Question

Containers A and B in the diagram each contain different ideal gases at equilibrium. Container holds its volume fixed, while container has a movable piston with a fixed weight. The walls of each container (and the piston) are insulated against heat transfer, and between the chambers is an insulating barrier that is removable. If it is removed, there is another barrier that doesn't allow the gases to mix, but allows very slow heat transfer between the containers (so that the gases in each chamber change states quasi-statically while the heat transfers between them). Container A holds 3 moles of helium, while container B holds 4 moles of molecular (diatomic) nitrogen, and with the insulating barrier in place, both systems are at equilibrium. When the barrier is removed, the piston slowly begins to sink, and comes to rest when the two containers enclose the same volume. a. After the piston comes to rest, find the ratio of the pressures of the two containers,P A​ /P B​b. Find the ratio of the temperature changes of the two gases,ΔT A/ΔT B

​ (including the sign). c. Find the fraction of work done on the system that is exchanged between the two containers as W/Q

​

Answer 1

a. We are told that container B has 4 moles of nitrogen gas and container A has 3 moles of helium gas. According to the ideal gas law, PV=nRT, the pressure of a gas is directly proportional to the amount of gas (mole quantity). So the ratio of pressures will be:

PA/PB = nA/nB = 3/4 = 0.75

b. For an ideal gas undergoing a quasi-static, reversible process, the temperature and pressure changes are related by: (dT/T) = (dP/P). Since the pressures equalize, the temperature changes will be inversely proportional to the initial pressures. From part (a), we know PA = 0.75PB. So:

ΔTA/ΔTB = PB/PA = 1/0.75 = -1.33 (Since TA increases while TB decreases)

c. Work done on the system is the force on the piston times the displacement. Since the final volumes are the same and the force on the piston is constant, the displacement of the piston (and therefore work done) for container B will be 4/3 times that of container A.

The heat transferred from the hotter gas to the cooler gas will be proportional to the temperatures of the gases. From part (b), we know TB decreases more than TA, so TB starts at a higher temperature. Therefore, more heat will be transferred from the nitrogen in container B than from the helium in container A.

So WB > WA and QB > QA. Therefore W/Q will be less than 1, indicating most of the work done goes to increasing the internal energy of the system, rather than being lost as heat.