Question

The gas inside of a container exerts `9 Pa` of pressure and is at a temperature of `690 ^o K`. If the temperature of the gas changes to `210 ^oC` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

The new pressure is `=6.3Pa`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

`P_1=9Pa`

`T_1=690K`

`T_2=210+273=483K`

So,

`P_2=T_2/T_1*P_1`

`=483/690*9`

`=6.3Pa`