749 torr (749 mmHg) Calculate the Pressure in pascals … 0.985 atm x 101,325 Pa/1 atm = 98,300 Pa  2009, Prentice-Hall, Inc.

Units of Pressure Pascals Bar 1 Pa = 1 N/m2 1 bar = 105 Pa = 100 kPa  2009, Prentice-Hall, Inc.

Units of Pressure mm Hg or torr These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. Atmosphere 1.00 atm = 760 torr  2009, Prentice-Hall, Inc.

Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.  2009, Prentice-Hall, Inc.

Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to 1.00 atm 760 torr (760 mm Hg) 101.325 kPa  2009, Prentice-Hall, Inc.

Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.  2009, Prentice-Hall, Inc.

As P and V are inversely proportional A plot of V versus P results in a curve. Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k  2009, Prentice-Hall, Inc.

Pressure x Volume = Constant Applications of Boyle’s Law Pressure x Volume = Constant  2009, Prentice-Hall, Inc.

Pressure x Volume = Constant Applications of Boyle’s Law Pressure x Volume = Constant P x V = k  2009, Prentice-Hall, Inc.

Pressure x Volume = Constant Applications of Boyle’s Law Pressure x Volume = Constant P x V = k If PV is constant for a gas at constant temperature, then …  2009, Prentice-Hall, Inc.

Pressure x Volume = Constant Applications of Boyle’s Law Pressure x Volume = Constant P x V = k If PV is constant for a gas at constant temperature, then … P1V1 = P2V2  2009, Prentice-Hall, Inc.

Pressure x Volume = Constant Applications of Boyle’s Law Pressure x Volume = Constant P x V = k If PV is constant for a gas at constant temperature, then … P1V1 = P2V2 P1 = Initial Pressure P2 = Final Pressure V1 = Initial Volume V2 = Final Volume  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature?  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known:  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known: P1 = 1.3 atm V1 = 27 L P2 = 3.9 atm  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known: Unknown: P1 = 1.3 atm V2 = ? L V1 = 27 L P2 = 3.9 atm  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known: Unknown: P1 = 1.3 atm V2 = ? L V1 = 27 L P2 = 3.9 atm Solution: P1V1 = P2V2  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known: Unknown: P1 = 1.3 atm V2 = ? L V1 = 27 L P2 = 3.9 atm Solution: P1V1 = P2V2 V2 = P1V1/P2  2009, Prentice-Hall, Inc.

V2 = ? L || (1.3 atm) (27 L) / (3.9 atm) = 9 L Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known: Unknown: P1 = 1.3 atm V2 = ? L V1 = 27 L P2 = 3.9 atm Solution: P1V1 = P2V2 V2 = P1V1/P2 V2 = ? L || (1.3 atm) (27 L) / (3.9 atm) = 9 L  2009, Prentice-Hall, Inc.

Applications of Boyle’s Law A gas which has pressure of 1.3 atm occupies a volume of 27 L. What volume will the gas occupy if the pressure is increased to 3.9 atm at constant temperature? Known: Unknown: P1 = 1.3 atm V2 = ? L V1 = 27 L P2 = 3.9 atm Solution: P1V1 = P2V2 V2 = P1V1/P2 V2 = ? L || (1.3 atm) (27 L) / (3.9 atm) = 9 L Does Your Answer Make Sense?  2009, Prentice-Hall, Inc.

The Units of P x V  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units …  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known:  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown:  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution:  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa (3.56 x 106 L Pa ) (1 x 103 m3 / 1 L)(1 kg m-1s-2) / 1 Pa = 3.56 x 103 kg m2 s-2  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa (3.56 x 106 L Pa ) (1 x 103 m3 / 1 L)(1 kg m-1s-2) / 1 Pa = 3.56 x 103 kg m2 s-2 3.56 x 103 kg m2 s-2 = 3.56 x 103 kg m s-2 x m  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa (3.56 x 106 L Pa ) (1 x 103 m3 / 1 L)(1 kg m-1s-2) / 1 Pa = 3.56 x 103 kg m2 s-2 3.56 x 103 kg m2 s-2 = 3.56 x 103 kg m s-2 x m = 3.56 x 103 N m  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa (3.56 x 106 L Pa ) (1 x 103 m3 / 1 L)(1 kg m-1s-2) / 1 Pa = 3.56 x 103 kg m2 s-2 3.56 x 103 kg m2 s-2 = 3.56 x 103 kg m s-2 x m = 3.56 x 103 N m = 3.56 x 103 J  2009, Prentice-Hall, Inc.

kg m2 s-2 is the SI unit of energy = I J The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa (3.56 x 106 L Pa ) (1 x 103 m3 / 1 L)(1 kg m-1s-2) / 1 Pa = 3.56 x 103 kg m2 s-2 3.56 x 103 kg m2 s-2 = 3.56 x 103 kg m s-2 x m = 3.56 x 103 N m = 3.56 x 103 J kg m2 s-2 is the SI unit of energy = I J  2009, Prentice-Hall, Inc.

The Units of P x V The PV constant in the previous example was 35.1 L atm (1.3 atm x 27 L). Convert this constant to: L Pa and the fundamental SI units … Known: PV = 35.1 L atm 101,325 Pa = 1 kg m-1 s-2 = atm 1 L = 1000 mL = 1000 cm3 = (10 cm)3 = (0.1 m)3 = 1.0 x 10-3m3 Unknown: 35.1 atm = ? L Pa = ? Kg m2s-2 Solution: (35.1 L atm x 101,325 Pa / atm = 3.56 x 106 L Pa (3.56 x 106 L Pa ) (1 x 103 m3 / 1 L)(1 kg m-1s-2) / 1 Pa = 3.56 x 103 kg m2 s-2 3.56 x 103 kg m2 s-2 = 3.56 x 103 kg m s-2 x m = 3.56 x 103 N m = 3.56 x 103 J kg m2 s-2 is the SI unit of energy = I J PV, then, is really a unit of energy!  2009, Prentice-Hall, Inc.

Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. i.e., V T = k A plot of V versus T will be a straight line.  2009, Prentice-Hall, Inc.

Charles’s Law At constant pressure, the volume of a confined gas is directly proportional to the temperature (K) of the gas.  2009, Prentice-Hall, Inc.

Volume / Temperature = Constant Charles’s Law At constant pressure, the volume of a confined gas is directly proportional to the temperature (K) of the gas. Volume / Temperature = Constant  2009, Prentice-Hall, Inc.

Volume / Temperature = Constant Charles’s Law At constant pressure, the volume of a confined gas is directly proportional to the temperature (K) of the gas. Volume / Temperature = Constant V/T = K  2009, Prentice-Hall, Inc.

Volume / Temperature = Constant Charles’s Law At constant pressure, the volume of a confined gas is directly proportional to the temperature (K) of the gas. Volume / Temperature = Constant V/T = K If the temperature is changed ( P = k), the volume will change so that the ratio of volume to temperature will remain constant.  2009, Prentice-Hall, Inc.

Volume / Temperature = Constant Charles’s Law At constant pressure, the volume of a confined gas is directly proportional to the temperature (K) of the gas. Volume / Temperature = Constant V/T = K If the temperature is changed ( P = k), the volume will change so that the ratio of volume to temperature will remain constant. V1/T1 = V2/T2 ( @ P = k)  2009, Prentice-Hall, Inc.

Volume / Temperature = Constant Charles’s Law At constant pressure, the volume of a confined gas is directly proportional to the temperature (K) of the gas. Volume / Temperature = Constant V/T = K If the temperature is changed ( P = k), the volume will change so that the ratio of volume to temperature will remain constant. V1/T1 = V2/T2 (@ P = k) V1 = Volumeinitial V2 = Volumefinal T1 = Temperatureinitial T2 = Temperaturefinal  2009, Prentice-Hall, Inc.

Charles’s Law  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm?  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known:  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: T1 = 30 ‘C + 273 K = 303 K  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: T1 = 30 ‘C + 273 K = 303 K V1 = 0.842 L  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: T1 = 30 ‘C + 273 K = 303 K V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K Solution:  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K Solution: V1 / T1 = V2 / T2  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K Solution: V1 / T1 = V2 / T2  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K Solution: V1 / T1 = V2 / T2 V2 = V1T2 / T1  2009, Prentice-Hall, Inc.

Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K Solution: V1 / T1 = V2 / T2 V2 = V1T2 / T1 V2 = ? L || (0.842 L) (333 K) / (303 K) = 0.925 L  2009, Prentice-Hall, Inc.

Does this answer make sense? Charles’s Law A gas at 30 ‘C and 1 atm occupies a volume of 0.842 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: Unknown: T1 = 30 ‘C + 273 K = 303 K V2 = ? L V1 = 0.842 L T1 = 60 ‘C + 273 K = 333 K Solution: V1 / T1 = V2 / T2 V2 = V1T2 / T1 V2 = (0.842 L) (333 K) / (303 K) = 0.925 L Does this answer make sense?  2009, Prentice-Hall, Inc.

Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn  2009, Prentice-Hall, Inc.

Volume = Constant x number of moles (constant T, P) Avogadro’s Law Volume = Constant x number of moles (constant T, P)  2009, Prentice-Hall, Inc.

Volume = Constant x number of moles (constant T, P) Avogadro’s Law Volume = Constant x number of moles (constant T, P) V = kn (constant T,P)  2009, Prentice-Hall, Inc.

Volume = Constant x number of moles (constant T, P) Avogadro’s Law Volume = Constant x number of moles (constant T, P) V = kn (constant T,P) If the moles of gas is tripled (@ constant T & P), the volume will also triple …  2009, Prentice-Hall, Inc.

Volume = Constant x number of moles (constant T, P) Avogadro’s Law Volume = Constant x number of moles (constant T, P) V = kn (constant T,P) If the moles of gas is tripled (@ constant T & P), the volume will also triple … V1 / n1 = V2 / n2 (constant T, P)  2009, Prentice-Hall, Inc.

Volume = Constant x number of moles (constant T, P) Avogadro’s Law Volume = Constant x number of moles (constant T, P) V = kn (constant T,P) If the moles of gas is tripled (@ constant T & P), the volume will also triple … V1 / n1 = V2 / n2 (constant T, P) V1 = Volumeinitial V2 = Volumeinitial n1 = number of molesinitial n2 = number of molesfinal  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known:  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known: V1 = 5.20 L  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known: V1 = 5.20 L n1 = 0.436 moles  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known: V1 = 5.20 L n1 = 0.436 moles n2 = 0.436 + 1.27 = 1.706 moles  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known: Unknown: V1 = 5.20 L V2 = ? L n1 = 0.436 moles n2 = 0.436 + 1.27 = 1.706 moles  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known: Unknown: V1 = 5.20 L V2 = ? L n1 = 0.436 moles n2 = 0.436 + 1.27 = 1.706 moles Solution:  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 0.436 moles of a gas. If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be? Known: Unknown: V1 = 5.20 L V2 = ? L n1 = 0.436 moles n2 = 0.436 + 1.27 = 1.706 moles Solution: V2 = ? L || (5.20 L) (1.706 moles) / (0.436 moles) = 20.3 L  2009, Prentice-Hall, Inc.

