Question

The gas inside of a container exerts `48 Pa` of pressure and is at a temperature of `320 ^o K`. If the pressure in the container changes to `60 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`T_2=400^o K`

Explanation:

`P_1/T_1=P_2/T_2`
`48/320=60/T_2`
`T_2=(320*60)/48`
`T_2=400^o K`