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Thermodynamics

# The work done in cyclic process is

$(a)\;\pi \;(\large\frac{P_{2}-P_{1}}{2})^2\qquad(b)\;\pi\;(\large\frac{V_{1}-V_{2}}{2})^2\qquad(c)\;\large\frac{\pi}{4}\;(P_{1}-P_{2})\;(V_{2}-V_{1})\qquad\;(D) $$\large\frac{\pi}{4}\;$$\large\frac{(V_{2}-V_{1})}{(P_{1}-P_{2})}$$Can you answer this question? ## 1 Answer 0 votes Answer : \;\large\frac{\pi}{4}\;$$(P_{1}-P_{2})\;(V_{2}-V_{1})$
Explanation :
Work done = area under ellipse = $\; \pi \times semi\;major\;axis\; \times semi\;minor\;axis$
answered Mar 21, 2014 by
edited Apr 2, 2014