Question

The gas inside of a container exerts `15 Pa` of pressure and is at a temperature of `180 ^o K`. If the pressure in the container changes to `32 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The new temperature is `=384K`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

Initial pressure is `P_1=15Pa`

Initial temperature is `T_1=180K`

Final pressure is `P_2=32Pa`

Final temperature is

`T_2=P_2/P_1*T_1`

`=32/15*180=384K`