Ideal-Gas Equation V  nT P So far we’ve seen that V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law) Combining these, we get V  nT P  2009, Prentice-Hall, Inc.

Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.  2009, Prentice-Hall, Inc.

Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship then becomes or PV = nRT  2009, Prentice-Hall, Inc.

Pressure x Volume = # moles of gas x a constant x temperature Ideal Gas Law Pressure x Volume = # moles of gas x a constant x temperature  2009, Prentice-Hall, Inc.

Pressure x Volume = # moles of gas x a constant x temperature Ideal Gas Law Pressure x Volume = # moles of gas x a constant x temperature PV = nRT  2009, Prentice-Hall, Inc.

Ideal Gas Law PV = nRT P = Pressure (atm) V = Volume (L) Pressure x Volume = # moles of gas x a constant x temperature PV = nRT P = Pressure (atm) V = Volume (L) N = number of moles (mol) R = 0.08206 L atm / mol K T = Temperature (K)  2009, Prentice-Hall, Inc.

Ideal Gas Law PV = nRT P = Pressure (atm) V = Volume (L) Pressure x Volume = # moles of gas x a constant x temperature PV = nRT P = Pressure (atm) V = Volume (L) N = number of moles (mol) R = 0.08206 L atm / mol K T = Temperature (K) The relationship assumes that a gas will behave ideally …  2009, Prentice-Hall, Inc.

Ideal Gas Law PV = nRT P = Pressure (atm) V = Volume (L) Pressure x Volume = # moles of gas x a constant x temperature PV = nRT P = Pressure (atm) V = Volume (L) N = number of moles (mol) R = 0.08206 L atm / mol K T = Temperature (K) The relationship assumes that a gas will behave ideally … @ Higher Temperatures and Lower Pressures  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert?  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: V = 12.9 L  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: V = 12.9 L T = 12 ‘C + 273 ‘C = 285 K  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: V = 12.9 L T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: V = 12.9 L T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: Unknown: V = 12.9 L P = ? atm T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: Unknown: V = 12.9 L P = ? atm T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol Solution:  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: Unknown: V = 12.9 L P = ? atm T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol Solution: PV = nRT  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: Unknown: V = 12.9 L P = ? atm T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol Solution: PV = nRT P = nRT / V  2009, Prentice-Hall, Inc.

P = ? at,m || (0.614 mol) (0.08206 L atm / mol K) (285 K) / 12.9 L Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: Unknown: V = 12.9 L P = ? atm T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol Solution: PV = nRT P = nRT / V P = ? at,m || (0.614 mol) (0.08206 L atm / mol K) (285 K) / 12.9 L  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 0.614 moles of gas at 12 ‘C occupies a volume of 12.9 L. What pressure does the gas exert? Known: Unknown: V = 12.9 L P = ? atm T = 12 ‘C + 273 ‘C = 285 K n = 0.614 mol R = 0.08206 L atm / K mol Solution: PV = nRT P = nRT / V P = ? atm || (0.614 mol) (0.08206 L atm / mol K) (285 K) / 12.9 L = 1.11 atm  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C?  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known:  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: P1 = 0.848 atm V1 = 7.0 L T1 = 4 ‘C + 273 ‘C = 277 K  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: P1 = 0.848 atm P2 = 1.52 atm V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution:  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R P1V1 = n1RT1 and P2V2 = n2RT2  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R P1V1 = n1RT1 and P2V2 = n2RT2 P1V1/T1 = n1R and P2V2/T2 = n2R2  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R P1V1 = n1RT1 and P2V2 = n2RT2 P1V1/T1 = n1R and P2V2/T2 = n2R So … P1V1/T1 = P2V2/T2  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R P1V1 = n1RT1 and P2V2 = n2RT2 P1V1/T1 = n1R and P2V2/T2 = n2R V2 = P1V1T1 / P2T1  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R P1V1 = n1RT1 and P2V2 = n2RT2 P1V1/T1 = n1R and P2V2/T2 = n2R V2 = ? L || P1V1T1 / P2T1 = (0.848 atm)(7.0 L)(284 K) / (1.52 atm) (277 K)  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 0.848 atm and 4 ‘C occupies a volume of 7.0 L. What volume will the gas occupy if the pressure is increased to 1.52 atm and the temperature increased to 11 ‘C? Known: Unknown: P1 = 0.848 atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C + 273 ‘C = 277 K Solution: n1 = n2 R = R P1V1 = n1RT1 and P2V2 = n2RT2 P1V1/T1 = n1R and P2V2/T2 = n2R V2 = ? L || P1V1T1 / P2T1 = (0.848 atm)(7.0 L)(284 K) / (1.52 atm) (277 K) = 4.0 L  2009, Prentice-Hall, Inc.

Molar Volume of an Ideal Gas 1 mole of any ideal gas @ Standard Temperature & Pressure (STP) occupies 22.4 L …  2009, Prentice-Hall, Inc.

Molar Volume of an Ideal Gas 1 mole of any ideal gas @ Standard Temperature & Pressure (STP) occupies 22.4 L … STP Conditions …  2009, Prentice-Hall, Inc.

Molar Volume of an Ideal Gas 1 mole of any ideal gas @ Standard Temperature & Pressure (STP) occupies 22.4 L … STP Conditions … 1.000 atm (standard pressure) & 0 ‘C (standard temperature)  2009, Prentice-Hall, Inc.

