# The gas inside of a container exerts 28 Pa of pressure and is at a temperature of 270 ^o K. If the pressure in the container changes to 60 Pa with no change in the container's volume, what is the new temperature of the gas?

##### 1 Answer
Apr 30, 2017

The new temperature is $= 578.6 K$

#### Explanation:

We apply Gay Lussac's Law

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

The initial pressure is ${P}_{1} = 28 P a$

The initial temperature is ${T}_{1} = 270 K$

The final pressure is ${P}_{2} = 60 P a$

The final temperature is

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

$= \frac{60}{28} \cdot 270 = 578.6 K$