John Bookstaver St. Charles Community College Cottleville, MO

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.

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

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.

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

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

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.

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.

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.

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

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.

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 = kg m/s2 = 110 N  2009, Prentice-Hall, Inc.

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 = kg m/s2 = 110 N Weight = 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.

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.

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.

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.

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 atm …  2009, Prentice-Hall, Inc.

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

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

Let’s Say the Pressure exerted by a gas is 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.

Let’s Say the Pressure exerted by a gas is 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) 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.

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 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 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 L. What volume will the gas occupy at 60 ‘C and 1 atm? Known: T1 = 30 ‘C K = 303 K  2009, Prentice-Hall, Inc.

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

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

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

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

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

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

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.

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.

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 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 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 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 = moles  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 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 = moles n2 = = moles  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 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 = moles n2 = = moles  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 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 = moles n2 = = moles Solution:  2009, Prentice-Hall, Inc.

Avogadro’s Law A 5.20 L sample at 18 ‘C and 2 atm pressure contains 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 = moles n2 = = 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 = L atm / mol K T = Temperature (K)  2009, Prentice-Hall, Inc.

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 = L atm / mol K T = Temperature (K) The relationship assumes that a gas will behave ideally …  2009, Prentice-Hall, Inc.

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 = 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 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 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 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 ‘C = 285 K  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol R = L atm / K mol  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol R = L atm / K mol  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol R = L atm / K mol Solution:  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol R = L atm / K mol Solution: PV = nRT  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol R = 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 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 ‘C = 285 K n = mol R = L atm / K mol Solution: PV = nRT P = nRT / V P = ? at,m || (0.614 mol) ( L atm / mol K) (285 K) / 12.9 L  2009, Prentice-Hall, Inc.

Ideal Gas Law A sample containing 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 ‘C = 285 K n = mol R = L atm / K mol Solution: PV = nRT P = nRT / V P = ? atm || (0.614 mol) ( 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 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 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 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 = atm V1 = 7.0 L T1 = 4 ‘C ‘C = 277 K  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 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 = atm P2 = 1.52 atm V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘C = 277 K  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘C = 277 K  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘C = 277 K Solution:  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘C = 277 K Solution: n1 = n2  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘C = 277 K Solution: n1 = n2 R = R  2009, Prentice-Hall, Inc.

Combined Gas Law A sample of methane gas (CH4) at 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘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 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘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 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘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 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘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 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘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 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 = atm P2 = 1.52 atm V2 = ? L V1 = 7.0 L T2 = 284 K T1 = 4 ‘C ‘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 Standard Temperature & Pressure (STP) occupies 22.4 L …  2009, Prentice-Hall, Inc.

1 mole of any ideal Standard Temperature & Pressure (STP) occupies 22.4 L … STP Conditions …  2009, Prentice-Hall, Inc.

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

1 mole of any ideal Standard Temperature & Pressure (STP) occupies 22.4 L … STP Conditions … 1.000 atm (standard pressure) & 0 ‘C (standard temperature) STP Means 0 ‘C & 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.

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

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

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

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

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

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

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

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

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

Ideal Gas Law & STP What volume will moles of O2 occupy at STP? STP Unknown: T = 273 K & P = atm V = ? L R = L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) ( 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.

Ideal Gas Law & STP What volume will moles of O2 occupy at STP? STP Unknown: T = 273 K & P = atm V = ? L R = L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) ( 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.

Ideal Gas Law & STP What volume will moles of O2 occupy at STP? STP Unknown: T = 273 K & P = atm V = ? L R = L atm / K mol n = 1.18 mol Solution: V = nRT / P V = ? L || (1.18 mol) ( 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.

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.

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.

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.

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.

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.

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.

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.

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.

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 ‘C = 295 K  2009, Prentice-Hall, Inc.

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 ‘C = 295 K P = 1.04 atm  2009, Prentice-Hall, Inc.

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 ‘C = 295 K P = 1.04 atm 1 mol CO2 = 44.0 g CO2  2009, Prentice-Hall, Inc.

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 ‘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.

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 ‘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) = mol CO2  2009, Prentice-Hall, Inc.

