The gas inside of a container exerts 42 Pa of pressure and is at a temperature of 270 ^o K. If the pressure in the container changes to 36 Pa with no change in the container's volume, what is the new temperature of the gas?

Nov 26, 2016

The new temperature is $= 231.4 K$

Explanation:

We use Gay Lussac's Law

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

${T}_{1} = 270 K$

${P}_{1} = 42 P a$

${P}_{2} = 36 P a$

${T}_{2} = {P}_{2} {T}_{1} / {P}_{1} = 36 \cdot \frac{270}{42}$

$= 231.4 K = - 41.6$ºC