Question

You use a canister to take an air sample. The atmospheric pressure and the temperature of sampling location are 680 mmHg and 14oC, respectively. You seal the canister and take it to the lab for analysis. The lab is kept at 25oC and 760 mmHg. What is the pressure inside the canister on the day of analysis? Show your work.

Answer 1

Let the number of moles of gas we took in the container to be n which is a constant as we sealed the container, volume of the container is V which is a constant. Pressure P1 at the time of sampling was 680mmHg and temperature T1 = 14 C = 273+14 K = 287 K

If we take the container to a location where T2 = 25 C = 273+25 K = 298 K

The volume of the container remains constant, number of moles of gas = constant, only temperature and pressure will vary.

Applying ideal gas equation PV = nRT where R is the gas constant.

Initially, we have

P1V = nRT1

P1 = 680 mmHg, T = 287K

so, 680 mmHg*V = nR*287 K ---- Eqn1

Finally, we have T2 = 298 K let the pressure is P2 so,

P2V = nRT2

P2V = nR*298 K ----- Eqn2

Dividing Eqn2 by Eqn1 we get

P2/680 = 298/287

so, P2 = (298/287)*680 mmHg = 706.06 mmHg