Gas Laws Chapter 11

11.1 Pressure and Force Pressure = Force per unit area Measure force in Newtons (N) 1N = force that will increase the speed of a 1 kg mass by one meter per second each second that the force is applied Force of gravity = 9.8m/s2

A 51 kg person (112 lb) exerts a force due to gravity of 500 N Pressure exerted on the floor depends on the area of contact

Barometer Force Area Pressure = Units of Pressure Memorize 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,300 Pa 1 atm = 101.3 kPa 1 atm = 14.7 lb/in2 Standard pressure is pressure exerted by a column of mercury 760 mm high

10 miles 0.2 atm 4 miles 0.5 atm Sea level 1 atm

Crushing a Coke Can Click on the picture

As P (h) increases V decreases

Graham’s Law Suppose 2 containers at the same temperature are opened across the room from you. One contains HCl & the other NH3. Which will you smell first?

What moves faster, a large particle or a small particle? The rate of diffusion will depend on both the average velocity and the mass of the particle

Graham’s Law T1 = T2  KE1 = KE2 Cancel ½ Move m1 to right Move v2 to left Take square root of both sides

or

11.2 The Gas Laws Measuring the density of a gas is harder than for liquids & solids Volume can change

Four quantities affect measuring gases: Volume: amount of space Pressure: # collisions with walls of container Temperature: average Kinetic Energy Quantity: # of moles We can make general statements about any 2 quantities if 2 are held constant

Boyle’s Law P a 1/V Constant temperature P x V = constant Constant amount of gas P x V = constant P1 x V1 = P2 x V2

P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

Learning Check 1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) 1) 3.2 L 2) 6.4 L 3) 12.8 L 4) I don’t understand Take a few minutes to work the problem before moving forward.

PV Calculation

Learning Check 2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg 4) I don’t understand

PV Calculation

As T increases V increases

Variation of gas volume with temperature at constant pressure. Charles’ Law V a T Temperature must be in Kelvin V = constant x T V1/T1 = V2/T2 T (K) = t (0C) + 273

Learning Check 3 Use Charles’ Law to complete the statements below: 1. If final T is higher than initial T, final V is 1) greater, or 2) less than the initial V. 2. If final V is less than initial V, final T is 1) higher, or 2) lower than the initial T.

Solution GL3

V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398 K T2 = ? V2 x T1 V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398 K T2 = ? V2 x T1 V1 1.54 L x 398 K 3.20 L = T2 = = 192 K

Learning Check 4 A sample of oxygen gas has a volume of 420. mL at a temperature of 18.0°C. What temperature (in °C) is needed to change the volume to 640. mL? 1) 443.°C 2) 170.°C 3) - 82.0°C

Solution GL4

Variation of gas pressure with temperature at constant volume. Gay-Lussac’s Law P a T Temperature must be in Kelvin P = constant x T P1/T1 = P2/T2 T (K) = t (0C) + 273

Learning Check 5 Use Gay-Lussac’s law to complete the statements below: 1. When temperature decreases, the pressure of a gas 1) decreases or 2) increases. 2. When temperature increases, the pressure of a gas 1) decreases or 2) increases).

P1 T1 P2 T2 = P2 = P1T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Learning Check 6 Complete with 1) Increases 2) Decreases 3) Does not change A. Pressure _____, when V decreases B. When T decreases, V _____. C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)

Homework Find the density of the following gases at STP: He, N2, Cl2, CO2 WS #1:18-23 WS #2: 10-14 #3: 5, 7 #4: 26, 36 #6: 2, 3

Combined Gas Law If it is true that: We can also say that: PV Which gives us: P1V1 P2V2 T1 T2 P x V = constant1 V1/T1 = constant2 P1/T1 = constant3 = constant T =

Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P1 V1 P2 Boyle’s Law Charles’ Law Gay-Lussac’s Law V2 T1 T2

Learning Check 7 A 28.5 mL sample of gas was collected at 751 mm Hg and 25.6 C. Find the volume at STP. A). 31.1 mL B). 25.7 mL C) 31.5 mL

Avogadro's Law: Volume and Moles In Avogadro’s Law the volume of a gas is directly related to the number of moles (n) of gas. T and P are constant. V1 = V2 n1 n2

Think of a balloon Pressure in the balloon stays equal to air pressure As you add more air molecules… the balloon expands (increases in volume)

