applied chapter 12.3 : molecular composition of gases

Chapter 12.3

Molecular Composition of Gases

1. Solve problems using ideal gas law.2. Describe the relationships between gas

behavior and chemical formulas, such as those expressed by Graham’s law of diffusion, law of combining gas volumes, and Dalton’s law of partial pressures.

Objectives:

The Ideal Gas LawThe mathematical relationship among pressure, volume,

temperature, and the number or moles of a gas.

Here’s how it works!!!

Deriving the Ideal gas lawo Boyle’s law PV = ko Charles’s law V/T = ko Avogadro’s law V/n = k

o V = 1/P x T x no V = R x 1/P x T x no V = nRT or PV = nRT P

The Ideal Gas ConstantBased on: 1 mol of gas at STP

R = PV = (1 atm)(22.4 L) = 0.0821 L*atm

nT (1 mol)(273.15 K) mol*K

ProblemsWhat is the pressure in atmospheres exerted by 0.500

mol sample of nitrogen gas in 10.0 L container at 298 K?

P =V =n =

R =T =

?

10.0 L0.500 molo.0821 mol*K

L*atm

298 K

PV = nRT

***Make sure all units match Gas Constant***Solve for P

V V

P =

nRT

V=

(0.500 mol)(0.0821 )(298 K)

mol*K

L*atm

10.0 L

1.22 atm=

Problem 2What is the volume, in liters, of 0.250 mol of

oxygen gas at 20.0 ⁰C and 0.974 atm pressure?P =V =n =

R =T =

0.974 atm?

0.250 molo.0821 mol*K

L*atm

20.0o

C

PV = nRT

P P

V =

nRT

P=

(0.250 mol)(0.0821 )(293 K)

mol*K

L*atm

0.974 atm

6.17 L of O2

=

***Temperature must be in Kelvin!!!

***Solve for V

+ 273 =

293 K

Problem 3How many moles of chlorine gas, Cl2, in grams, is contained in a

10.0 L tank at 27 ⁰C and 3.50 atm of pressure?

P =V =n =

R =T =

3.50 atm

10.0 L?

o.0821 mol*K

L*atm

27 oC

PV = nRTRT

***Temperature must be in Kelvin!!!

***Solve for n

+ 273 =

300. K

RT

n =PVRT

=(3.50 atm)(10.0 L)

mol*K

L*atm(0.0821 )(300.k)

=1.42 mol Cl2

EffusionProcess whereby the molecules of a gas

confined in a container randomly pass through a tiny opening in the container.

Diffusion The gradual mixing of two gases due to

their sp0ontaneous, random motion

Effusion

Graham’s Law of EffusionRate of effusion:

Depends on:Velocity of gas moleculesMass of molecules

12

MAvA2

=1

2MBvB2

MAvA2 = MBvB

2

vA2

vB2

=MA

MB

vA

vB

=MA

MB

SO:

Rate of effusion of ARate of effusion of B

=MA

MB

Defined as: The rates of effusion of gases at

the same temperature and pressure are inversely proportional to the square roots of their molar masses.

Application of Graham’s Law1. Lighter gases (lower Molar mass or

densities) diffuse faster than heavier gases.

2. Also provides a method for determining molar masses.

a) Rates of effusion of known and unknown gases can be compared to one another

Rates of effusion of different gases

Problem 1Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

Rate of effusion of H2Rate of effusion of O2

=MH2

MO2 =32.00 g/mol

2.02 g/mol= 3.9

8

***Remember that the molar masses are inversely related ***Find the molar masses of each

***Expressed like this

Hydrogen effuses 3.98 times faster than oxygen

Problem 2A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Rate of effusion of H2Rate of effusion of

unknown

=MH2

Munknown

2.02 g/mol

Munknown=9

22

81

=2.02 g/mol

Munknownx 2.o2 g/mol2.o2 g/mol

x

Munknown =160 g/mol

Dalton’s Law of Partial PressuresStates that the total pressure of a mixture of gases

is equal to the sum of the partial pressures of the component gases. Partial pressure: pressure of each gas in a mixture

PT = p1 + p2 + p3 + ……

PT = Total Pressurep1 + p2 + p3 = partial pressures

Dalton’s Law of Partial Pressures

PT = p1 + p2 + p3 + ……

So: Patm = pgas + pH2O

Gas collected by water displacement. Must include the pressure exerted by water

vapor

Dalton’s Law of Partial PressuresSample Problem 5Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0oC, respectively. What was the partial pressure of the oxygen collected?Patm = pO2 +

pH2O Patm = 731.0 torr

PO2 = ?

PH2O = 17.5 torr (from appendix in table A-

8, pg. 899) pO2 = Patm -

pH2O pO2 = 731.0 torr – 17.5

torr = 713.5 torr