# A container with a volume of 5 L contains a gas with a temperature of 360 K. If the temperature of the gas changes to 120 K without any change in pressure, what must the container's new volume be?

Feb 14, 2016

Start with the 'combined gas law': $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$. Pressure is constant, so we can ignore it and be left with $\frac{{V}_{1}}{T} _ 1 = \frac{{V}_{2}}{T} _ 2$.
Rearranging, ${V}_{2} = \frac{{V}_{1} {T}_{2}}{T} _ 1 = \frac{5 \cdot 120}{360} = \frac{5}{3}$ $L = 1.67$ $L$.

#### Explanation:

There are a number of different gas laws that link the pressure, temperature and volume of gases.

You could memorise Boyle's Law, Charles' Law and Gay Lussac's Law. I find it simpler, though, to just learn the Combined Gas Law that combines all of them, and then eliminate the things I don't need.

More understanding, less memory: my approach to learning physics. ;-)