# A container with a volume of 14 L contains a gas with a temperature of 160^o K. If the temperature of the gas changes to 80 ^o K without any change in pressure, what must the container's new volume be?

Mar 25, 2016

$7 \setminus \textrm{L}$

#### Explanation:

Assuming the gas is ideal, this can be calculated in a few different ways. The Combined Gas Law is more appropriate than the Ideal Gas Law, and more general (so being familiar with it will benefit you in future problems more frequently) than Charles' Law, so I'll use it.

$\setminus \frac{{P}_{1} {V}_{1}}{{T}_{1}} = \setminus \frac{{P}_{2} {V}_{2}}{{T}_{2}}$

Rearrange for ${V}_{2}$
${V}_{2} = \setminus \frac{{P}_{1} {V}_{1}}{{T}_{1}} \setminus \frac{{T}_{2}}{{P}_{2}}$

Rearrange to make proportional variables obvious
${V}_{2} = \setminus \frac{{P}_{1}}{{P}_{2}} \setminus \frac{{T}_{2}}{{T}_{1}} {V}_{1}$

Pressure is constant, so whatever it is, it divided by itself will be $1$. Substitute in values for temperature and volume.
${V}_{2} = \left(1\right) \left(\setminus \frac{80}{160}\right) \left(14\right)$

Simplify
${V}_{2} = \setminus \frac{14}{2}$

End with the same units you started with
${V}_{2} = 7 \setminus \textrm{L}$

This answer makes intuitive sense. If the pressure is constant, decreasing the temperature should decrease the volume, since less energetic particles will take up a smaller amount of room.

Note that $\setminus \textrm{L}$ is not an SI unit, so it would usually be bad practice to not convert it to $\setminus {\textrm{m}}^{3}$ before doing any calculations with it. If I had tried to use volume in liters to calculate pressure, for example, the units for pressure that would result would be non-standard and the value would be difficult to compare to anything.

It worked here because this equation was based on how all the same variables varied with respect to each other, and I started with volume in a non-standard unit and ended with volume a non-standard unit.