Looking for classes? Ksquare Career Institute, Bengaluru →
A thermodynamic system undergoes cyclic process ABCDA as shown in fig. The work done by the system in the cycle is:
A
$P_0V_0$
B
$2P_0V_0$
C
$\frac{P_0V_0}{2}$
D
Zero
Detailed Solution
Work done in the cycle = area enclosed by the P–V loops; the two triangular loops are traversed in opposite senses.
$W = \frac{1}{2}(2P_0 - P_0)(2V_0 - V_0) + \left[-\frac{1}{2}(3P_0 - 2P_0)(2V_0 - V_0)\right]$
$W = \frac{P_0V_0}{2} - \frac{P_0V_0}{2} = 0$
$W = \frac{1}{2}(2P_0 - P_0)(2V_0 - V_0) + \left[-\frac{1}{2}(3P_0 - 2P_0)(2V_0 - V_0)\right]$
$W = \frac{P_0V_0}{2} - \frac{P_0V_0}{2} = 0$