# The gas inside of a container exerts 12 Pa of pressure and is at a temperature of 35 ^o K. If the temperature of the gas changes to 20 ^oK with no change in the container's volume, what is the new pressure of the gas?

##### 1 Answer
Dec 31, 2016

${P}_{2} = 6.86 P a$

#### Explanation:

The ideal gas laws (Charles' Law) state:

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

Where ${P}_{1} {,}_{2}$ are pressures – units don't matter in this case as long as they are consistent, because this is a ratio.
${V}_{1} {,}_{2}$ are the corresponding volumes in Liters
${T}_{1} {,}_{2}$ are the temperatures in degrees

With a constant container volume, Charles' Law becomes simply:
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

Rearrange for your known values:
${P}_{2} = {P}_{1} \cdot {T}_{2} / {T}_{1}$

${P}_{2} = 12 \cdot \frac{20}{35}$

${P}_{2} = 6.86 P a$