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11.Thermodynamics
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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
$\frac{{{P_0}{V_0}}}{2}$
D
Zero
(AIPMT-2014)
Solution
(d) $W_{BCOB} = -$ Area of triangle $BCO = – \frac{{{P_0}{V_0}}}{2}$
$W_{AODA} = + $ Area of triangle $AOD = + \frac{{{P_0}{V_0}}}{2}$
Standard 11
Physics
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