Question

The gas inside of a container exerts `27 Pa` of pressure and is at a temperature of `250 ^o K`. If the pressure in the container changes to `30 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`T_2=277,78 ^o K`

Explanation:

`"we must use Gay-Lussac's Law"`

`P_1/T_1=P_2/T_2`

`"where; " P_1=27 Pa" "T_1=250^ o K" "P_2=30Pa`

`27/250=30/T_2`

`T_2=(250*30)/27`

`T_2=277,78 ^o K`