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 $10 \mathrm{~J}$. The internal energy decreases by $20 \mathrm{~J}$ in the isochoric process. The work done by the gas in the adiabatic process is $-20 \mathrm{~J}$. The heat added to the system in the isothermal process is

• A

  $0 \mathrm{~J}$
• B

  $10 \mathrm{~J}$
• C

  $20 \mathrm{~J}$
• Correct answer
  D

  $30 \mathrm{~J}$

Solution

$\Delta \mathrm{Q}_{1 \rightarrow 2}=\Delta \mathrm{W}_{12}$ $\mathrm{w}_{\text {Total }}=\Delta \mathrm{W}_{12}+\Delta \mathrm{W}_{31}$ $10=\Delta \mathrm{w}_{12}-20$ $\Delta \mathrm{Q}_{12}=\Delta \mathrm{w}_{12}=30 \mathrm{~J}$

Hence, the correct answer is option: 'D'.