Molar Volume of an Ideal Gas 1 mole of any ideal gas @ Standard Temperature & Pressure (STP) occupies 22.4 L … STP Conditions … 1.000 atm (standard pressure) & 0 ‘C (standard temperature) STP Means 0 ‘C & 1.000 atm  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP?  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known:  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP T = 273 K & P = 1.000 atm  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP T = 273 K & P = 1.000 atm R = 0.08206 L atm / K mol  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP T = 273 K & P = 1.000 atm R = 0.08206 L atm / K mol n = 1.18 mol  2009, Prentice-Hall, Inc.

What volume will 1.18 moles of O2 occupy at STP? Ideal Gas Law & STP What volume will 1.18 moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution:  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P  2009, Prentice-Hall, Inc.

Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) (0.08026 L atm / mol K) (273) / (1.00 atm)  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) (0.08026 L atm / mol K) (273) / (1.00 atm) = 26.4 L  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) (0.08026 L atm / mol K) (273) / (1.00 atm) = 26.4 L Alternative Approach …  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) (0.08026 L atm / mol K) (273) / (1.00 atm) = 26.4 L Alternative Approach … 22.4 L / 1 mol = V / 1.18 mol  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) (0.08026 L atm / mol K) (273) / (1.00 atm) = 26.4 L Alternative Approach … 22.4 L / 1 mol = V / 1.18 mol V = (22.4 L)(1.18 mol) / (1 mol)  2009, Prentice-Hall, Inc.

What volume will moles of O2 occupy at STP? Ideal Gas Law & STP What volume will moles of O2 occupy at STP? Known: @ STP Unknown: T = 273 K & P = 1.000 atm V = ? L R = 0.08206 L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) (0.08026 L atm / mol K) (273) / (1.00 atm) = 26.4 L Alternative Approach … 22.4 L / 1 mol = V / 1.18 mol V = (22.4 L)(1.18 mol) / (1 mol) = 26.4 L  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law Idea … Many gas law problems involve calculating the volume of a gas produced by the reaction of volumes of other gases …  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law Idea … Many gas law problems involve calculating the volume of a gas produced by the reaction of volumes of other gases … Solution … Relate moles of reactants to moles of products ...  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law Idea … Many gas law problems involve calculating the volume of a gas produced by the reaction of volumes of other gases … Solution … Relate moles of reactants to moles of products ... The ideal gas law will allow to use the following strategy …  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law Idea … Many gas law problems involve calculating the volume of a gas produced by the reaction of volumes of other gases … Solution … Relate moles of reactants to moles of products ... The ideal gas law will allow to use the following strategy … Strategy …  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law Idea … Many gas law problems involve calculating the volume of a gas produced by the reaction of volumes of other gases … Solution … Relate moles of reactants to moles of products ... The ideal gas law will allow to use the following strategy … Strategy … Ideal Gas Law Stoichiometry Ideal Gas Law Vreactants ------------> molesreactants -------------> molesproducts -------------> Vproducts  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g)  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed?  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known:  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Mass dry ice = 15.0 g CO2(s)  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Mass dry ice = 15.0 g CO2(s) T = 22’C + 273 ‘C = 295 K  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Mass dry ice = 15.0 g CO2(s) T = 22’C + 273 ‘C = 295 K P = 1.04 atm  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Unknown: Mass dry ice = 15.0 g CO2(s) Vballon = ? L T = 22’C + 273 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Unknown: Mass dry ice = 15.0 g CO2(s) Vballon = ? L T = 22’C + 273 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2 molar mass Ideal Gas Law Solution: (g CO2(s) -----------> mol CO2(g) --------------> Volume CO2(g))  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Unknown: Mass dry ice = 15.0 g CO2(s) Vballon = ? L T = 22’C + 273 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2 molar mass Ideal Gas Law Solution: (g CO2(s) -----------> mol CO2(g) --------------> Volume CO2(g))  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Unknown: Mass dry ice = 15.0 g CO2(s) Vballon = ? L T = 22’C + 273 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2 molar mass Ideal Gas Law Solution: (g CO2(s) -----------> mol CO2(g) --------------> Volume CO2(g)) moles CO2(s) = ? mol CO2(s) || (15.0 g CO2) (1 mol CO2 / 44.0 g CO2) = 0.341 mol CO2  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Unknown: Mass dry ice = 15.0 g CO2(s) Vballon = ? L T = 22’C + 273 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2 molar mass Ideal Gas Law Solution: (g CO2(s) -----------> mol CO2(g) --------------> Volume CO2(g)) moles CO2(s) = ? mol CO2(s) || (15.0 g CO2) (1 mol CO2 / 44.0 g CO2) = 0.341 mol CO2 Vballoon = ? L || nRT/P = (0.341 mol) (0.08206 L atm / K mol) (295 K) / 1.04 atm  2009, Prentice-Hall, Inc.

