A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

Question

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 P0 V0 (C) (P0 V0/2) (D) zero

Tardigrade Admin · Accepted Answer

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_{A E D A} = +

Area of

△ A E D = + \frac{1}{2} P_{0} V_{0}

W_{BCEB} = -

Area of

△ BCE = - \frac{1}{2} P_{0} V_{0}

The net work done by the system is

W_{net} = W_{A E D A} + W_{BCEB}

= + \frac{1}{2} P_{0} V_{0} - \frac{1}{2} P_{0} V_{0} =

zero

Q. A thermodynamic system undergoes cyclic process $A BC D A$ as shown in figure. The work done by the system in the cycle is

$P_{0} V_{0}$

$2 P_{0} V_{0}$

$\frac{P _{0} V _{0}}{2}$

zero

Q. A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is

P0​V0​

2P0​V0​

2P0​V0​​

zero

Q. A thermodynamic system undergoes cyclic process $A BC D A$ as shown in figure. The work done by the system in the cycle is

$P_{0} V_{0}$

$2 P_{0} V_{0}$

$\frac{P _{0} V _{0}}{2}$