The Ideal Gas Lawand

Stoichiometry

Chemistry 142 B

Autumn Quarter, 2004

J. B. Callis, Instructor

Lecture #14

Ideal Gas Law• An ideal gas is defined as one for which both the volume of

molecules and forces between the molecules are so small that they have no effect on the behavior of the gas.

• The ideal gas equation is:

PV=nRT

R = Ideal gas constant = 8.314 J / mol K = 8.314 J mol-1 K-1

• R = 0.08206 L atm mol-1 K-1

The Ideal Gas Law Subsumes the Other Gas Laws

• During chemical and physical processes, any of the four variables in the ideal gas equation may be fixed.

• Thus, PV=nRT can be rearranged for the fixed variables:– for a fixed amount at constant temperature

• PV = nRT = constant Boyle’s Law– for a fixed amount at constant volume

• P/T = nR/V = constant Amonton’s Law– for a fixed amount at constant pressure

• V/T = nR/P = constant Charles’s Law– for a fixed volume and temperature

• P/n = RT/V = constant Avogadro’s Law

Many gas law problems involve a change of conditions, with no change in the amount of gas.

= constant Therefore, for a change

of conditions :

T1 T2

P x V

T

P1 x V1=

P2 x V2

Problem 14-1: Change of Three Variables - I

A gas sample in the laboratory has a volume of 45.9 L at 25 oC and a pressure of 743 mm Hg. If the temperature is increased to 155 oC by compressing the gas to a new volume of 31.0 L what is the pressure?

P1=

P2 =

V1 = V2 =

T1 =

T2 =

Problem 14-1: Change of Three Variables - II

=T1 T2

P1 x V1 P2 x V2

=

P2 =

Problem 14-2: Gas Law

Problem: Calculate the pressure in a container whose Volume is 87.5 Land it is filled with 5.038kg of Xenon at a temperature of 18.8 oC.Plan: Convert all information into the units required, and substitute into the Ideal Gas equation ( PV=nRT ).Solution:

nXe =

T =

P =

Problem 14-3: Ideal Gas Calculation - Nitrogen

Calculate the pressure in a container holding 375 g of Nitrogen gas. The volume of the container is 0.150 m3 and the temperature is 36.0 oC.

Problem 14-4: Sodium Azide Decomposition - I

Sodium Azide (NaN3) is used in some air bags in automobiles. Calculate the volume of Nitrogen gas generated at 21 oC and 823 mm Hg by the decomposition of 60.0 g of NaN3 .

2 NaN3 (s) 2 Na (s) + 3 N2 (g)

Problem 14-4: Sodium Azide Decomposition - II

Problem 14-5: Ammonia Density

Calculate the Density of ammonia gas (NH3) in grams per liter at 752 mm Hg and 55 oC.

Density = mass per unit volume = g / L

P =

T =

n = mass / Molar mass = g / M d =

Calculation of Molar Mass

n =

n = =

Mass

Molar Mass

P x V

R x T

Mass

Molar Mass

Molar Mass = MM = Mass x R x T

P x V

Problem 14-6: Dumas Method of Molar Mass

Problem: A volatile liquid is placed in a flask whose volume is 590.0 mland allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass of the flask before and after the experiment was 148.375g and 149.457 g,what is the molar mass of the liquid?Plan: Use the ideal gas law to calculate the molar mass of the liquid.Solution:

Problem 14-7: Calculation of Molecular Weight of a Natural Gas - Methane

Problem: A sample of natural gas is collected at 25.0 oC in a 250.0 ml flask. If the sample had a mass of 0.118 g at a pressure of 550.0 torr,what is the molecular weight of the gas?Plan: Use the ideal gas law to calculate n, then calculate the molar mass.Solution:

Gas Mixtures

• Gas behavior depends on the number, not the identity, of gas molecules.

• The ideal gas equation applies to each gas individually and to the mixture as a whole.

• All molecules in a sample of an ideal gas behave exactly the same way.

