A thermodynamic system undergoes cyclic process ABCDA as shown in figure. work done by system i

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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

Solution

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

As is clear from figure,

${W_{AEDA}} = + area\,of\,\Delta AED = + \frac{1}{2}{P_0}{V_0}$

${W_{BCEB}} = – Area\,of\,\Delta BCE\, – – \frac{1}{2}\,{P_0}{V_0}$

The net work done by the system is

${W_{net}} = {W_{AEDA}} + {W_{BCEB}}$

$ = + \frac{1}{2}{P_0}{V_0} – \frac{1}{2}{P_0}{V_0} = zero$

Standard 11

Physics

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