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A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in figure. Find heat exchanged by the engine, with the surroundings for eachsection of the cycle. C_V=3/2R (a) A to B : constant volume (b) B to C: constant pressure (c) C to D: adiabatic (d) D to A: constant pressure |
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Answer» Solution :(a)For process A to B volume is CONSTANT hence WORK done dW=0 `rArr` From first LAW of thermodynamics , dQ=dU+dW `therefore` dQ=dU+0 `therefore` dQ=dU `=nC_V DT` `=nC_V(T_B-T_A)` `=(1)(3/2R)(T_B-T_A)` `=3/2(RT_B-RT_A)` `=3/2(P_B V_B -P_A V_A)` because in EQUATION of gas PV = RT at point A, `P_A V_A =RT_A` and at point B, `P_B V_B=RT_B` `therefore` Heat enchanged `=3/2 (P_B V_B-P_A V_A)` (b) For process B to C pressure is constant hence work done, `dW=PDeltaV` `therefore dW=P_B(V_C-V_B)` `rArr` From first law of thermodynamics, dQ=dU+dW `=3/2R(T_C-T_B)+P_B(V_C+V_B)` `therefore dQ=3/2 (RT_C -RT_B) +P_B (V_C -V_B)` `=3/2(P_C V_C- P_B V_B) + P_B (V_C-V_B)` `=5/2 P_B (V_C -V_B)` [ `because P_B=P_C` and `P_B=P_A` ] `therefore` Heat exchanged `=5/2P_B (V_C-V_A)` (d) For process C to D , it is adiabatic process so heat exchanged dQ=0 (d) For process D to A at constant pressure `P_A` the gas compressed from volume `V_D` to `V_A` The heat exchanged will be as shown in (b) `P_A` instead of `P_B`, `V_A` instead of `V_C` and `V_A` instead of `V_D` have to be taken. `therefore`Heat exchanged `dQ=5/2 P_A(V_A-V_D)` |
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