# The gas inside of a container exerts 9 Pa of pressure and is at a temperature of 240 ^o K. If the pressure in the container changes to 3 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 12, 2016

${T}_{2} = {80}^{o} K$

#### Explanation:

$\text{use The Gay-Lussac's law}$
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$
$\frac{9}{240} = \frac{3}{T} _ 2$
${T}_{2} = \frac{240 \cdot \cancel{3}}{\cancel{9}}$
${T}_{2} = \frac{\cancel{240}}{\cancel{3}}$
${T}_{2} = {80}^{o} K$