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A cycle followed by an engine is shown in the figure.

A to B: volume constant

B to C: adiabatic

C to D: volume constant

D to A: adiabatic

$V_C = V_D = 2V_A = 2V_B$

a) in which part of the cycle heat is supplied to the engine from outside?

b) in which part of the cycle heat is being given to the surroundings by the engine?

c) what is the work done by the engine in one cycle in terms of $P_A, P_B, V_A$ ?

Answers (1)

(a) Heat is supplied to the engine in part AB where dV=0

dW=PdV=P(0)

dW=0

dQ=dU+dW(First law of thermodynamics)

dQ=dU

$P=\frac{nRT}{V}$

V=constant

$p\propto T$

(b) Heat is given out by the system in a part CD where dV=0 and pressure and temperature decrease

(c ) WD by system= $\int_{B}^{A}PdV+\int_{B}^{C}PdV+\int_{C}^D{}PdV+\int_{D}^{A}PdV$

$\int_{B}^{A}PdV=0 \text{and}\int_{C}^D{}PdV=0$

For adiabatic change

$PV^{\gamma}=K$

$P=\frac{K}{V^{\gamma}}$

$\int_{B}^{C}PdV=\int_{A}^{B}\frac{K}{V^{\gamma }}dV=K\left [ \frac{V^{-\gamma +1}}{1-\gamma } \right ]^{V_{C}}_{V_{B}}$

$\int_{B}^{C}PdV=\frac{K}{1-\gamma}\left [ V_{C}^{1-\gamma}-V_{B}^{1-\gamma} \right ]$

$\int_{D}^{A}PdV=\frac{K}{1-\gamma}\left [ V_{A}^{1-\gamma}-V_{D}^{1-\gamma} \right ]$

$\int_{D}^{A}PdV=\frac{P_{A}V_{A}^{\gamma}V_{A}^{1-\gamma}-P_{D}V_{D}^{\gamma}V_{D}^{1-\gamma}}{1-\gamma}$

$\int_{D}^{A}PdV=\frac{P_{A}V_{A}-P_{D}V_{D}}{1-\gamma}$

$\int_{D}^{A}PdV=\frac{P_{A}V_{A}-P_{D}V_{D}}{1-\gamma}$

$\int_{B}^{C}PdV=\frac{P_{C}V_{C}-P_{B}V_{B}}{1-\gamma}$

Total $WD=\frac{[P_{C}V_{C}-P_{B}V_{B}+P_{A}V_{A}-P_{D}V_{D}]}{1-\gamma}$

For Adiabatic change= $P_{B}V_{B}^\gamma =P_{C}V_{C}^{\gamma}$

$P_C=\frac{P_BV_B^{\gamma}}{V_C^\gamma}=P_B\left (\frac{V_B}{V_C} \right )^\gamma=P_B\left (\frac{V_B}{2V_B} \right )^\gamma=\frac{P_B}{2^\gamma}$
$P_{D}=\frac{P_{A}}{2^\gamma }$

Net Work Done = $\frac{1}{1-\gamma} [P_BV_C2-\gamma-P_BV_B+P_AV_A-P_AV_D2-\gamma ]$

$=\frac{1}{1-\gamma}[P_B2V_B2^{-\gamma}-P_BV_B+P_AV_A-P_A2V_A2^{-\gamma}]$

$=\frac{1}{1-\gamma}[-P_{B}V_{B}[-2^{1-\gamma }+1]P_{A}V_{A}[-2^{1-\gamma }+1]]$

$=\frac{1}{1-\gamma}[P_{A}V_{A}-P_{B}V_{B}][1-2^{1-\gamma}]$

$2V_B=2V_A\Rightarrow V_B=V_A=\frac{1}{1-\frac{5}{3}}\left [ -2^{1-\frac{5}{3}}+1 \right ][-P_B+P_A]V_A$

$=\frac{3}{2}\left [ 1-\left ( \frac{1}{2} \right )^{\frac{2}{3}} \right ][P_B-P_A]V_A$

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