# The gas inside of a container exerts 24 Pa of pressure and is at a temperature of 180 ^o K. If the pressure in the container changes to 30 Pa with no change in the container's volume, what is the new temperature of the gas?

May 31, 2016

${225}^{0} K$

#### Explanation:

The combined gas law states that if $P$ is the pressure, $V$ is the volume and $T$ is the temperature, then $\frac{P V}{T} = k$ where $k$ is a constant.
In this case the volume is constant so this can be restated as
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} = k$ (Gay-Lussac's Law)

$\frac{24}{180} = \frac{30}{T} _ 2$

$\therefore {T}_{2} = \frac{30 \cdot \cancel{180} 15}{\cancel{24} 2} = 15 \cdot 15 = 225$