Reactions & the Ideal Gas Law A sample containing 15.0 g of dry ice (CO2(s)) is put into a balloon and allowed to sublime according to the following reaction: CO2(s) ----> CO2(g) What will the volume of the balloon be at 22 ‘C and 1.04 atm, after all of the dry ice has sublimed? Known: Unknown: Mass dry ice = 15.0 g CO2(s) Vballon = ? L T = 22’C + 273 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2 molar mass Ideal Gas Law Solution: (g CO2(s) -----------> mol CO2(g) --------------> Volume CO2(g)) moles CO2(s) = ? mol CO2(s) || (15.0 g CO2) (1 mol CO2 / 44.0 g CO2) = 0.341 mol CO2 Vballoon = ? L || nRT/P = (0.341 mol) (0.08206 L atm / K mol) (295 K) / 1.04 atm = 7.94 L  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation …  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g)  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C?  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP V H2(g) = 0.500 L H2(g)  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g)  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct)  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct) Ideal Mole Rato: mol H2 / mol O2 = 2/1 = 2  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct) IMR: mol H2 / mol O2 = 2/1 = 2 AMR: mol H2 / mol O2 = (0.500 L) (1 mol / 22.4 L) / (0.600 L) (1 mol / 22.4 L)  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct) IMR: mol H2 / mol O2 = 2/1 = 2 AMR: mol H2 / mol O2 = (0.500 L) (1 mol / 22.4 L) / (0.600 L) (1 mol / 22.4 L) = .022/.026 = .85  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct) IMR: mol H2 / mol O2 = 2/1 = 2 AMR: mol H2 / mol O2 = (0.500 L) (1 mol / 22.4 L) / (0.600 L) (1 mol / 22.4 L) = .022/.026 = .85 LR: H2  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct) IMR: mol H2 / mol O2 = 2/1 = 2 AMR: mol H2 / mol O2 = (0.500 L) (1 mol / 22.4 L) / (0.600 L) (1 mol / 22.4 L) = .022/.026 = .85 LR: H2 V H2O = ? L H2O || nRT/P = (0.0223 mol)(0.08206 L atm/mol K)(623 K)/(1.00 atm)  2009, Prentice-Hall, Inc.

Limiting Reagents and the Ideal Gas Law 0.500 L of H2(g) are reacted with 0.600 L of O2(g) @ STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? Known: @ STP Unknown: V H2(g) = 0.500 L H2(g) ; V O2(g) = 0.600 L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C + 273 ‘C = 623 K ; P = 1 atm Ideal Gas Law Limiting Reagent Ideal Gas Law Solution: (Vreactant --------------> molreactant --------------> molproduct --------------> Vproduct) IMR: mol H2 / mol O2 = 2/1 = 2 AMR: mol H2 / mol O2 = (0.500 L) (1 mol / 22.4 L) / (0.600 L) (1 mol / 22.4 L) = .022/.026 = .85 LR: H2 V H2O = ? L H2O || nRT/P = (0.0223 mol)(0.08206 L atm/mol K)(623 K)/(1.00 atm) = 1.41 L H2O  2009, Prentice-Hall, Inc.

Densities of Gases n P V = RT If we divide both sides of the ideal-gas equation by V and by RT, we get n V P RT =  2009, Prentice-Hall, Inc.

Densities of Gases n   = m P RT m V = We know that moles  molecular mass = mass n   = m So multiplying both sides by the molecular mass ( ) gives P RT m V =  2009, Prentice-Hall, Inc.

Densities of Gases P RT m V = d = Mass  volume = density So, P RT m V = d = Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.  2009, Prentice-Hall, Inc.

Molecular Mass P d = RT dRT P  = We can manipulate the density equation to enable us to find the molecular mass of a gas: P RT d = Becomes dRT P  =  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas …  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known:  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: T = 34 ‘C + 273 ‘C = 307 K  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: T = 34 ‘C + 273 ‘C = 307 K P = 1.75 atm  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: T = 34 ‘C + 273 ‘C = 307 K P = 1.75 atm D = 3.4 g/L  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: Unknown: T = 34 ‘C + 273 ‘C = 307 K MM = ? g/mol P = 1.75 atm D = 3.4 g/L  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: Unknown: T = 34 ‘C + 273 ‘C = 307 K MM = ? g/mol P = 1.75 atm D = 3.4 g/L Solution: MM = dRT/P  2009, Prentice-Hall, Inc.

MM = ? g/mol || (3.4 g/L) (0.08206 L atm/mol K)(307 K) / (1.75 atm) Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: Unknown: T = 34 ‘C + 273 ‘C = 307 K MM = ? g/mol P = 1.75 atm D = 3.4 g/L Solution: MM = dRT/P MM = ? g/mol || (3.4 g/L) (0.08206 L atm/mol K)(307 K) / (1.75 atm)  2009, Prentice-Hall, Inc.

Density & Molar Mass A gas at 34 ‘C and 1.75 atm has a density of 3.4 g/L. Calculate the molar mass of the gas … Known: Unknown: T = 34 ‘C + 273 ‘C = 307 K MM = ? g/mol P = 1.75 atm D = 3.4 g/L Solution: MM = dRT/P MM = ? g/mol || (3.4 g/L) (0.08206 L atm/mol K)(307 K) / (1.75 atm) = 48.9 g/mol  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, Ptotal = P1 + P2 + P3 + Pn … = (n1 + n2 + … + nn) [ RT/V]  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, Ptotal = P1 + P2 + P3 + Pn … = (n1 + n2 + … + nn) [ RT/V] The Key Problem Solving Strategy …  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, Ptotal = P1 + P2 + P3 + Pn … = (n1 + n2 + … + nn) [ RT/V] The Key Problem Solving Strategy … Use the Ideal Gas Law to Interconvert between Pressure and Moles of Each Gas!  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP?  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known:  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Initial … VHe = 2.0 LHe ;  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Find the nHe …  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Find the nHe … n = PV / RT n = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K)  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Find the nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe …  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe …  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)(0.08026 L atm / mol K)(273 K) / (4.5 L)  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)(0.08026 L atm / mol K)(273 K) / (4.5 L) = 0.457 atm  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)(0.08026 L atm / mol K)(273 K) / (4.5 L) = 0.457 atm 3rd: Calculate PN2: @ STP … PN2 = 1.00 atm  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)(0.08026 L atm / mol K)(273 K) / (4.5 L) = 0.457 atm 3rd: Calculate PN2: @ STP … PN2 = 1.00 atm 4th: Calculate the Ptot …  2009, Prentice-Hall, Inc.

