For the given cyclic process CAB as shown for a gas, the work done is- [JEE Main 12-Jan-2019 Morning] |
A) 10 J
B) 1 J
C) 5 J
D) 30 J
Correct Answer: A
Solution :
[a] CA = change in volume \[\Delta V=(5-1)=4{{m}^{3}}AB=\]change in pressure \[\Delta P=(6-1)=5{{m}^{3}}\] |
Work done = Area of \[\Delta ABC=\frac{1}{2}\times AB\times CA\] |
\[W=\frac{1}{2}\times 4\times 5=10J\] |
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