Question

The gas inside of a container exerts `12 Pa` of pressure and is at a temperature of `35 ^o K`. If the temperature of the gas changes to `20 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

`P_2 = 6.86 Pa`

Explanation:

The ideal gas laws (Charles' Law) state:

`(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2`

Where `P_1,_2` are pressures – units don't matter in this case as long as they are consistent, because this is a ratio.
`V_1,_2` are the corresponding volumes in Liters
`T_1,_2` are the temperatures in degrees

With a constant container volume, Charles' Law becomes simply:
`P_1/T_1 = P_2/T_2`

Rearrange for your known values:
`P_2 = P_1 * T_2/T_1`

`P_2 = 12 * 20/35`

`P_2 = 6.86 Pa`