Question

A container with a volume of `6 L` contains a gas with a temperature of `280^o K`. If the temperature of the gas changes to `320 ^o K` without any change in pressure, what must the container's new volume be?

Answer 1

The new volume will be `6.86L`

Explanation:

We start with the ideal gas law:

`PV = nRT`

and move all of the constant terms to one side. We know from the question that the temperature and the volume are changing - everything else is constant, in other words:

`V/T = (nR)/P = "constant"`

This means that we can set the two different sets of volumes and temperatures equal to each other using this equation, since we know they are constant:

`V_2/T_2 = V_1/T_1 = "constant"`

We can now solve for the new volume we are trying to find and plug in the numbers from the question:

`V_2 = (V_1*T_2)/T_1 = (6L*320K)/(280K) = 6.86L`