# A container with a volume of 36 L contains a gas with a temperature of 280^o K. If the temperature of the gas changes to 160 ^o K without any change in pressure, what must the container's new volume be?

Dec 30, 2016

The new volume is $\approx 21 L$.

#### Explanation:

I answered a similar question recently, so I will use the same explanation.

Given that the pressure of the gas remains constant, we can identify this as an isobaric process. This is a process is one in which pressure remains constant, but temperature and volume change. We can use Charles' Law to find an unknown temperature or volume value for an isobaric process, given that we know something about the other three values.

$\frac{{V}_{i}}{{T}_{i}} = \frac{{V}_{f}}{{T}_{f}}$

This law describes how gas expands as temperature increases. The inverse is also true. You don't need to memorize Charles' Law (or Boyle's' or Gay-Lussac's) if you remember the combined gas law, which relates all of these simple gas laws:

$\frac{P V}{T} =$constant

or, equivalently:

$\frac{{P}_{i} {V}_{i}}{{T}_{i}} = \frac{{P}_{f} {V}_{f}}{{T}_{f}}$

( also written $P V = n R T$ )

We can see that if pressure is constant, ${P}_{i} = {P}_{f}$ and the terms drop away, leaving us with Charles' Law. This works for any of the simple gas laws: the term which remains constant will drop away.

We want to solve for ${V}_{f}$, so we can rearrange the equation to isolate this variable:

${V}_{f} = \frac{{V}_{i} {T}_{f}}{T} _ i$

Given ${V}_{i} = 36 L$, ${T}_{i} = 280 K$, and ${T}_{f} = 160 K$:

${V}_{f} = \frac{36 L \cdot 160 \cancel{K}}{280 \cancel{K}}$

${V}_{f} = 20.57 \ldots L \approx 21 L$