JEE Main Online Paper (Held On 12 May 2012)
A) \[\frac{8{{P}_{0}}{{V}_{0}}}{nR}\]
B) \[\frac{3{{P}_{0}}{{V}_{0}}}{2nR}\]
C) \[\frac{9{{P}_{0}}{{V}_{0}}}{2nR}\]
D) \[\frac{9{{P}_{0}}{{V}_{0}}}{4nR}\]
Correct Answer: B
Solution :
Work done during the process\[A\to B\] = Area of trapezium (= area bounded by indicator diagram with F-axis) \[=\frac{1}{2}\left( 2{{P}_{0}}+{{P}_{0}} \right)\left( 2{{V}_{0}}-{{V}_{0}} \right)=\frac{3}{2}{{P}_{0}}{{V}_{0}}\] Ideal gas eqn : PV= nRT \[\Rightarrow \]\[T=\frac{PV}{nR}=\frac{3{{P}_{0}}{{V}_{0}}}{2nR}\]
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