# The gas inside of a container exerts 12 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 2, 2018

The new temperature is $= 280 K$

#### Explanation:

Apply Gay Lussac's Law

$\text{Pressure"/"Temperature"="Constant}$

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$ at $\text{ Constant Volume }$

The initial pressure is ${P}_{1} = 12 P a$

The initial temperature is ${T}_{1} = 70 K$

The final pressure is ${P}_{2} = 48 P a$

The final temperature is

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

$= \frac{48}{12} \cdot 70$

$= 280 K$