Initially, a gas in a closed container has a temperature of 30°C and a pressure of 760 mm Hg. We’re being asked to determine the **pressure of the gas sample **if the temperature changes to 1155°C.

Recall that the * ideal gas law *is:

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}$

The * pressure, volume, and temperature of a gas *are related to the number of moles of gas and the universal gas constant:

$\frac{\mathbf{PV}}{\mathbf{T}}\mathbf{=}\mathbf{nR}$

The value **nR is constant**. For a given moles of gas, the initial and final pressure, volume, and temperature of the gas are related by the * combined gas law*:

$\overline{)\frac{{\mathbf{P}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{2}}{\mathbf{V}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}}}$

Since the container is closed, the **volume of the gas remains constant** and the combined gas law becomes:

$\overline{)\frac{{\mathbf{P}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}}}$

Aerosol cans carry clear warnings against incineration because of the high pressures that can develop upon heating. Suppose that a can contains a residual amount of gas at a pressure of 760 mmHg and a temperature of 30 ^{o}C.

What would the pressure be if the can were heated to 1155 ^{o}C?