Question

The gas inside of a container exerts `12 Pa` of pressure and is at a temperature of `250 ^o C`. If the temperature of the gas changes to `220 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

`P_2~=5,048 Pa`

Explanation:

`"use Gay-Lussac's Law"`
`P_1/T_1=P_2/T_2`
`250 ^o C=250+273=523 ^o K`
`12/523=P_2/220`
`P_2=(12*220)/523`
`P_2~=5,048 Pa`