# The gas inside of a container exerts 18 Pa of pressure and is at a temperature of 360 ^o K. If the pressure in the container changes to 16 Pa with no change in the container's volume, what is the new temperature of the gas?

Mar 2, 2017

The new temperature is $= 320 K$

#### Explanation:

We apply Gay Lussac 's Law

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

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

${T}_{1} = 360 K$

${P}_{2} = 16 P a$

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

$= \frac{16}{18} \cdot 360$

$= 320 K$