Gas Densities, Partial Pressures, and

Kinetic-Molecular Theory

Sections 10.5-10.8

Objectives

Apply the ideal-gas equation to real gas situations.

Interpret the kinetic-molecular theory of gases

Key Terms Partial pressures Dalton’s Law of Partial Pressures Mole fraction Kinetic-molecular theory Root-mean-square speed Effusion Graham’s Law Diffusion Mean free path

Gas Densities and Molar Mass Rearrange the ideal-gas equation :

n = PV RT

Multiply both sides by molar mass, MnM = PMV RT

Product of n/V and M = density in g/LMoles x grams = gramsLiter mole liter

Gas Densities and Molar Mass

Density is expressed:

d = PM RT

Density depends on pressure, molar mass, and temperature

Example

Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21 ºC and 740.0 torr.

Gas Mixtures and Partial Pressure

Dalton’s Law of Partial Pressures:– Total pressure of a mixture equals

sum of the pressures that each would exert if present alone.

Pt = P1 + P2 + P3 + ….

Gas Mixtures and Partial Pressures

P1 = n1 (RT); P2 = n2 (RT); P3 = n3

(RT);… V V V

AndPt = (n1 + n2 + n3 + ….) RT = nt (RT)

V V

Example 1

A gaseous mixture made from 6.00 g oxygen and 9.00 g methane is placed in a 15.0 L vessel at 0 C. What is the partial pressure of each gas, and what is the total pressure of the vessel?

Example 2

What is the total pressure exerted by a mixture of 2.00g hydrogen and 8.00 g nitrogen at 273 K in a 10.0 L vessel?

Mole Fraction, X

P1 = n1 RT/ V = n1

Pt = nt RT/ V = nt

Thus…

P1 = (n1/nt)Pt = X1Pt

Partial press = mole frac x total press

Example 3

Mole fraction of N2 in air is 0.78 (78%). If the total pressure is 760 torr, what is the partial pressure of N2?

Homework

44, 48, and 60-68 even only

Sections 10.7 & 10.8

Kinetic-Molecular Theory

AndEffusion/Diffusion

Objectives

Understand why gas behave as they do

Apply the Kinetic-Molecular Theory to the Gas Laws

Define molecular effusion and diffusion

Solve problems using Graham’s Law of Effusion

Key Terms

Kinetic-Molecular Theory Root-Mean-Square Speed Effusion Diffusion Graham’s Law of Effusion Mean Free Path

Kinetic-Molecular Theory•Explains why gases behave as they do

•Developed over 100 year period

•Published in 1857 by Rudolf Clausius

Kinetic Molecular Theory

* Theory of moving molecules

You Must Know the 5 Postulates(page 421).

Five Postulates 1) Gases consist of large numbers of molecules that are

in continuous, random motion.2) The combined volume of the molecules is negligible

relative to the total volume in which the gas is contained.

3) Attractive and repulsive forces between gas molecules are negligible.

4) Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as T is constant

5) The average kinetic energy of the molecules is proportional to T. At any given T, all molecules have same avg. kinetic energy

Root-mean-square speed, u Speed of a molecule possessing

average kinetic energy

Є = ½ mu2

Є is average kinetic energym is mass of molecule

Both Є and u increase as temperature increases

Application to Gas Laws

1. Effect of a V increase at constant T:- Є does not change when T is constant. Thus u is unchanged. With V increase, there are fewer collisions with container walls, and pressure decreases (Boyle’s Law).

Application to Gas Laws

2. Effect of a T increase at constant V:

- Increase T means increase of Є and u. No change in V means there will be more collisions with walls (P increase).

Learning Check

A sample of carbon dioxide initially at STP is compressed into a smaller volume at constant temperature. How does this effect:

(a)Average kinetic energy(b)rms speed(c)Total number of collisions(d)Pressure

Molecular Effusion & Diffusion

u = 3RT

M

*Derived equation from the k-m theory

**Less massive gas molecules have higher rms speed

***Use R in units of J/mol-K

Example

Calculate the rms speed of a nitrogen molecule at 298K.

Effusion

Escape of gas molecules through a tiny hole into an evacuated space

Diffusion

Spread of one substance throughout a space or throughout a second substance

Graham’s Law of Effusion

Effusion rate of a gas is inversely proportional to the square root of its molar mass.

Rates of effusion of two gases under identical conditions*:

* At same T and P in containers with identical pinholes

Graham’s Law of Effusion

Rate directly proportional to u:

u1 = 3RT/M1

u2 3RT/M2

Graham’s Law of Effusion

Diffusion and Mean Free Path

Similar to Effusion (faster for lower mass molecules)

BUT diffusion is slower than molecular speeds because of molecular collisions

Mean Free Path: average distance traveled by a molecule between collisions– For air molecules at seal level = 6 x 10-8

m– At about 100 km in altitude = 10 cm

Homework

69, 70, 73, 76, 77, 79