# The gas inside of a container exerts 18 Pa of pressure and is at a temperature of 60 ^o K. If the pressure in the container changes to 32 Pa with no change in the container's volume, what is the new temperature of the gas?

Dec 29, 2016

The temperature is $= 106.7 K$

#### Explanation:

We use Gay Lussac's Law

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

${P}_{1} = 18 P a$

${T}_{1} = 60 K$

${P}_{2} = 32 P a$

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

$= 32 \cdot \frac{60}{18} = 106.7 K$