MCQ
A thermodynamic system undergoes cyclic process $ABCDA $ as shown in figure. The work done by the system in the cycle is
- A${P_0}{V_0}$
- B$2{P_0}{V_0}$
- C$\frac{{{P_0}{V_0}}}{2}$
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Answer
Correct option: D.
Zero
d
(d) $W_{BCOB} = -$ Area of triangle $BCO = - \frac{{{P_0}{V_0}}}{2}$
(d) $W_{BCOB} = -$ Area of triangle $BCO = - \frac{{{P_0}{V_0}}}{2}$
$W_{AODA} = + $ Area of triangle $AOD = + \frac{{{P_0}{V_0}}}{2}$
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