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 27 Pa with no change in the container's volume, what is the new temperature of the gas?

Feb 3, 2017

The temperature is $= 810 K$

Explanation:

We apply Gay Lussac's Law

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

${P}_{1} = 15 P a$

${T}_{1} = 450 K$

${P}_{2} = 27 P a$

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

$= \frac{27}{15} \cdot 450 = 810 K$