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 ‘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) = mol CO2 Vballoon = ? L || nRT/P = (0.341 mol) ( L atm / K mol) (295 K) / 1.04 atm  2009, Prentice-Hall, Inc.

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 ‘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) = mol CO2 Vballoon = ? L || nRT/P = (0.341 mol) ( 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.

0.500 L of H2(g) are reacted with L of STP according to the equation …  2009, Prentice-Hall, Inc.

0.500 L of H2(g) are reacted with L of STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g)  2009, Prentice-Hall, Inc.

0.500 L of H2(g) are reacted with L of 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.

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

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

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

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

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

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

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

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

0.500 L of H2(g) are reacted with L of STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? STP Unknown: V H2(g) = L H2(g) ; V O2(g) = L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C ‘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.

0.500 L of H2(g) are reacted with L of STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? STP Unknown: V H2(g) = L H2(g) ; V O2(g) = L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C ‘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.

0.500 L of H2(g) are reacted with L of STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? STP Unknown: V H2(g) = L H2(g) ; V O2(g) = L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C ‘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.

0.500 L of H2(g) are reacted with L of STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? STP Unknown: V H2(g) = L H2(g) ; V O2(g) = L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C ‘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.

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

0.500 L of H2(g) are reacted with L of STP according to the equation … 2H2(g) + O2(g) ----> 2H2O(g) What volume will the H2O occupy at 1.00 atm and 350 ‘C? STP Unknown: V H2(g) = L H2(g) ; V O2(g) = L O2(g) V H2O = ? L T1 = O ‘C ; T2 = 350 ‘C ‘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 = ( mol)( 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 ‘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 ‘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 ‘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 ‘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 ‘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 ‘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) ( 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 ‘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) ( 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.

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.

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.

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 STP. What is the total pressure and the partial pressure of each STP?  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known:  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Initial … VHe = 2.0 LHe ;  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Find the nHe …  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Find the nHe … n = PV / RT n = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K)  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Find the nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe 2nd: Calculate PHe …  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)( L atm / mol K)(273 K) / (4.5 L)  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)( L atm / mol K)(273 K) / (4.5 L) = atm  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)( L atm / mol K)(273 K) / (4.5 L) = atm 3rd: Calculate STP … PN2 = 1.00 atm  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)( L atm / mol K)(273 K) / (4.5 L) = atm 3rd: Calculate STP … PN2 = 1.00 atm 4th: Calculate the Ptot …  2009, Prentice-Hall, Inc.

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 STP. What is the total pressure and the partial pressure of each STP? Known: Unknown: Initial … VHe = 2.0 LHe ; T = 46 ‘C ‘C = 319 K ; P = 1.2 atmHe Ptot = ? atm Final … VN2 = 4.5 STP PHe = ? atm PN2 = ? atm Solution: 1st: Calculate nHe … n = PV / RT nHe = (1.2 atm)(2.0 L) / ( L atm/mol K)(319 K) = molHe 2nd: Calculate PHe … P = nRT/V PHe = (0.020 mol)( L atm / mol K)(273 K) / (4.5 L) = atm 3rd: Calculate STP … PN2 = 1.00 atm 4th: Calculate the Ptot … Ptot = ? atm || PN2 + PHe = 1.00 atm 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.

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.

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.

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 … STP  2009, Prentice-Hall, Inc.

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 … STP 1 mol N2 = 22.4 L N2  2009, Prentice-Hall, Inc.

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 … STP 1 mol N2 = 22.4 L N2 VN2 = 4.5 LN2  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 VN2 = 4.5 LN2  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ?  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? XHe = ?  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? PN2 = 1.00 atm; PHe = atm XHe = ?  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2:  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN2 XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot:  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = molN molHe = moltot  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = molN molHe = moltot 3rd: Calculate XN2 & XHe: (Using no. of moles)  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = molN molHe = moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = molN2 / moltot = 0.69  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = molN molHe = moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = molN2 / moltot = 0.69 XHe = ? || nHe/ntot = molHe / moltot = 0.31  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = molN molHe = moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = molN2 / moltot = 0.69 XHe = ? || nHe/ntot = molHe / moltot = 0.31 4th: Calculate XN2 & XHe: (Using partial pressure data) XN2 = ? || PN2 / Ptot = 1.00 atm / atm = 0.69  2009, Prentice-Hall, Inc.

