An ideal gas is taken through a cyclic thermodynamic process thro… - Vidyadip

Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are $Q_1=5960 \mathrm{~J}, Q_2=-5585 \mathrm{~J} . Q_3=-2980 \mathrm{~J}$ and $Q_4=3645$ respectively. The corresponding quantities of work involved are $W_1=200 \mathrm{~J}, \mathrm{~W}_2=-825 \mathrm{~J}, \mathrm{~W}_3=-1100 \mathrm{~J}$ and $\mathrm{W}_4$ respectively. Find the value of $\mathrm{W}_4$. What is the efficiency of the cycle?

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Answer

As the process is cyclic, therefore, $\Delta\text{U}=0$ According to first law of thermodynamics $\Delta\text{Q}=\Delta\text{U}+\Delta\text{W}=\Delta\text{W}$
$\Delta\text{W}=\Delta\text{Q}$ or $\text{W}_1+\text{W}_2+\text{W}_3+\text{W}_4$
$=\text{Q}_1+\text{Q}_2+\text{Q}_3+\text{Q}_4$
$\text{W}_4(\text{Q}_1+\text{Q}_2+\text{Q}_3+\text{Q}_4)-(\text{W}_1+\text{W}_2+\text{W}_3)$
$=(5960-5585-2980+3645)$
$-(-200-825-1100)$
$=1040-275=765\text{J}$
$\text{Effeciency}=\frac{\text{Net work done}}{\text{total heat absorbed}}$
$=\frac{\text{W}_1+\text{W}_2+\text{W}_3+\text{W}_4}{\text{Q}_1+\text{Q}_4}$
$\eta=\frac{2200-825-1100+765}{5960+3645}=\frac{1040}{9605}=0.1083$
$\eta=0.1083\times100\%=10.83\%$

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