Constant-Pressure Heating Followed by Isochoric Cooling
A cylinder with a frictionless piston contains 0.240 mol of an ideal monatomic gas. The gas is initially at 300 K and 1.0×10^5 Pa. It is heated at constant pressure until its temperature is 360 K; during this process the gas does 120 J of work on the piston. The piston is then locked in place, and the gas is cooled at constant volume back to 300 K. The surroundings are at 300 K. A student writes Q = ΔU + W and uses W = 120 J for the first process, then claims that because the gas returns to its starting temperature the net heat added over the two processes must be zero. You are asked to check the energy accounting and the sign conventions.
1. Calculate the heat absorbed by the gas during the constant-pressure heating.
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The gas is monatomic, so Cv = (3/2)R and Cp = (5/2)R. At constant pressure, W_by_gas = nRΔT = 0.240×8.31×60 = 120 J, matching the stated value. The internal-energy change is ΔU = nCvΔT = (3/2)(120) = 180 J. First law in the form Q = ΔU + W_by_gas gives Q = 180 + 120 = 300 J. Equivalently Q = nCpΔT = (5/2)(120) = 300 J. The positive sign means heat is absorbed by the gas. The initial volume, if needed, is V = nRT/P ≈ 6.0×10^-3 m³; the constant-pressure expansion increases it by nRΔT/P ≈ 1.2×10^-3 m³.
2. Calculate the heat exchanged during the constant-volume cooling and the net work done over the two processes.
Read the worked answer
During the locked-piston cooling, the volume is constant, so W = 0. The temperature falls by 60 K, so ΔU = nCvΔT = (3/2)(0.240)(8.31)(−60) = −180 J. Therefore Q = ΔU + W = −180 J: 180 J of heat is released. Over the two processes, the net work done by the gas is 120 J and the net heat absorbed is 300 − 180 = 120 J. The gas returns to 300 K, so its net internal-energy change is zero, and Qnet = Wnet is satisfied. The student's claim that the net heat must be zero confuses ΔU = 0 with Q = 0; the gas has expanded, so it has done net work and must have taken in net heat.
3. Identify the flaw in the student's reasoning and state what additional process would return the gas to its initial state.
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The flaw is that internal energy for an ideal gas depends on temperature, but heat and work are path-dependent. A zero net ΔU only requires Qnet = Wnet; it does not force either one to be zero. The gas ends at 300 K but at a larger volume, so it has converted 120 J of heat into work. To return to the original state, the gas must be compressed isothermally at 300 K back to its initial volume. During that compression, work is done on the gas and an equal amount of heat is rejected to the surroundings at 300 K. The full cycle would then have zero net change in internal energy, and the net work would equal the net heat over the complete loop.