IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas properties.properties.

Can be derived from Can be derived from experiment and theory.experiment and theory.

BE SURE YOU KNOW THIS BE SURE YOU KNOW THIS EQUATION!EQUATION!

P V = n R TP V = n R T

The Ideal Gas Law

PV = nRTP = pressure (atm, mmHg, torr, KPa)V = volume (in Liters)n = number of moles (mol)R = Universal Gas Law Constant (on STAAR chart)T = Temperature (in Kelvins)

CHEMSAVER P 31

The Gas Constant R The Gas Constant R

Repeated experiments show that at standard Repeated experiments show that at standard temperature (273 K) and pressure (1 atm), one temperature (273 K) and pressure (1 atm), one mole (n = 1) of gas occupies 22.4 L volume. mole (n = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate Using this experimental value, you can evaluate the the gas constant gas constant R = R = PV/nT = (1 atm* 22.4 L)/(1 mol*273 K) (1 atm* 22.4 L)/(1 mol*273 K)

RR = 0.0821 L ·atm / mol·K = 0.0821 L ·atm / mol·K = 8.31 L = 8.31 L ··kPa / mol·K kPa / mol·K

= 62.4 L = 62.4 L ··mmHg/ mol·K mmHg/ mol·K = 62.4 L= 62.4 L·· torr/ mol·K torr/ mol·K

CHEMSAVER P 31

Using PV = nRTUsing PV = nRTHow much NHow much N22 is required to fill a small room with is required to fill a small room with

a volume of 960 cubic feet (27,000 L) to a volume of 960 cubic feet (27,000 L) to 0.98atm at 25 0.98atm at 25 ooC?C?

SolutionSolution

1. Get all data into proper units1. Get all data into proper units

V = 27,000 LV = 27,000 L

T = 25 T = 25 ooC + 273 = 298 KC + 273 = 298 K

P = 0.98 atm P = 0.98 atm

R = 0.0821 L ·atm / mol·K R = 0.0821 L ·atm / mol·K

Using PV = nRTUsing PV = nRTHow much NHow much N22 is required to fill a small room with a is required to fill a small room with a

volume of 960 cubic feet (27,000 L) to 0.98atm at volume of 960 cubic feet (27,000 L) to 0.98atm at 25 25 ooC?C?

SolutionSolution

2. Now plug in those values and solve for the unknown.2. Now plug in those values and solve for the unknown.

PV = nRTPV = nRT

n = (0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)n =

(0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)

n = 1100 mol or 1.1 x 10n = 1100 mol or 1.1 x 1033 mol mol

RT RTRT RT

Using Ideal Gas Using Ideal Gas LawLawExample

What is the volume of 2.30 moles of hydrogen gas at a pressure of 122 kPa and temperature of 20.0oC?

Ans: V = nRT/P

V = (2.30 mol)(8.31 L ·kPa / mol·K8.31 L ·kPa / mol·K)(293K) 122 kPa = 46.0 L

Deviations from Deviations from Ideal Gas LawIdeal Gas Law

Real molecules have volume.Real molecules have volume.The ideal gas consumes the entire The ideal gas consumes the entire

amount of available volume. It amount of available volume. It does not account for the volume does not account for the volume of the molecules themselves.of the molecules themselves.There are intermolecular forces.There are intermolecular forces.

An ideal gas assumes there are no An ideal gas assumes there are no attractions between molecules. attractions between molecules. Attractions slow down the Attractions slow down the molecules and reduce the molecules and reduce the amount of collisions.amount of collisions.– Otherwise a gas could not Otherwise a gas could not

condense to become a liquid.condense to become a liquid.

This implies:This implies:

If the volume of space occupied is If the volume of space occupied is large and the pressure is low, the large and the pressure is low, the behavior of a gas is very close to that behavior of a gas is very close to that of an ideal gas.of an ideal gas.

We will not deal with gases at We will not deal with gases at conditions that make them non-ideal conditions that make them non-ideal in this class.in this class.