Dalton’s Law of Partial Pressures A volume of 2.0 L of He at 46 ‘C, and 1.2 atm of pressure, was added to a vessel that contained 4.5 L of N2 @ STP. What is the total pressure and the partial pressure of each gas @ STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C + 273 ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 L @ STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / (0.08206 L atm/mol K)(319 K) = 0.020 molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)(0.08026 L atm / mol K)(273 K) / (4.5 L) = 0.457 atm 3rd: Calculate PN2: @ STP … PN2 = 1.00 atm 4th: Calculate the Ptot … Ptot = ? atm || PN2 + PHe = 1.00 atm + 0.46 atm = 1.46 atm  2009, Prentice-Hall, Inc.

Mole Fraction  2009, Prentice-Hall, Inc.

Mole Fraction Xi = ni / ntot = Pi / Ptot  2009, Prentice-Hall, Inc.

Mole Fraction Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system …  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction …  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data …  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known:  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP 1 mol N2 = 22.4 L N2  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP 1 mol N2 = 22.4 L N2 VN2 = 4.5 LN2  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 VN2 = 4.5 LN2  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ?  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? XHe = ?  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? PN2 = 1.00 atm; PHe = 0.457 atm XHe = ?  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2:  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot:  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot 3rd: Calculate XN2 & XHe: (Using no. of moles)  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = 0.201 molN2 / 0.293 moltot = 0.69  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = 0.201 molN2 / 0.293 moltot = 0.69 XHe = ? || nHe/ntot = 0.0917 molHe / 0.293 moltot = 0.31  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = 0.201 molN2 / 0.293 moltot = 0.69 XHe = ? || nHe/ntot = 0.0917 molHe / 0.293 moltot = 0.31  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = 0.201 molN2 / 0.293 moltot = 0.69 XHe = ? || nHe/ntot = 0.0917 molHe / 0.293 moltot = 0.31 4th: Calculate XN2 & XHe: (Using partial pressure data) XN2 = ? || PN2 / Ptot = 1.00 atm / 1.457 atm = 0.69  2009, Prentice-Hall, Inc.

you can calculate mole fraction … Xi = ni / ntot = Pi / Ptot Key Idea … If you know either the n or P of each component of your system … you can calculate mole fraction … Calculate the number of moles of N2 present in the previous example. Also, calculate the mole fractions of N2 and the He given the following mole and pressure data … Known: @ STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = 0.457 atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = 0.201 molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = 0.201 molN2 + 0.0917 molHe = 0.293 moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = 0.201 molN2 / 0.293 moltot = 0.69 XHe = ? || nHe/ntot = 0.0917 molHe / 0.293 moltot = 0.31 4th: Calculate XN2 & XHe: (Using partial pressure data) XN2 = ? || PN2 / Ptot = 1.00 atm / 1.457 atm = 0.69 Xhe = ? || PHe / Ptot = 0.457 atm / 1.457 atm = 0.31  2009, Prentice-Hall, Inc.

Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.  2009, Prentice-Hall, Inc.

Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to an ideal gas particles as environmental conditions change.  2009, Prentice-Hall, Inc.

Main Tenets of Kinetic-Molecular Theory The volume of individual particles of a gas can be assumed to be negligible.  2009, Prentice-Hall, Inc.

Main Tenets of Kinetic-Molecular Theory - The volume of the individual particles of a gas can be assumed to be negligible. - The particles of a gas are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by a gas.  2009, Prentice-Hall, Inc.

Main Tenets of Kinetic-Molecular Theory - The volume of the individual particles of a gas can be assumed to be negligible. - The particles of a gas are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by a gas. - The particles are assumed to exert no forces on each other.  2009, Prentice-Hall, Inc.

Main Tenets of Kinetic-Molecular Theory - The volume of the individual particles of a gas can be assumed to be negligible. - The particles of a gas are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by a gas. - The particles are assumed to exert no forces on each other. - The KEavg of a collection of gas particles is assumed to be directly proportional to the T(K) of the gas.  2009, Prentice-Hall, Inc.

Temperature is a Measure of KEavg of a Gas KEavg = 3/2RT  2009, Prentice-Hall, Inc.

Temperature is a Measure of KEavg of a Gas KEavg = 3/2RT KE = 1/2mv2 = 3/2RT  2009, Prentice-Hall, Inc.

Temperature is a Measure of KEavg of a Gas KEavg = 3/2RT KE = 1/2mv2 = 3/2RT Velocity Increases with Higher Temperature!  2009, Prentice-Hall, Inc.

Main Tenets of Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.  2009, Prentice-Hall, Inc.

Main Tenets of Kinetic-Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature.  2009, Prentice-Hall, Inc.

Root Mean Square Velocity µrms = (3RT/M)1/2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity µrms = (3RT/M)1/2 R = 8.3145 J / K mol  2009, Prentice-Hall, Inc.

Root Mean Square Velocity µrms = (3RT/M)1/2 R = 8.3145 J / K mol = 8.3145 N m / K mol  2009, Prentice-Hall, Inc.

Root Mean Square Velocity µrms = (3RT/M)1/2 R = 8.3145 J / K mol = 8.3145 N m / K mol = 8.3145 Kg m2/s2 / K mol  2009, Prentice-Hall, Inc.

Root Mean Square Velocity µrms = (3RT/M)1/2 R = 8.3145 J / K mol = 8.3145 N m / K mol = 8.3145 Kg m2/s2 / K mol T = temperature (Kelvins)  2009, Prentice-Hall, Inc.

