Question

The gas inside of a container exerts `8 Pa` of pressure and is at a temperature of `350 ^o K`. If the temperature of the gas changes to `190 ^oK` with no change in the container's volume, what is the new pressure of the gas?

Answer 1

The new pressure is `4.3 Pa`

Explanation:

We use Gay-Lussac's Law

`P_1/T_1=P_2/T_2`

`P_1=8 Pa`

`T_1=350K`

`T_2=190K`

So,

`P_2=P_1*T_2/T_1`

`=8*190/350=4.3 Pa`