# The gas inside of a container exerts 18 Pa of pressure and is at a temperature of 70 ^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?

Feb 19, 2017

The new trmperature is $= 186.7 K$

#### Explanation:

We apply Gay Lussac's Law

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

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

${T}_{1} = 70 K$

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

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

$= \frac{48}{18} \cdot 70 = 186.7 K$