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 40 Pa with no change in the container's volume, what is the new temperature of the gas?

Feb 27, 2016

$300 \text{K}$

Explanation:

First the gas to be ideal. Then, from Gay-Lussac's Law, we know that

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

So,

$\left(180 \text{K")/(24"Pa") = T_2/(40"Pa}\right)$

${T}_{2} = 300 \text{K}$