If a gas is cooled from 222 K to 125 K and the volume is kept constant, what would be the final pressure if the original pressure was 760 mmHg?

Assuming the mass of the gas also remains constant, this problem is an straightforward application of Gay-Lussac’s gas law: P1/T1 = P2/T2. That is, at a constant volume, the pressure and temperature of a gas are directly proportional.

We know the initial and final temperatures of the gas (T1 = 222 K; T2 = 125 K) and the initial pressure (760 mmHg). And we want to know what the final pressure would be. Since pressure and temperature are directly proportional and the final temperature is less than the initial temperature, we should expect the final pressure (P2) to be less than 760 mmHg.

Rearranging Gay-Lussac’s law to solve for P2, we obtain P2 = P1T2/T1. Plugging in our quantities, we get our final pressure: P2 = (760 mmHg)(125 K)/(222 K) = 428 mmHg.