# The gas inside of a container exerts 18 Pa of pressure and is at a temperature of 360 ^o K. If the pressure in the container changes to 27 Pa with no change in the container's volume, what is the new temperature of the gas?

May 23, 2017

The final temperature, ${T}_{2}$, will be $\text{540 K}$.

#### Explanation:

This question is an example of Gay-Lussac's gas law, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that if the pressure increases, so does the temperature, and vice-versa. The equation to use in order to answer your question is:

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

Given

${P}_{1} = \text{18 Pa}$
${T}_{1} = \text{360 K}$ $\Leftarrow$ The degree symbol $\left(\text{^@}\right)$ is not used with Kelvins.
${P}_{2} = \text{27 Pa}$

Unknown: ${T}_{2}$

Solution
Rearrange the equation to isolate ${T}_{2}$. Insert your data into the equation and solve.

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

T_2=(27color(red)cancel(color(black)("Pa"))xx360"K")/(18color(red)cancel(color(black)("Pa")))="540 K"

As you can see, as the pressure increased, so did the temperature.

May 23, 2017

$540 \text{K}$

#### Explanation:

We can use the temperature-pressure relationship of gases illustrated by Gay-Lussac's law:

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

At constant volume, the temperature and pressure of a fixed quantity of gas are directly proportional to each other, which is explained by the kinetic-molecular theory.

We can rearrange this equation to solve for the final temperature ${T}_{2}$:

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

and finally, plug in the known variables to find the temperature.

T_2 = ((27"Pa")(360K))/(18"Pa") = color(red)(540"K"