A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific
heat at constant volume, Cv = 2R. Here, R is the gas constant. Initially, each side has a volume Vo and
temperature To. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to Vo/2. Consequently, the gas temperatures on the left and the right sides become $T_{L}$ and $T_{R}$ respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

Question

A thermally insulating cylinder has a thermally insulating and frictionless movable partition in the middle, as shown in the figure below. On each side of the partition, there is one mole of an ideal gas, with specific

heat at constant volume, Cv = 2R. Here, R is the gas constant. Initially, each side has a volume Vo and

temperature To. The left side has an electric heater, which is turned on at very low power to transfer heat Q to the gas on the left side. As a result the partition moves slowly towards the right reducing the right side volume to Vo/2. Consequently, the gas temperatures on the left and the right sides become $T_{L}$ and $T_{R}$ respectively. Ignore the changes in the temperatures of the cylinder, heater and the partition.

G1 Physics · Accepted Answer

Use the first law Q=ΔU+W (W by the gas), ΔU=nCᵥΔT for an ideal gas, and the appropriate process relation. For Carnot devices, let T₁ be the cold-reservoir temperature and T₂ be the hot-reservoir temperature; η=1-T₁/T₂ and COP=T₁/(T₂-T₁).