01The question
Q.7 A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (ab), followed by an isochoric process (bc) and an adiabatic process (ca) as shown in the figure. The volumes of the gas are V1 and V2 at a and b, respectively. If the cycle has heat input Qin and output Qout, then the efficiency of the cycle is defined as η = (Qin − Qout)/Qin. The correct statement(s) is/are:
[Given: ln 2 ≈ 0.7]
[Given: ln 2 ≈ 0.7]
- (A)If V2/V1 = 8, the heat released in the process bc is smaller than the heat absorbed in the process ab.
- (B)For a given value of V2/V1, η does not depend on the temperature of the isothermal process.
- (C)If V2/V1 = 8, then the temperature of the gas at a is 4 times the temperature of the gas at c.
- (D)If V2/V1 = 8, then the pressure of the gas at a is 4 times the pressure of the gas at b.
02The experiment
Answer—
Press START: the engine runs the cycle, measuring every heat flow, temperature and pressure.
Ready.
V₂/V₁
T of ab
Speed
03Detailed solution
Answer: —
- The three states. Take n moles with CV = (3/2)R and γ = 5/3, and let r = V2/V1. Then a = (V1, Ta) and b = (V2, Ta), since ab is isothermal. On the adiabat, TVγ−1 is constant, so Tc = Ta(V1/V2)2/3.
- (C) is true. For r = 8: Ta/Tc = 82/3 = 4.
- (A) is true. Heat absorbed on ab: Qab = nRTa ln 8 = 3 ln 2 · nRTa ≈ 2.1 nRTa. Heat released on bc: Qbc = nCV(Ta − Tc) = (3/2)(3/4) nRTa = 1.125 nRTa, which is smaller. No heat flows on the adiabat.
η = 1 − Qbc/Qab = 1 − (3/2)(1 − r−2/3)/ln r ≈ 0.46 for r = 8 (0.464 with ln 2 = 0.7).
- (B) is true. Both heats are proportional to Ta, so it cancels in η. The efficiency depends only on r. The experiment runs the cycle at 300 K and at 600 K and measures the same η.
- (D) is false. On the isotherm PV is constant, so Pa/Pb = V2/V1 = 8, not 4. The 4 is the temperature ratio Ta/Tc.
| V2/V1 | Tc/Ta | Qab in | Qbc out | η |
|---|---|---|---|---|
| 2 | 0.630 | 0.693 | 0.555 | 0.199 |
| 4 | 0.397 | 1.386 | 0.905 | 0.347 |
| 8 | 0.250 | 2.079 | 1.125 | 0.459 |
Understand the concept
Heat enters only on the isotherm: the gas does work while its temperature, and so its internal energy, stays fixed. It leaves only on the isochore, where the gas cools at constant volume. The adiabat exchanges no heat and simply returns the gas to a. Every heat in the cycle is proportional to the temperature scale, so the efficiency is a pure function of the volume ratio. The ratios are easy to mix up. On the isotherm, pressure scales with 1/V (a factor of 8). Along the adiabat, temperature scales with V−2/3 (a factor of 4).
Key takeaways
- Isothermal: ΔU = 0, so Q = W = nRT ln(V2/V1), and P ∝ 1/V.
- Isochoric: W = 0, so Q = nCVΔT.
- Adiabatic: Q = 0 and TVγ−1 is constant; for a monoatomic gas, T ∝ V−2/3.
- η = 1 − (3/2)(1 − r−2/3)/ln r, independent of T.
Verification: an independent derivation, a sympy route, a numerical integration of the first law along each leg (with no process formula assumed), and the page's own engine, run by tests/verify_p1phyq07.py. The official key (A, B, C) and the MathonGo solution (A, B, C) agree. Status: script-verified by the PrayogX simulation factory; no review by a person has been recorded. The model is 1 mol of ideal gas. The molecules' speed is drawn proportional to √T.
How to use this simulation
- Press ▶ START EXPERIMENT. The cylinder expands on the hot bath (ab), is locked and cooled on the cold bath (bc), then is insulated and compressed (ca).
- Watch the heat arrows: in on ab, out on bc, none on ca. The plotter traces the cycle on the P–V diagram.
- The readouts give T, V/V1, P/Pa and the heat so far. The last stage checks options A to D against the measurements.
- Change V₂/V₁ or T of ab and run again. The efficiency changes with the volume ratio but not with the temperature.
- Replay runs it again. Reset clears it. Classroom fills the screen for a projector.