# The gas inside of a container exerts 15 Pa of pressure and is at a temperature of 450 ^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?

Mar 14, 2018

${T}_{2} = 960 K$

#### Explanation:

Lussac's Law:
${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

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

${T}_{2} = \frac{32 P a \cdot 450 K}{15 P a}$

${T}_{2} = 960 K$