Root Mean Square Velocity µrms = (3RT/M)1/2 R = 8.3145 J / K mol = 8.3145 N m / K mol = 8.3145 Kg m2/s2 / K mol T = temperature (Kelvins) M = mass of a mole of the gas in kilograms (kg/mol)  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C.  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known:  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms)  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) T1 = 0 ‘C + 273 ‘C = 273 K  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) T1 = 0 ‘C + 273 ‘C = 273 K T2 = 300 ‘C + 273 ‘C = 573 K  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) T1 = 0 ‘C + 273 ‘C = 273 K T2 = 300 ‘C + 273 ‘C = 573 K R = 8.3145 Kg m2 / s2 K mol  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) T1 = 0 ‘C + 273 ‘C = 273 K T2 = 300 ‘C + 273 ‘C = 573 K R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K T2 = 300 ‘C + 273 ‘C = 573 K R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution:  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C …  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 =  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate µrms @ 300 ‘C …  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate µrms @ 300 ‘C … µrms = ((3(8.3145 kg m2/s2 / K mol)(573 K) / (0.0320 kg/mol))1/2 =  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate µrms @ 300 ‘C … µrms = ((3(8.3145 kg m2/s2 / K mol)(573 K) / (0.0320 kg/mol))1/2 = ((4.467 x 105 m2)/s2)1/2  2009, Prentice-Hall, Inc.

Root Mean Square Velocity Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 0 ‘C and 300 ‘C. Known: (Increased T ---> Increased KE ---> Increased rms) Unknown: T1 = 0 ‘C + 273 ‘C = 273 K µrms = ? m/s @ 0 ‘C T2 = 300 ‘C + 273 ‘C = 573 K µrms = ? m/s @ 300 ‘C R = 8.3145 Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution: 1st: Calculate the MO2: M = ? Kg/mol || (32.0 gO2 / molO2) (1 kg / 1000 g) = 0.0320 kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3(8.3145 kg m2/s2 / K mol)(273 K) / (0.0320 kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate µrms @ 300 ‘C … µrms = ((3(8.3145 kg m2/s2 / K mol)(573 K) / (0.0320 kg/mol))1/2 = ((4.467 x 105 m2)/s2)1/2 = 668 m/s  2009, Prentice-Hall, Inc.

Mean Free Path - Particles collide with one another and exchange energy after traveling a very short distance.  2009, Prentice-Hall, Inc.

Mean Free Path - Particles collide with one another and exchange energy after traveling a very short distance. - The distance between collisions is called the Mean Free Path. - Exchange of energy between particles happens at different times … particles are speeding up and slowing down …  2009, Prentice-Hall, Inc.

Mean Free Path - Particles collide with one another and exchange energy after traveling a very short distance. - The distance between collisions is called the Mean Free Path. - Exchange of energy between particles happens at different times … particles are speeding up and slowing down … - Particles have an avg. µrms velocity, but, rarely precisely that velocity …  2009, Prentice-Hall, Inc.

Mean Free Path - Particles collide with one another and exchange energy after traveling a very short distance. - The distance between collisions is called the Mean Free Path. - Exchange of energy between particles happens at different times … particles are speeding up and slowing down … - Particles have an avg. µrms velocity, but, rarely precisely that velocity … - Particles have an average range of velocities, the average of which is the µrms velocity …  2009, Prentice-Hall, Inc.

Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.  2009, Prentice-Hall, Inc.

Effusion The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles …  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22)  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22) If M, molar mass = m/NA  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22) If M, molar mass = m/NA 1/2 M1µ12 = ½ M2µ22  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22) If M, molar mass = m/NA 1/2 M1µ12 = ½ M2µ22 M1/M2 = µ22/µ12  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22) If M, molar mass = m/NA 1/2 M1µ12 = ½ M2µ22 M1/M2 = µ22/µ12 OR:  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22) If M, molar mass = m/NA 1/2 M1µ12 = ½ M2µ22 M1/M2 = µ22/µ12 OR: (M1/M2)1/2 = µrms2/µrms1 = rate of effusion2 / rate of effusion1  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion If the KE of two gases, 1 and 2, in a system are the same at a given temperature, then for a mole of the particles … KEavg = NA (1/2 m1µ12) = NA (1/2 m2µ22) If M, molar mass = m/NA 1/2 M1µ12 = ½ M2µ22 M1/M2 = µ22/µ12 OR: (M1/M2)1/2 = µrms2/µrms1 = rate of effusion2 / rate of effusion1 The heavier the molecule, the slower the rate of effusion through a small orifice …  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse?  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: MNO2 = 46.01 g/mol  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: MNO2 = 46.01 g/mol Mhe = 4.003 g/mol  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = 46.01 g/mol Mhe = 4.003 g/mol  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = 46.01 g/mol rateHe/rateNO2 = ? Mhe = 4.003 g/mol  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = 46.01 g/mol rateHe/rateNO2 = ? Mhe = 4.003 g/mol Solution: (MNO2/Mhe)1/2 = rateHe/rateNO2  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = 46.01 g/mol rateHe/rateNO2 = ? Mhe = 4.003 g/mol Solution: (MNO2/Mhe)1/2 = rateHe/rateNO2 (46.01/4.003)1/2 = rateHe/rateNO2  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = 46.01 g/mol rateHe/rateNO2 = ? Mhe = 4.003 g/mol Solution: (MNO2/Mhe)1/2 = rateHe/rateNO2 (46.01/4.003)1/2 = rateHe/rateNO2 rateHe/rateNO2 = 3.39  2009, Prentice-Hall, Inc.

