# The gas inside of a container exerts 15 Pa of pressure and is at a temperature of 45 ^o K. If the temperature of the gas changes to 90 ^oK with no change in the container's volume, what is the new pressure of the gas?

Feb 27, 2016

$30 \text{Pa}$

#### Explanation:

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

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

So

$\left(15 \text{Pa")/(45"K") = p_2/(90"K}\right)$

${p}_{2} = 30 \text{Pa}$