# The gas inside of a container exerts 18 Pa of pressure and is at a temperature of 450 ^o K. If the pressure in the container changes to 25 Pa with no change in the container's volume, what is the new temperature of the gas?

Dec 25, 2017

The new temperature will be $\approx$$\text{630 K}$.

#### Explanation:

This is an example of Guy-Lussac's law which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. So if the pressure increases, so will the temperature and vice versa.

The equation is:

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

where:

$P$ is pressure and $T$ is temperature.

Organize the data.

Known

${P}_{1} = \text{18 Pa}$

${T}_{1} = \text{450 K}$

${P}_{2} = \text{25 Pa}$

Unknown

${T}_{2}$

Solution

Rearrange the equation to isolate ${T}_{2}$. Plug in the known values and solve.

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

T_2=(25color(red)cancel(color(black)("Pa"))xx450"K")/(18color(red)cancel(color(black)("Pa")))="630 K" rounded to two significant figures due to 18 Pa and 25 Pa