Question

The gas inside of a container exerts `32 Pa` of pressure and is at a temperature of `50 ^o K`. If the temperature of the gas changes to `80 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

The new pressure is `=51.2Pa`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

The initial pressure is `P_1=32Pa`

The initial temperature is `T_1=50K`

The final temperature is `T_2=80K`

The final pressure is

`P_2=T_2/T_1*P_1`

`=80/50*32`

`=51.2Pa`