Question

A thermodynamic system is taken through the cycle $ABCD$ as shown in the figure. Heat rejected by the gas during the cycle is:

• (A) $PV$
• (B) $2PV$
• (C) $4PV$
• (D) $\frac{1}{2}PV$

Answer 1

The Correct Option is B

Explanation:
The following figure shows the cyclic process of gas. If an object returns to its initial position after one or more processes it went through.$ABCDA$ is the cycle. The pressure at the points both $D$ and $C$ remain the same, which is $2P$. The volume at the point $D$ is $V$ and at the point $C$ is $3V$. So, the work done by the gas from point $D$ to point $C,W_{DC}=2P(3V-V)=4PV$The pressures at the points $C$ and $B$ are $2P$ and $P$ respectively. The volume at the points both $C$ and $B$ remain the same, which is $3V$. So, the work done by the gas from point $C$ to point $B,$$W_{CB}=P(3V-3V)=0$The pressure at the points both $B$ and $A$ remain the same, which is $P$. The volume at the point $B$ is $3V$ and at the point $A$ is $V$So, the work done by the gas from point $B$ to point $A,W_{BA}=P(V-3V)=-2PV$The pressures at the points $A$ and $D$ are $P$ and $2P$ respectively. The volume at the points both $A$ and $D$ remain the same, which is $V$.So, the work done by the gas from point $A$ to the point $D,W_{AD}=P(V-V)=0$Hence the total work done in the whole cycle,$W=4PV-2PV=2PV$We know the heat rejected from the cycle is equal to the amount of total work done by the gas, so $Q=W$$Q=2PV$Hence, the correct option is (B).