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 … STP Unknown: 1 mol N2 = 22.4 L N2 molesN2 = ? molN2 VN2 = 4.5 LN XN2 = ? ; XHe = ? PN2 = 1.00 atm; PHe = atm Solution: 1st: Calculate nN2: nN2 = ? molN2 || (4.5 LN2)(1molN2 / 22.4 LN2) = molN2 2nd: Calculate ntot: ntot = ? moltot || nN2 + nHe = molN molHe = moltot 3rd: Calculate XN2 & XHe: (Using no. of moles) XN2 = ? || nN2 /ntot = molN2 / moltot = 0.69 XHe = ? || nHe/ntot = molHe / moltot = 0.31 4th: Calculate XN2 & XHe: (Using partial pressure data) XN2 = ? || PN2 / Ptot = 1.00 atm / atm = 0.69 Xhe = ? || PHe / Ptot = atm / 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.

- 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.

- 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.

- 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.

KEavg = 3/2RT KE = 1/2mv2 = 3/2RT  2009, Prentice-Hall, Inc.

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

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.

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.

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

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

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

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

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

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.

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.

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.

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 ‘C = 273 K  2009, Prentice-Hall, Inc.

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 ‘C = 273 K T2 = 300 ‘C ‘C = 573 K  2009, Prentice-Hall, Inc.

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 ‘C = 273 K T2 = 300 ‘C ‘C = 573 K R = Kg m2 / s2 K mol  2009, Prentice-Hall, Inc.

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 ‘C = 273 K T2 = 300 ‘C ‘C = 573 K R = Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K T2 = 300 ‘C ‘C = 573 K R = Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K R = Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = Kg m2 / s2 K mol gmmO2 = 32.0 gO2/molO2 Solution:  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C …  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 =  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate 300 ‘C …  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate 300 ‘C … µrms = ((3( kg m2/s2 / K mol)(573 K) / ( kg/mol))1/2 =  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate 300 ‘C … µrms = ((3( kg m2/s2 / K mol)(573 K) / ( kg/mol))1/2 = ((4.467 x 105 m2)/s2)1/2  2009, Prentice-Hall, Inc.

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 ‘C = 273 K µrms = ? 0 ‘C T2 = 300 ‘C ‘C = 573 K µrms = ? 300 ‘C R = 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) = kgO2 / molO2 2nd: Calculate µrms at 0 ‘C … µrms = (3( kg m2/s2 / K mol)(273 K) / ( kg/mol))1/2 = ((2.128 x 105 m2)/s2)1/2 = 461 m/s 3rd: Calculate 300 ‘C … µrms = ((3( kg m2/s2 / K mol)(573 K) / ( 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.

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.

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.

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.

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.

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.

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.

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.

How many times faster than He would NO2 gas effuse?  2009, Prentice-Hall, Inc.

How many times faster than He would NO2 gas effuse? Known: MNO2 = g/mol  2009, Prentice-Hall, Inc.

How many times faster than He would NO2 gas effuse? Known: MNO2 = g/mol Mhe = g/mol  2009, Prentice-Hall, Inc.

How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = g/mol Mhe = g/mol  2009, Prentice-Hall, Inc.

How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = g/mol rateHe/rateNO2 = ? Mhe = g/mol  2009, Prentice-Hall, Inc.

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

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

How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = g/mol rateHe/rateNO2 = ? Mhe = 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.

How many times faster than He would NO2 gas effuse? Known: Uknown: MNO2 = g/mol rateHe/rateNO2 = ? Mhe = 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.

Volume Correction:  2009, Prentice-Hall, Inc.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

John Bookstaver St. Charles Community College Cottleville, MO

Similar presentations

Presentation on theme: "John Bookstaver St. Charles Community College Cottleville, MO"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

John Bookstaver St. Charles Community College Cottleville, MO

Similar presentations

Presentation on theme: "John Bookstaver St. Charles Community College Cottleville, MO"— Presentation transcript:

Similar presentations

About project

Feedback