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Question: Answered & Verified by Expert
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
PhysicsThermodynamicsKVPYKVPY 2013 (SB/SX)
Options:
  • A $0 \mathrm{~J}$
  • B $10 \mathrm{~J}$
  • C $20 \mathrm{~J}$
  • D $30 \mathrm{~J}$
Solution:
1482 Upvotes Verified Answer
The correct answer is: $30 \mathrm{~J}$


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

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