Learning Check 8 If 0.75 mole helium gas occupies a volume of 1.5 L, what volume will 1.2 moles helium occupy at the same temperature and pressure? 1) 0.94 L 2) 1.8 L 3) 2.4 L

Solution

No Name Law Relates number of moles to pressure The pressure of a gas is directly related to the number of moles (n) of gas. T and V are constant. P1 = P2 n1 n2

Density of Gases m D= Mass depends on the number of particles and identity of the gas Volume depends on # moles, temperature and pressure Density changes as volume changes v

Avogadro’s Hypothesis At the same temperature and pressure, equal volumes of gases contain equal numbers of molecules (moles) Size of molecules very small in relation to overall volume Molar volume of a gas: The volume of one mole of any gas at a particular temperature and pressure.

Molar Volume of a Gas Each temp and pressure combination has its own molar volume 1 mole occupies 30.6L @ 100 C and 1 atm 1 mole occupies 24.5L @ 25 C and 1 atm At STP, standard molar volume of any gas is 22.4 L (memorize)

A chemical reaction produces 0. 35 moles of a gas A chemical reaction produces 0.35 moles of a gas. What volume will it occupy at STP? Use molar volume as conversion factor: 1 mole = 22.4 L 0.35 mol 22.4L =7.8 L 1 mol

65. 0 mL of hydrogen sulfide gas is produced at STP 65.0 mL of hydrogen sulfide gas is produced at STP. How many moles are produced? 65.0 mL 1.00 L 1 mol =2.90 x 10-3 mol 1000 mL 22.4L

Finding Density D= or D= At STP: D= m Molar mass v Molar volume

Finding Density Given the volume & mass of any sample 0.519 g in 200.mL From formula at STP F2 at STP

Finding Molar Mass from Density Can use density formula as a proportion:

Sample problem What is the molar mass of a gas with a density of 1.28 g/L at STP?

Sample problem Find the mass of 2.50L of ammonia at STP.

Wording Molecular weight = molar mass (WS1) Review finding the empirical and molecular formula Find the mass of exactly 1 L means find the Density (WS2) Quiz on Friday: Gas Laws and Density with Empirical/molecular Formula

Homework Find the density of the following gases at STP: He, N2, Cl2, CO2 WS #1:18-27 WS #2: 10-14 #3: 5, 7, 8 #4: 26, 36 #5: 3 #6: 2, 3 #8: 12

Homework Eudiometer Problems Dalton’s Law Problems WS#2: 15-20; 25

Dalton’s Law of Partial Pressures He CO2 H2O 1.00 atm 1.00 atm 1.00 atm 3.00 atm

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases

Dalton’s Law of Partial Pressures Based on Avogadro’s hypothesis: equal # particles occupy equal volumes Total pressure of a gas mixture is the sum of the partial pressures of the component gases PT = P1 + P2 + P3 + …..

Ex 1: A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container? Ptotal = P1 + P2 + P3 + … Ptotal = PO2 + + PCO2 + PHe Ptotal = 2.00 atm + 3.00 atm + 4.00 atm Ptotal = 9.00 atm

A sample of natural gas contains 8. 24 moles of CH4, 0 A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Eudiometer Tubes (mercury or water displacement) Eudiometer – a tube used to collect a gas. It is closed at one end and is graduated. Reaction = 2HCl (aq) + Ca (s)  H2 (g) + CaCl2 (aq)

Eudiometer with mercury Situation #1 If there is just enough gas to make the mercury level inside the tube the same as the level of the mercury in the bowl then the pressure of the hydrogen gas is the same as that of the atmosphere! Pgas = Pair

Eudiometer, Situation #2 Suppose enough hydrogen gas is added to make the level inside the tube lower than the level of the mercury in the bowl. The P of the hydrogen gas is greater than the atmospheric pressure. (To determine the hydrogen gas pressure one must add the level difference to the barometric reading.) Pgas = Pair + d d

Eudiometer with mercury, situation #3 Suppose there was not enough gas to make the mercury level the same. Then, the pressure of hydrogen gas would not be the same as the pressure of the air outside the tube. (So, we must subtract the level difference from the barometric reading.) Pgas = Pair - d d

Problem 1 The volume of oxygen in a eudiometer is 37.0 mL. The mercury level inside the tube is 25.00mm lower than the outside. The barometric reading is 742.0 mm Hg. What is the pressure of the gas? 717.0 mm 767.0 mm I don’t know

Solving Problem 1

Problem 2 What is the P of the gas in an eudiometer, when the mercury level in the tube is 14mm higher, than that outside? That barometer reads 735mm Hg. 749 mm 721 mm

Answer to Problem 2

Water in Place of Mercury Calculations are done the same way but the difference in water levels must first be divided by 13.6 to convert it to its equivalent height in terms of a column of mercury since water is 1/13.6 as dense as mercury.

