Six moles of an ideal gas performs a cycle shown in figure. If temperatures are T_A = 600, K, T

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11.Thermodynamics

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Six moles of an ideal gas performs a cycle shown in figure. If the temperatures are $T_A = 600\, K,$ $T_B = 800\,K,$ $T_C = 2200\,K$ and $T_D = 1200\,K,$ then the work done per cycle is approximately ...... $kJ$

$20$

$30$

$40$

$60$

Solution

$\mathrm{w}=\mathrm{w}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CD}}+\mathrm{W}_{\mathrm{DA}}$

$\mathrm{W}_{\mathrm{AB}}=\mathrm{W}_{\mathrm{CD}}=0(\text { isochoric process })$

$\mathrm{W}_{\mathrm{BC}}=\mathrm{P}_{2}\left(\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}\right)=\mathrm{nR}\left(\mathrm{T}_{\mathrm{C}}-\mathrm{T}_{\mathrm{B}}\right)$

$=6 \mathrm{R}(2200-800)$

$=6 \mathrm{R} \times 1400$

$\mathrm{W}_{\mathrm{DA}}=\mathrm{P}_{1}\left(\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{D}}\right)=\mathrm{nR}\left(\mathrm{T}_{\mathrm{A}}-\mathrm{T}_{\mathrm{D}}\right)$

$=-6 R \times 600$

$\mathrm{W}=6 \mathrm{R} \times 800 \approx 40 \mathrm{kJ}$

$\mathrm{nR}\left(\mathrm{T}_{\mathrm{A}}-\mathrm{T}_{\mathrm{D}}\right)$

$6 \mathrm{R}(600-1200)$

$=-3600 \mathrm{R}$

$\mathrm{W}=8400 \mathrm{R}-3600 \mathrm{R}$

$=4800 \times \frac{25}{3}=40000=40 \mathrm{kJ}$

Standard 11

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Six moles of an ideal gas performs a cycle shown in figure. If the temperatures are $T_A = 600\, K,$ $T_B = 800\,K,$ $T_C = 2200\,K$ and $T_D = 1200\,K,$ then the work done per cycle is approximately ...... $kJ$

Solution

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