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The figure expansion of an ideal gas through two processes A & B . Let $\;\bigtriangleup Q_{A}\;$ & $\;\bigtriangleup Q_{B}\;$ be the heat given to the system in processes A & B respectively . Then

$(a)\;\bigtriangleup Q_{A} > \bigtriangleup Q_{B}\qquad(b)\;\bigtriangleup Q_{A} = \bigtriangleup Q_{B}\qquad(c)\;\bigtriangleup Q_{A} < \bigtriangleup Q_{B}\qquad(d)\;\bigtriangleup Q_{A} \geq \bigtriangleup Q_{B}$

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Answer : $\;\bigtriangleup Q_{A} > \bigtriangleup Q_{B}$
Explanation :
For the two processes , internal energy change is same . But $\;\bigtriangleup W_{A} >\bigtriangleup W_{B}\;$ as the area under curve for A is more than that for B
Therefore , $\bigtriangleup Q_{A} > \bigtriangleup Q_{A}$
answered Mar 12, 2014 by yamini.v
edited Mar 25, 2014 by yamini.v

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