Question

The gas inside of a container exerts `12 Pa` of pressure and is at a temperature of `360 ^o K`. If the pressure in the container changes to `16 Pa` with no change in the container's volume, what is the new temperature of the gas?

Answer 1

`T_2=480 ^o K`

Explanation:

`P_1/T_1=P_2/T_2`
`cancel(12)/cancel(360)=16/T_2`
`1/30=16/T_2`
`T_2=16*30`
`T_2=480 ^o K`