Efficiency of a Three-Step Ideal-Gas Cycle
Two moles of a monatomic ideal gas are taken through a three-step cycle. State A has volume 10.0 L and temperature 300 K. Step A→B is isochoric heating to 600 K. Step B→C is isothermal expansion to 20.0 L. Step C→A is isobaric compression back to 10.0 L. The gas is assumed to be ideal and the steps are quasi-static. R = 8.31 J/(mol·K). A student computes the efficiency and compares it with the Carnot efficiency between 600 K and 300 K. She is surprised that her result is much lower and claims she must have made an arithmetic error, because each step is reversible. She asks you to check the calculation and to explain the comparison. The molar heat capacity at constant volume is (3/2)R; at constant pressure it is (5/2)R.
1. Calculate the net work done per cycle and the efficiency of this cycle. Show the heat exchange for each step.
Read the worked answer
Use R = 8.31 J/(mol·K). A→B is isochoric: W_AB = 0, Q_AB = n C_V ΔT = 2 × (3/2 × 8.31) × 300 = 7.48 × 10³ J. B→C is isothermal at 600 K: W_BC = Q_BC = nRT ln(V_C/V_B) = 2 × 8.31 × 600 × ln2 = 6.91 × 10³ J. C→A is isobaric: W_CA = P ΔV = nR ΔT = 2 × 8.31 × (300 − 600) = −4.99 × 10³ J; Q_CA = n C_P ΔT = 2 × (5/2 × 8.31) × (−300) = −1.25 × 10⁴ J.
Net work W = 0 + 6.91 × 10³ − 4.99 × 10³ = 1.93 × 10³ J per cycle. Heat added Q_in = 7.48 × 10³ + 6.91 × 10³ = 14.39 × 10³ J. Efficiency η = W/Q_in = 1.93/14.39 = 0.134, about 13%.
2. Evaluate the student’s claim that the efficiency should equal the Carnot efficiency because each step is reversible.
Read the worked answer
The claim is false. Reversibility of each step does not make the cycle a Carnot cycle. The Carnot value 1 − T_C/T_H is the maximum efficiency for any cycle exchanging heat only with two reservoirs at fixed temperatures T_H and T_C. Here heat is added during A→B while the temperature rises from 300 K to 600 K, not at a single T_H, and heat is rejected during C→A while the temperature falls from 600 K to 300 K.
The cycle is not bounded by two constant-temperature reservoirs, so its efficiency is lower. The calculation is consistent: the Carnot value between 600 K and 300 K would be 1 − 300/600 = 0.50, much larger than 0.134.
3. Would the efficiency be higher or lower if the gas were diatomic with C_V = (5/2)R? Justify with a calculation or clear proportionality.
Read the worked answer
For a diatomic gas, C_V = (5/2)R and C_P = (7/2)R. The isothermal work and the isobaric work are unchanged because they depend on R, not on C_V or C_P: W_BC = 6.91 × 10³ J and W_CA = −4.99 × 10³ J, so net work is still 1.93 × 10³ J.
But Q_AB becomes 2 × (5/2 × 8.31) × 300 = 1.25 × 10⁴ J instead of 7.48 × 10³ J, so Q_in increases to about 1.94 × 10⁴ J. Therefore η ≈ 1.93/19.4 = 0.099, about 9.9%, lower than the monatomic value. Adding the same net work costs more heat input.