Question

The gas inside of a container exerts `12 Pa` of pressure and is at a temperature of `250 ^o K`. If the temperature of the gas changes to `200 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

The pressure is `=9.6K`

Explanation:

We apply Gay Lussac's Law

`P_1/T_1=P_2/T_2`

`P_1=12Pa`

`T_1=250K`

`T_2=200Pa`

`P_2=(P_1*T_2)/T_1`

`=12*200/250=9.6Pa`