# A container with a volume of 4 L contains a gas with a temperature of 270^o K. If the temperature of the gas changes to 420^o K without any change in pressure, what must the container's new volume be?

Jan 14, 2017

The new volume is $\approx 6 L$.

#### Explanation:

Given that the pressure of the gas remains constant, we can identify this as an isobaric process. This is a process 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 ${V}_{i} = 4 L$, ${T}_{i} = 270 K$, and ${T}_{f} = 420 K$:

${V}_{f} = \frac{4 L \cdot 420 \cancel{K}}{270 \cancel{K}}$

${V}_{f} = 6.22 \ldots L \approx 6 L$