# 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?

##### 1 Answer
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