Question

The gas inside of a container exerts `15 Pa` of pressure and is at a temperature of `450 ^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

`T_2= 960 K`

Explanation:

Lussac's Law:
`P_1/T_1=P_2/T_2`

`T_2=(P_2*T_1)/P_1`

`T_2=(32Pa*450K)/(15 Pa)`

`T_2= 960 K`