# 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?

Apr 11, 2016

The new volume will be $6.86 L$

#### Explanation:

We start with the ideal gas law:

$P V = n R T$

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:

$\frac{V}{T} = \frac{n R}{P} = \text{constant}$

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:

${V}_{2} / {T}_{2} = {V}_{1} / {T}_{1} = \text{constant}$

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

${V}_{2} = \frac{{V}_{1} \cdot {T}_{2}}{T} _ 1 = \frac{6 L \cdot 320 K}{280 K} = 6.86 L$