Graham’s Law of Effusion How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = 46.01 g/mol rateHe/rateNO2 = ? Mhe = 4.003 g/mol Solution: (MNO2/MHe)1/2 = rateHe/rateNO2 (46.01/4.003)1/2 = rateHe/rateNO2 rateHe/rateNO2 = 3.39 He would effuse 3.39 times as fast as NO2 …  2009, Prentice-Hall, Inc.

Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.  2009, Prentice-Hall, Inc.

Graham’s Law of Diffusion Distance Traveled2 / Distance Traveled1 = (M1/M2)1/2  2009, Prentice-Hall, Inc.

Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.  2009, Prentice-Hall, Inc.

Real Gases Even the same gas will show wildly different behavior under high pressure at different temperatures.  2009, Prentice-Hall, Inc.

Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.  2009, Prentice-Hall, Inc.

Corrections for Nonideal Behavior The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. The corrected ideal-gas equation is known as the van der Waals equation.  2009, Prentice-Hall, Inc.

The van der Waals Equation ) (V − nb) = nRT n2a V2 (P +  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction:  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty …  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer - nb  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as …  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb)  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) …  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) … Pobserved = Pideal – a correction factor  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) … Pobserved = Pideal – a correction factor The correction factor that accounts for the decrease in pressure due to intermolecular attraction …  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) … Pobserved = Pideal – a correction factor The correction factor that accounts for the decrease in pressure due to intermolecular attraction … Pobs = Pideal – a[n/V]2 Where a = a constant which depends on the gas … n = no. of moles of gas  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) … Pobserved = Pideal – a correction factor The correction factor that accounts for the decrease in pressure due to intermolecular attraction … Pobs = Pideal – a[n/V]2 Where a = a constant which depends on the gas … n = no. of moles of gas Combining both correction factors into the Van der Wall’s equation …  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) … Pobserved = Pideal – a correction factor The correction factor that accounts for the decrease in pressure due to intermolecular attraction … Pobs = Pideal – a[n/V]2 Where a = a constant which depends on the gas … n = no. of moles of gas Combining both correction factors into the Van der Wall’s equation … [P + a[n/V]2] (V-nb) = nRT  2009, Prentice-Hall, Inc.

The van der Waals Equation Volume Correction … Because gas molecules do not take up space, the free volume of the container is not as large as it would be if it were empty … Vavailable = Vcontainer – nb Where … n = #moles of gas & b = correction factor Pressure can be expressed as … P = nRT / (V-nb) Pressure Correction … Because gas molecules can interact with each other, they do not collide with the walls of the container (“exert pressure”) to as great an extent as when there were no intermolecular attractions (as with an ideal gas) … Pobserved = Pideal – a correction factor The correction factor that accounts for the decrease in pressure due to intermolecular attraction … Pobs = Pideal – a[n/V]2 Where a = a constant which depends on the gas … n = no. of moles of gas Combining both correction factors into the Van der Wall’s equation … [P + a[n/V]2] (V-nb) = nRT Interactions between molecules are greatest at low Temperatures (low rms velocities), high pressures and low volumes. Deviations from ideality are expected to be greatest under these conditions.  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation.  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known:  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: nHe = 0.3000 molHe  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: nHe = 0.3000 molHe VHe = 0.2000 LHe  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: nHe = 0.3000 molHe VHe = 0.2000 LHe T = -25 ‘C + 273 ‘C = 248 K  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: nHe = 0.3000 molHe VHe = 0.2000 LHe T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution:  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution: 1st: Calculate PHe using the ideal gas law …  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution: 1st: Calculate PHe using the ideal gas law … PHe = nRT/V  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution: 1st: Calculate PHe using the ideal gas law … PHe = ? Atm || nRT/V = (0.3000 mol)(0.08206 L atm / mol K)(248 K) / (0.20000 L) = 30.6 atm  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution: 1st: Calculate PHe using the ideal gas law … PHe = ? Atm || nRT/V = (0.3000 mol)(0.08206 L atm / mol K)(248 K) / (0.20000 L) = 30.6 atm 2nd: Calculate PHe using the van der Walls Equation …  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution: 1st: Calculate PHe using the ideal gas law … PHe = ? Atm || nRT/V = (0.3000 mol)(0.08206 L atm / mol K)(248 K) / (0.20000 L) = 30.6 atm 2nd: Calculate PHe using the van der Walls Equation … PHe = ? atm || nRT / (V – nb) – a[n/V]2 = (0.3000 mol) (0.08206 L atm / mol K)(248 K) / (0.2000 L – (0.3000 mol)(0.0237 L/mol)) - (0.034 atm L2 / mol2) [0.03000 mol/0.2000 L]2  2009, Prentice-Hall, Inc.

The van der Waals Equation Calculate the pressure exerted by 0.3000 molHe in a 0.2000 L container at -25 ‘C using the ideal gas law as well as the van der Waals equation. Known: Unknown: nHe = 0.3000 molHe Pideal = ? atm VHe = 0.2000 LHe Pnonideal = ? atm T = -25 ‘C + 273 ‘C = 248 K a = 0.034 atm L2 / mol2 b = 0.0237 L mol Solution: 1st: Calculate PHe using the ideal gas law … PHe = ? Atm || nRT/V = (0.3000 mol)(0.08206 L atm / mol K)(248 K) / (0.20000 L) = 30.6 atm 2nd: Calculate PHe using the van der Walls Equation … PHe = ? atm || nRT / (V – nb) – a[n/V]2 = (0.3000 mol) (0.08206 L atm / mol K)(248 K) / (0.2000 L – (0.3000 mol)(0.0237 L/mol)) - (0.034 atm L2 / mol2) [0.03000 mol/0.2000 L]2 = 31.67 atm – 0.077 atm = 31.6 atm  2009, Prentice-Hall, Inc.