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A cycle followed by an engine is shown in the figure. Find heat exchanged by the engine, with the surroundings for each section of the cycle considering $C_{v} = \left (\frac{3}{2} \right )R$ .

AB: constant volume

BC: constant pressure

CD: adiabatic

DA: constant pressure

Answers (1)

Ans.

(a) For $A\rightarrow B$ , dV=0

$dW=\int PdV=\int P\times 0=0$

$dQ=dU+dW=dU+0$

dQ=dU

$dQ=nC_vdT$

$n=1;C_v=\frac{3}{2}R$

$dQ=1\left (\frac{3}{2} R \right )(T_B-T_A)$

$dU=dQ_1=\frac{3}{2}\left (RT_B-RT_A \right )=\frac{3}{2}(P_BV_B-P_AV_A)$

(b) $B\rightarrow C$

$dQ_2=dU+dW=C_vdT+P_BdV$

$dQ_2=\frac{3}{2}R\left (T_C-T_B \right )+P_B\left (V_C-V_B \right )$

$=\frac{3}{2}[P_CV_C]-\frac{3}{2}{P_BV_B}-P_BV_B+P_BV_C$

$V_A=V_B and P_B=P_{C}$

$dQ_2=\frac{5}{2}P_BV_C-\frac{5}{2}P_BV_A=\frac{5}{2}P_B[V_C-V_A ]$

(c ) $C\rightarrow D$ , adiabatic change

$dQ_{3}=0$

In diagram $D\rightarrow A$

$\Delta P=0$

Compression of gas at constant pressure takes place. Therefore, heat exchange is similar to part (b)

$dQ_{4}=\frac{5}{2}P_A(V_A-V_D)$

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