Question

The gas inside of a container exerts `32 Pa` of pressure and is at a temperature of `90 ^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 pressure is `=28.44 Pa`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

`P_1=32Pa`

`T_1=90K`

`T_2=80K`

`P_2=T_2/T_1*P_1`

`=80/90*32=28.44Pa`