Question

A 650,000 mL volume of monatomic ideal gas inside of a heat engine starts at room temperature (20.0^o C) and pressure (101.325 kPa).  At first, the volume is compressed without changing the temperature.  Then, 3.7 kJ of heat is added while the volume remains constant.  Finally, the volume is allowed to expand adiabatically back to its starting conditions. Assume all processes are reversible and ideal. Prove that the entropy and efficiency for the three-stroke heat engine are consistent with the second law of thermodynamics.

Answer 1

Cycle of the three stroke heat engine is shown below +3.7 Isochore Q=0 Adiabatic - Isotherm - -- 101,325 kla VB=VC VA AB process is isothermal T= constant AT = 0 AUAB = LnRAT = 0 from first law of thermodynamics AB = WAB + AU AB QAB = WAB +0 QAB = WAB Number of moles of monoatomic ideal gas PV = nRT

.. (101.325) X (10') (650000) X (10-) (10") - 27037 (8.314) X (20+273) BC process is isochoric V= constant dV = 0 W= [PdV WBC = 0 from first law of thermodynamics lbc = WBC + AU AB lbc = 0) + AUBC QBC = AUBC = 3.7 x 10' J AU Bc = 3.7 x 10' J = {nr(T. -Tp) 3.7 x 10° ~ x (27.037) (8.314) (T. – 20°C) 3700 - + 20 = 30.973° C x (27.037) X (8.314) CA process is isothermal lca = 0 from first law of thermodynamics QcA = WCA + AUCA 0 = WCA + AUCA WCA = - AUCA In a cyclic process AU ABCA = 0 AU ABCA = 4U AB + AU BC + AUCA

0= 0 + 3700 + AUCA AUCA = - 3700J Now Wca = - AUCA WA = 3700J In process CA TV =T.V.-1 TV,' =TV- (20 + 273) X (650000) – 1 = (30.973 + 273) X(V)-1 (20 + 273) (30.973 + 273) x (650000) = (V) = 615123.434 V. = 615123.4mL VB = VC VB = 615123.4mL Now work done in isothermal process AB WAB = nRTINOVA WAB = (27.037) X (8.314) X (20+273) x In 615123.4 650000 WAB = - 3632.26J Net work done in the cycle W=WAB + WBC + WCA W= - 3632.26 +0+ 3700 = 67.74J Now efficiency of the engine is given as net work done in the cycle divided by the

heat absorbed by the system Here in process AB → heat rejected (Q-) BC — heat absorbed (Q+) CA → No heat exchange n= W n 94,14 -0.0183 ne 1.83 % This is the real efficiency of the engine If the engine were an ideal carnot engine then efficiency would have been » = 1 - Tamen 121-(302973 27373) = 0.0361 n 3.61% Entropy change in AB AS AB =948 = - WAB AS AB = - 363229 = – 12.4x1 Entropy change in BC 48 bc = nC, INCE 45 pc = ne?) x in China AS gc = (27.037) > @ < (8.314) In (302073 :12373) = + 12.4 x 1

Entropy change in CA ASCA = A = 0 (AS)net = 0 According to second law of thermodynamics 100 percent conversion of heat into work is not possible ie. efficiency of an engine can not be 100 percent and efficiency of a real engine is always less than carnot efficiency. n = n = n<n' 3.61 % 1.83%