Question

Three processes form a thermodynamic cycle as shown on $P-V$ diagram for an ideal gas. Process $1\to 2$ takes place at constant temperature $\left(300K\right)$. Process $2\to 3$ takes place at constant volume. During this process $40J$ of heat leaves the system. Process $3\to 1$ is adiabatic and temperature $T_3$ is $275K$. Work done by the gas during the process $3\to 1$ is

• A. $-40J$
• B. $-20J$
• C. $+40J$
• D. $+20J$

Answer 1

Answer: (A)

$-40J$

In the process $1\to 2,\Delta Q_{1\to 2}=\Delta W_{1\to 2}$
In the process $2\to 3,\Delta W_{2\to 3}=0$
In the process $3\to 1,\Delta Q_{3\to 1}=0$
The complete process being cyclic, $\Delta U=0$
Hence $\Delta Q=\Delta W\Rightarrow \Delta Q_{1\to 2}+\Delta Q_{2\to 3}+\Delta Q_{3\to 1}=\Delta W_{1\to 2}+\Delta W_{2\to 3}+\Delta W_{3\to 1}\Rightarrow \Delta Q_{2\to 3}=\Delta W_{3\to 1}=-40J$