# The gas inside of a container exerts "24 Pa" of pressure and is at a temperature of "320 K". If the pressure in the container changes to "32 Pa" with no change in the container's volume, what is the new temperature of the gas?

Apr 2, 2016

The final temperature is 430 K.

#### Explanation:

This is an example of Guy-Lussac's law, which states that as long as volume is constant, the pressure and temperature of a gas are directly proportional. This means that if the pressure goes up, so does the temperature and vice-versa. The equation for Gay-Lussac's law is ${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$, where ${P}_{1}$ is the initial temperature, ${T}_{1}$ is the initial temperature, ${P}_{2}$ is the final temperature, and ${T}_{2}$ is the final temperature.

Given
${P}_{1} = \text{24 Pa}$
${T}_{1} = \text{320 K}$
${P}_{2} = \text{32 Pa}$

Unknown
${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$. Substitute the given values into the equation and solve.

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

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

T_2=(32cancel"Pa"xx320"K")/(24cancel"Pa")="430 K" rounded to two significant figures