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A gas is expanded from volume $\;V_{0}\;$ to $\;2V_{0}\;$ under three different processes . Process 1 is isobaric , process 2 is isothermal and process 3 is adiabatic . Let $\;\bigtriangleup u_{1}\;, \bigtriangleup u_{2}\;$ and $\;\bigtriangleup u_{3}\;$ be the change in internal energy of the gas in the three processes . Then :

$(a)\;\bigtriangleup u_{1} > \bigtriangleup u_{2} > \bigtriangleup u_{3}\qquad(b)\; \bigtriangleup u_{1} < \bigtriangleup u_{2} < \bigtriangleup u_{3}\qquad(c)\;\bigtriangleup u_{2} < \bigtriangleup u_{1} < \bigtriangleup u_{3}\qquad(d)\;\bigtriangleup u_{2} < \bigtriangleup u_{3} < \bigtriangleup u_{1}$

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Answer : $\;\bigtriangleup u_{1} > \bigtriangleup u_{2} > \bigtriangleup u_{3}$
Explanation :
Process (1) : isobaric : $\;\bigtriangleup w= P (\bigtriangleup V)$
$P \bigtriangleup V = nR \bigtriangleup T$
$\bigtriangleup T = \large\frac{\bigtriangleup w}{nR}$
$\bigtriangleup u = \large\frac{nR}{(r-1)}\;. \bigtriangleup T= \large\frac{nR}{(r-1)}\;.\large\frac{\bigtriangleup w}{nR}$
$\bigtriangleup u = \large\frac{\bigtriangleup w}{r-1}$
process (2) : isothermal : $\;\bigtriangleup T =0$
$\bigtriangleup u=0$
Process (3) : Adiabatic ; $\;\bigtriangleup Q=0$
$\bigtriangleup w = - \bigtriangleup u$
$\bigtriangleup u=-\bigtriangleup $w
$\large\frac{nR}{(r-1)} \bigtriangleup T = \bigtriangleup w$
$\bigtriangleup T = \large\frac{\bigtriangleup w (r-1)}{nR}\;.$
answered Mar 21, 2014 by yamini.v

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