11.Thermodynamics
medium

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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