Dalton’s Law of Partial Pressures - I

• Definition: In a mixture of gases, each gas contributes to the total pressure: the pressure it would exert if the gas were present in the container by itself.

• To obtain a total pressure, add all of the partial pressures: Ptotal = P1+P2+P3+…PN

Dalton’s Law of Partial Pressure - II

• Pressure exerted by an ideal gas mixture is determined by the total number of moles:

P=(ntotal RT)/V• ntotal = sum of the amounts of each gas pressure• the partial pressure is the pressure of gas if it was

present by itself.• P = (n1 RT)/V + (n2 RT)/V + (n3RT)/V + ...• the total pressure is the sum of the partial

pressures.

Problem 14-8: Dalton’s Law of Partial Pressures

A 2.00 L flask contains 3.00 g of CO2 and 0.10 g of Helium at a temperature of 17.0 oC.

What are the Partial Pressures of each gas, and the total Pressure?

Problem 14-8: Dalton’s Law of Partial Pressurescont.

Problem 14-9: Dalton’s Law using mole fractions

A mixture of gases contains 4.46 mol Ne, 0.74 mol Ar and 2.15 mol Xe. What are the partial pressures of the gases if the total pressure is 2.00 atm ?

Total # moles =

XNe =

PNe = XNe PTotal

XAr =

PAr =

XXe =

Relative Humidity

Rel Hum = x 100%

Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the relative humidity?

Pressure of Water in Air

Maximum Vapor Pressure of Water

Relative Humidity

Rel Hum = x 100%

Example : the partial pressure of water at 15oC is 6.54 mm Hg, what is the relative humidity?

Rel Hum =(6.54 mm Hg/ 12.788 mm Hg )x100%

= 51.1 %

Pressure of Water in Air

Maximum Vapor Pressure of Water

Problem 14-10: Collection of Hydrogen gas over Water - Vapor pressure - I

2 HCl (aq) + Zn(s) ZnCl2 (aq) + H2 (g)

Calculate the mass of Hydrogen gas collected over water if 156 ml of gas is collected at 20oC and 769 mm Hg.

Problem 14-10: Collection of Hydrogen gas over Water - Vapor pressure - II

PV = nRT n = PV / RT

n =

n =

mass =

Chemical Equation Calc - III

Reactants ProductsMolecules

Moles

MassMolecularWeight g/mol

Atoms (Molecules)Avogadro’sNumber

6.02 x 1023

Solutions

Molaritymoles / liter

Gases

PV = nRT

Problem 14-11: Gas Law Stoichiometry

Problem: A slide separating two containers is removed, and the gases are allowed to mix and react. The first container with a volume of 2.79 Lcontains Ammonia gas at a pressure of 0.776 atm and a temperature of 18.7 oC. The second with a volume of 1.16 L contains HCl gas at a pressure of 0.932 atm and a temperature of 18.7 oC. What mass of solid ammonium chloride will be formed, and what will be remaining in the container, and what is the pressure?Plan: This is a limiting reactant problem, so we must calculate the molesof each reactant using the gas law to determine the limiting reagent. Then we can calculate the mass of product, and determine what is left in the combined volume of the container, and the conditions.Solution:

Equation: NH3 (g) + HCl (g) NH4Cl (s)

TNH3 = 18.7 oC + 273.15 = 291.9 K

Problem 14-11: Gas Law Stoichiometry

n = PVRT

RRNH3 =

RRHCl =

Therefore the product will be

Answers to Problems in Lecture #141. 2.08 atm

2. 10.5 atm

3. 2.26 atm

4. 30.8 liters

5. 0.626 g / L

6. 58.03 g/mol

7. 15.9 g/mol

8. PCO2 = 0.812 atm, PHe = 0.30 atm, PTotal = 1.11 atm

9. 1.21 atm for Ne, 0.20 atm for Ar, 0.586 atm for Xe

10. 0.0129 g hydrogen

11. 2.28 g NH4Cl made; remaining NH3 at a pressure of 0.274 atm