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Thermodynamics · Paper 2 — Free Physics HL Practice Questions

1ChallengeSAQ-LHeat engines and efficiency7 marksPaper 2~11 min
A Stirling engine operates with n=0.040 moln = 0.040\,\text{mol} of an ideal monatomic gas undergoing four processes per cycle: - 1→21 \to 2: Isothermal compression at Tc=350 KT_c = 350\,\text{K}, volume from V1=2.0×10−3 m3V_1 = 2.0 \times 10^{-3}\,\text{m}^3 to V2=1.0×10−3 m3V_2 = 1.0 \times 10^{-3}\,\text{m}^3 - 2→32 \to 3: Isochoric heating from 350 K350\,\text{K} to Th=650 KT_h = 650\,\text{K} - 3→43 \to 4: Isothermal expansion at Th=650 KT_h = 650\,\text{K}, volume from 1.0×10−3 m31.0 \times 10^{-3}\,\text{m}^3 to 2.0×10−3 m32.0 \times 10^{-3}\,\text{m}^3 - 4→14 \to 1: Isochoric cooling from 650 K650\,\text{K} to 350 K350\,\text{K} Use R=8.31 J mol−1K−1R = 8.31\,\text{J\,mol}^{-1}\text{K}^{-1} and isothermal work W=nRTln⁡ ⁣(VfVi)W = nRT\ln\!\left(\dfrac{V_f}{V_i}\right).
(a)
Calculate the net work done per cycle by the engine. [3 marks]
(b)
The Stirling cycle is represented on a pp–VV . Explain, by reference to the areas enclosed on this , why the net work output per cycle is positive. [2 marks]
(c)
The engine is reversed to operate as a refrigerator between the same two temperatures. Calculate the maximum (ideal) coefficient of performance COPref\text{COP}_{\text{ref}} of this reversed cycle, and evaluate one thermodynamic and one engineering reason why the actual COP would be lower than this ideal value. [2 marks]
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