Question

The gas inside of a container exerts `24 Pa` of pressure and is at a temperature of `180 ^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

`225^0K`

Explanation:

The combined gas law states that if `P` is the pressure, `V` is the volume and `T` is the temperature, then `(PV)/T = k` where `k` is a constant.
In this case the volume is constant so this can be restated as
`P_1/T_1 = P_2/T_2 = k` (Gay-Lussac's Law)

`24/180=30/T_2`

`:.T_2=(30*cancel(180)15)/(cancel(24)2)=15*15 = 225`