An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat

Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are $Q_{1}=5960 \mathrm{~J}, Q_{2}=-5585 \mathrm{~J}, Q_{3}=-2908 \mathrm{~J}$ and $Q_{4}=3645 \mathrm{~J}$, respectively. The corresponding quantities of work involved are $W_{1}=2200 \mathrm{~J}, W_{2}=-825 \mathrm{~J}$, $W_{3}=-1100 \mathrm{~J}$ and $W_{4}$ respectively.\ (i) Find the value of $W_{4}$ in J\

Professor PC Thomas Classes · Answer

(i) : To calculate $W_{4}$.\ $\Delta U=0$ as the process is given to be cyclic.\ According to first law of thermodynamics,\ $\Delta Q=\Delta U+\Delta W=0+\Delta W$ or $\Delta Q=\Delta W$\ or $Q_{1}+Q_{2}+Q_{3}+Q_{4}=W_{1}+W_{2}+W_{3}+W_{4}$\ or 5960-5585-2980+3645\ $=2200-825-1100+W_{4}$\ or $\quad W_{4}=765 \mathrm{~J}$\ (ii) To calculate efficiency of cycle $\eta$.\ $\eta=\frac{	ext { Work done }}{	ext { Heat supplied }}$\ or $\eta=\frac{W_{1}+W_{2}+W_{3}+W_{4}}{Q_{1}+Q_{4}}=\frac{1040}{9605}=10.82 \%$\ Hence $W_{4}=765 \mathrm{~J}$ and $\eta=10.82 \%$.