A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is [NEET 2014] |
A) \[{{p}_{0}}{{V}_{0}}\]
B) \[2{{p}_{0}}{{V}_{0}}\]
C) \[\frac{{{p}_{0}}{{V}_{0}}}{2}\]
D) zero
Correct Answer: D
Solution :
Work done in the cyclic process = Area bound by the closed configuration |
= Area of closed configuration |
\[=\left\{ 2\left[ \frac{1}{2}({{v}_{0}}/2)\times {{p}_{0}} \right]+\left\{ -2\left[ \frac{1}{2}({{v}_{0}}/2){{p}_{0}} \right] \right. \right\}\] |
= zero. |
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