Question
A thermodynamic system undergoes cyclic process $$ABCDA$$ as shown in fig. The work done by the system in the cycle is
A thermodynamic system undergoes cyclic process $$ABCDA$$ as shown in fig. The work done by the system in the cycle is
A.
$${P_0}{V_0}$$
B.
$$2{P_0}{V_0}$$
C.
$$\frac{{{P_0}{V_0}}}{2}$$
D.
Zero
Answer :
Zero
Solution :
Work done by the system in the cycle $$=$$ Area under $$P-V$$ curve and $$V$$-axis
$$\eqalign{ & = \frac{1}{2}\left( {2{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right) + \left[ { - \left( {\frac{1}{2}} \right)\left( {3{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right)} \right] \cr & = \frac{{{P_0}{V_0}}}{2} - \frac{{{P_0}{V_0}}}{2} = 0 \cr} $$
Work done by the system in the cycle $$=$$ Area under $$P-V$$ curve and $$V$$-axis
$$\eqalign{ & = \frac{1}{2}\left( {2{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right) + \left[ { - \left( {\frac{1}{2}} \right)\left( {3{P_0} - {P_0}} \right)\left( {2{V_0} - {V_0}} \right)} \right] \cr & = \frac{{{P_0}{V_0}}}{2} - \frac{{{P_0}{V_0}}}{2} = 0 \cr} $$
