InterviewSolution
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InterviewSolution
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A cyclic process abcd is given for a monoatomic gas (`C_V=3/2R` and `C_p=5/2R`) as shown in figure. Find Q, W and `DeltaU` in each of the four processes separately. Also find the effieciency of cycle. |
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Answer» Correct Answer - A::C Process ab `V=` constant (`:.` Isochoric process) `W_(ab)=0` `:.` `Q_(ab)=DeltaU_(ab)=nC_VDeltaT` `=n(3/2R)(T_b-T_a)` `=3/2(nRT_b-nRT_a)` `=3/2(p_bV_b-p_aV_a)` `=3/2(2p_0V_0-p_0V_0)` `=1.5p_0V_0` Process bc `p=` constant (`:.` Isobaric process) `Q_(bc)=nC_pDeltaT` `=n(5/2R)(T_c-T_b)` `=5/2(nRT_c-nRT_b)` `=5/2(p_cV_c-p_bV_b)` `=5/2(4p_0V_0-2p_0V_0)` `=5p_0V_0` `DeltaU_(bc)=nC_VDeltaT` `=n(3/2R)(T_c-T_b)` `=3/2(nRT_c-nRT_b)` `=3/2(p_cV_c-p_bV_b)` `=3/2(4p_0V_0-2p_0V_0)` `=3p_0V_0` `W_(bc)=Q_(bc)-DeltaU_(bc)=2p_0V_0` Process cd Again an isochoric process. `:.` `W_(cd)=0` `Q_(cd)=DeltaU_(cd)=nC_VDeltaT` `=n(3/2R)(T_d-T_c)` `=3/2(nRT_d-nRT_c)` `=3/2(p_dv_d-p_cV_c)` `=3/2(2p_0V_0-4p_0V_0)` `=-3p_0V_0` Process da This is an isobaric process. `:.` `Q_(da)=nC_pDeltaT` `=n(5/2R)(T_a-T_d)` `=5/2(nRT_a-nRT_d)` `=5/2(p_aV_a-p_dV_d)` `=5/2(p_0V_0-2p_0V_0)` `=-2.5 p_0V_0` `DeltaU_(da)=nC_VDeltaT` `=n(3/2R)(T_a-T_d)` `=3/2(nRT_a-nRT_d)` `=3/2(p_aV_a-p_dV_d)` `=3/2(p_0V_0-2p_0V_0)` `=-1.5 p_0V_0` `W_(da)=Q_(da)-DeltaU_(da)` `=-p_0V_0` Efficiency of cycle In the complete cycle, `W_(n et)=W_(ab)+W_(bc)+W_(cd)+W_(da)` `=0+2p_0V_0+0-p_0V_0` `=p_0V_0` |
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