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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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During a process , p & V to be related by the equation - $\;p=\large\frac{p_{0}}{1+(\large\frac{V_{0}}{V})^2}\;$ , $\;p_{0}\;$ & $\;V_{0}\;$ are constants . Change in temperature of the gas . When volume is changed from $\;V=V_{0}\;$ to $\;V=5V_{0}$

$(a)\;\large\frac{6}{13}\;\large\frac{p_{0}V_{0}}{nR}\qquad(b)\;-\large\frac{6}{13}\;\large\frac{p_{0}V_{0}}{nR}\qquad(c)\;\large\frac{1}{13}\;\large\frac{p_{0}V_{0}}{nR}\qquad(d)\;-\large\frac{1}{13}\;\large\frac{p_{0}V_{0}}{nR}$

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1 Answer

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Answer : $\;-\large\frac{6}{13}\;\large\frac{p_{0}V_{0}}{nR}$
Explanation :
Let $\;T_{1}\;$ be the initial temperature
$T_{1}=(\large\frac{p_{0}}{1+1})\;(\large\frac{V_{0}}{nR})$
Let $\;T_{2}\;$ be the final temperature
Therefore , $\;T_{2}=(\large\frac{p_{0}}{1+25})\;(\large\frac{V_{0}}{nR})$
$\bigtriangleup T=T_{2}-T_{1}\large\frac{p_{0}V_{0}}{nR}\;(\large\frac{1}{26}-\large\frac{1}{2})=-\large\frac{6p_{0}V_{0}}{13nR}$
answered Mar 13, 2014 by yamini.v
 

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