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The Ideal Gas Lawand

Stoichiometry

Chemistry 142 B

Autumn Quarter, 2004J. B. Callis, Instructor

Lecture #14

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Ideal Gas Law

An ideal gas is defined as one for which both the volume

of molecules and forces between the molecules are sosmall 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

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The Ideal Gas Law Subsumesthe Other Gas Laws

During chemical and physical processes, any of the fourvariables 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 Boyles Law for a fixed amount at constant volume

P/T = nR/V = constant Amontons Law

for a fixed amount at constant pressure

V/T = nR/P = constant Charless Law for a fixed volume and temperature

P/n = RT/V = constant Avogadros Law

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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

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Problem 14-1: Change of Three Variables - I

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

P1=

P2 =

V1 = V2 =T1 =

T2 =

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Problem 14-1: Change of Three Variables - II

=T1 T2

P1 x V1 P2 x V2

=

P2 =

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Problem 14-2: Gas Law

Problem: Calculate the pressure in a container whose Volume is 87.5 L

and 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 =

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Problem 14-3: Ideal Gas Calculation - Nitrogen

Calculate the pressure in a container holding 375 g ofNitrogen gas. The volume of the container is 0.150 m3and the temperature is 36.0 oC.

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Problem 14-4: Sodium Azide Decomposition - I

Sodium Azide (NaN3) is used in some air bags inautomobiles. Calculate the volume of Nitrogen gasgenerated at 21 oC and 823 mm Hg by the decompositionof 60.0 g of NaN3 .

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

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Problem 14-4: Sodium Azide Decomposition - II

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Problem 14-5: Ammonia Density

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

Density = mass per unit volume = g / L

P =

T =

n = mass / Molar mass = g / M

d =

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Calculation of Molar Mass

n =

n = =

Mass

Molar Mass

P x VR x T

MassMolar Mass

Molar Mass = MM =

Mass x R x T

P x V

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Problem 14-6: Dumas Method of Molar Mass

Problem: A volatile liquid is placed in a flask whose volume is 590.0 ml

and 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:

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Problem 14-7: Calculation of Molecular Weight of aNatural 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:

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Gas Mixtures

Gas behavior depends on the number, not

the identity, of gas molecules.

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

All molecules in a sample of an ideal gas

behave exactly the same way.

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Daltons Law of Partial Pressures - I

Definition: In a mixture of gases, each gascontributes 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

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Daltons Law of Partial Pressure - II

Pressure exerted by an ideal gas mixture isdetermined 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 waspresent by itself.

P = (n1 RT)/V + (n2 RT)/V + (n3RT)/V + ...

the total pressure is the sum of the partialpressures.

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Problem 14-8: Daltons Law of Partial Pressures

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

What are the Partial Pressures of each gas, and the totalPressure?

P bl 14 8 D lt L f P ti l P

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Problem 14-8: Daltons Law of Partial Pressures

cont.

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Problem 14-9: Daltons Law using mole fractions

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

Total # moles =

XNe =PNe = XNe PTotal

XAr =

PAr =

XXe =

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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

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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

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

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

Calculate the mass of Hydrogen gas collected over water if156 ml of gas is collected at 20oC and 769 mm Hg.

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

PV = nRT n = PV / RT

n =

n =

mass =

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Chemical Equation Calc - III

Reactants ProductsMoleculesMoles

MassMolecular

Weight g/molAtoms (Molecules)Avogadros

Number6.02 x 1023

Solutions

Molaritymoles / liter

Gases

PV = nRT

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Problem 14 11: Gas Law Stoichiometry

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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 L

contains 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 moles

of 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

P bl 14 11 G L St i hi t

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Problem 14-11: Gas Law Stoichiometry

n =PV

RT

RRNH3 =

RRHCl =

Therefore the product will be

A P bl i L 14

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Answers to Problems in Lecture #14

1. 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 = 1.11atm

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 of0.274 atm