Question

The gas inside of a container exerts `18 Pa` of pressure and is at a temperature of `70 ^o K`. If the pressure in the container changes to `48 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

The new trmperature is `=186.7K`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

`P_1=18Pa`

`T_1=70K`

`P_2=48Pa`

`T_2=P_2/P_1*T_1`

`=48/18*70=186.7K`