# The gas inside of a container exerts 7 Pa of pressure and is at a temperature of 90 ^o K. If the pressure in the container changes to 48 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 7, 2017

The new temperature is $= 617.1 K$

#### Explanation:

We apply Gay Lussac's Law

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

${P}_{1} = 7 P a$

${T}_{1} = 90 K$

${P}_{2} = 48 P a$

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

$= \frac{48}{7} \cdot 90 = 617.1 K$