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 cm^{3} 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 cm^{3}. 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:

P_{1} = P, V_{1} = 100 cm^{3}, n_{1} = n, T_{1} = 0 + 273 = 273 K

PV_{1} = nRT_{1}

Finally:

P_{2} = P, V_{2} = 87.2 cm^{3}, n_{2} = n, T_{2} = ?

PV_{2} = nRT_{2}

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

P/nR = T_{1}/V_{1} = T_{2}/V_{2}

and T_{2} = V_{2}T_{1}/V_{1}

Plugging in the values we know:

T_{2} = 87.2 cm^{3} x 273 K / 100.0 cm^{3}

T_{2} = 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