Question

In a cycle $ABCA$ consisting of isothermal expansion $AB$, isobaric compression $BC$ & adiabatic
[Given:$T_A=T_B=400K;\gamma =1.5,ln2=0.693,2\frac{1}{3}=1.26$]

Answer 1

$V_A=V_0,T_A=400k$

$P_A=\frac{400nR}{V_0}$

At $B$

$V_B=2V_0,T_B=400k{P}_B=\frac{400nR}{2V_0}$

At $C$

$V_c=?,T_c=?,P_c=\frac{400nR}{2V_0}$

$AC$ is adiabatic process

$p{v}^r=constant$

$V_c={\left(\frac{P_A}{P_C}\right)}^{\frac{1}{r}}$ $V_A=(2{)}^{\frac{1}{1.5}}{V}_0$

$V_c=2^{\frac{2}{3}}{V}_0$

$T_c=\frac{P_c{V}_c}{nR}=400\times (2{)}^{\frac{-1}{3}}$

For $AB$ $△U=0$ $△Q=△W=nR\times \left(400\right)\times ln\left(\frac{2V_0}{V_0}\right)$

$=400nRl{n}_2$

For $BC△U=n{C}_v△T=\frac{nr}{R-1}\times \left(400\right(2{)}^{\frac{-1}{3}}-400)$

$=-2\times 400nR(1-2^{\frac{-1}{3}})$

$△Q=n{C}_p△T=\frac{-nrR}{r-1}\times 400(1-2^{\frac{-1}{3}})$

$=-3\times 400nR(1-2^{\frac{-1}{3}})$

For $AC△Q=O$ $△U=n{C}_v△T=2\times 400nR(1-2^{\frac{-1}{3}})$

$△w=-△V=-2\times 400nR(1-2^{\frac{-1}{3}})$

Efficiency (e) $=1-\frac{Q_{released}}{Q_{absorbed}}=-1-\frac{1-3(1-2^{\frac{-1}{3}})}{ln2}$