Also, water evaporates much more readily than mercury and so is in with the collected gas. So, you need to determine the partial pressure of the dry gas (unmixed with water vapor). The vapor pressure of water, at the given temperature, must be subtracted from the total pressure of the gas within the tube.

Water in Place of Mercury Bottle full of oxygen gas and water vapor 2KClO3 (s) 2KCl (s) + 3O2 (g) PT = PO + PH O 2

d Problem : Oxygen is collected using the water displacement method. The water level inside the tube is 27.2mm higher than that outside. The temp. is 25.0 C. The barometric pressure is 741.0mm. What is the partial pressure of the dry oxygen in the eudiometer?

Pgas = Pair ± Step 1: 27.2mm/13.6 = 2.00mm Step 2: 741.0 mm - 2.00mm = 739.0 mm So: 739.0 mm = P total = Poxygen + Pwater Step 3: To correct for water vapor pressure, you need to know the pressure of water vapor at 25.0C and it is 23.8mm (always given - see handout). Subtract this water partial pressure from the pressure total found in step 2. 739.0mm - 23.8mm = 715.2 mm = partial pressure of the dry oxygen. d H2O Pgas = Pair ± - P H2O 13.6

A eudiometer contains 38.4 mL of air collected by water displacement at a temperature of 20.0C. The water level inside the eudiometer is 140. mm higher than that outside. The barometric reading is 740.0mm. Water vapor pressure at 20.0C is 17.5mm. Calculate the volume of dry air at STP.

Problem 3: The pressure of the dry gas is 732.8 mm 712.2 mm 747.2 mm I have no idea

Problem 4: The volume at STP is 33.5 mL 38.6 mL I forgot the combined gas law

Solution Problem 4

Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R is the gas constant PV = nRT

PV = nRT PV (1 atm)(22.4L) R = = nT (1 mol)(273 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.4 L. PV = nRT R = PV nT = (1 atm)(22.4L) (1 mol)(273 K) R = 0.0821 L • atm / (mol • K)

PV = nRT nRT V = P 1.37 mol x 0.0821 x 273 K V = 1 atm V = 30.6 L What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273 K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273 K L•atm mol•K V = 30.6 L

PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

d is the density of the gas in g/L Density (d) Calculations m is the mass of the gas in g m V = PM RT d = M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) Gas Stoichiometry What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 180 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x 5.60 g C6H12O6 = 0.187 mol CO2 0.187 mol x 0.0821 x 310.15 K L•atm mol•K 1.00 atm = nRT P V = = 4.76 L

PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB Consider a case in which two gases, A and B, are in a container of volume V. PA = nART V nA is the number of moles of A PB = nBRT V nB is the number of moles of B XA = nA nA + nB XB = nB nA + nB PT = PA + PB PA = XA PT PB = XB PT Pi = Xi PT

Kinetic Molecular Theory of Gases A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. Gas molecules exert neither attractive nor repulsive forces on one another. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

Kinetic theory of gases and … Compressibility of Gases Boyle’s Law P a collision rate with wall Collision rate a number density Number density a 1/V P a 1/V Charles’ Law Collision rate a average kinetic energy of gas molecules Average kinetic energy a T P a T

Kinetic theory of gases and … Avogadro’s Law P a collision rate with wall Collision rate a number density Number density a n P a n Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas Ptotal = SPi

Apparatus for studying molecular speed distribution

 3RT urms = M The distribution of speeds of three different gases at the same temperature The distribution of speeds for nitrogen gas molecules at three different temperatures urms = 3RT M 

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. NH4Cl NH3 17 g/mol HCl 36 g/mol

Deviations from Ideal Behavior 1 mole of ideal gas Repulsive Forces PV = nRT n = PV RT = 1.0 Attractive Forces

Effect of intermolecular forces on the pressure exerted by a gas.