Ideal Gas Law: Worked Chemistry Problems

The ideal gas law relates the pressure, volume, quantity, and temperature of an ideal gas. At ordinary temperatures, you can use the ideal gas law to approximate the behavior of real gases. Here are examples of how to use the ideal gas law. You may wish to refer to the general properties of gases to review concepts and formulae related to ideal gasses.

Ideal Gas Law Problem #1

Problem

A hydrogen gas thermometer is found to have a volume of 100.0 cm³ when placed in an ice-water bath at 0°C. When the same thermometer is immersed in boiling liquid chlorine, the volume of hydrogen at the same pressure is found to be 87.2 cm³. What is the temperature of the boiling point of chlorine?

Solution

For hydrogen, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

Initially:

P₁ = P, V₁ = 100 cm³, n₁ = n, T₁ = 0 + 273 = 273 K

PV₁ = nRT₁

Finally:

P₂ = P, V₂ = 87.2 cm³, n₂ = n, T₂ = ?

PV₂ = nRT₂

Note that P, n, and R are the same. Therefore, the equations may be rewritten:

P/nR = T₁/V₁ = T₂/V₂

and T₂ = V₂T₁/V₁

Plugging in the values we know:

T₂ = 87.2 cm³ x 273 K / 100.0 cm³

T₂ = 238 K

Answer

238 K (which could also be written as -35°C)

Ideal Gas Law Problem #2

Problem

2.50 g of XeF4 gas is placed into an evacuated 3.00 liter container at 80°C. What is the pressure in the container?

Solution

PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

P=?
V = 3.00 liters
n = 2.50 g XeF4 x 1 mol/ 207.3 g XeF4 = 0.0121 mol
R = 0.0821 l·atm/(mol·K)
T = 273 + 80 = 353 K

Plugging in these values:

P = nRT/V

P = 00121 mol x 0.0821 l·atm/(mol·K) x 353 K / 3.00 liter

P = 0.117 atm

Answer

0.117 atm