The gas inside of a container exerts 24 Pa of pressure and is at a temperature of 90 ^o K. If the pressure in the container changes to 42 Pa with no change in the container's volume, what is the new temperature of the gas?

Sep 19, 2017

The temperature is $= 157.5 K$

Explanation:

We apply GayLussac's Law

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ at constant Volume

The initial Pressure is ${P}_{1} = 24 P a$

The initial Temperature is ${T}_{1} = 90 K$

The final Pressure is ${P}_{2} = 42 P a$

The final temperature is

${T}_{2} = {P}_{2} / {P}_{1} \cdot {T}_{1} = \frac{42}{24} \cdot 90 = 157.5 K$