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The three processes in a thermodynamic cycle shown in the figure are : Process $1 \rightarrow 2$ is isothermal; Process $2 \rightarrow 3$ is isochoric (volume remains constant); Process $3 \rightarrow 1$ is adiabatic. The total work done by the ideal gas in this cycle is $10 \,J$. The internal energy decreases by $20 \,J$ in the isochoric process. The work done by the gas in the adiabatic process is $-20 \,J$. The heat added to the system in the isothermal process is .............. $J$
$0$
$10$
$20$
$30$
Solution
(d)
Work done in complete cycle,
$W=W_{12}+W_{23}+W_{31}$
Given $W=10 \,J , W_{23}=0$ and $W_{31}=-20 \,J$
So, from Eq. $(i)$, we have $W_{12}=30 \,J$
As in isothermal process,
Heat absorbed $=$ Work done
$\therefore$ Heat absorbed in process $1 \rightarrow 2=30 \,J$
