Two cylinders A and B fitted with pistons contain equal number of moles of an ideal monoatomic gas at 400K. The piston of A is free to move while that of B is held fixed. The same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in A is 42K, the rise in temperature of the gas in B is

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answer is 70.

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Detailed Solution

$P i s t o n o f A i s f r e e t o m o v e, i m p l i e s h e a t i s a d d e d a t c o n s \tan t p r e s s u r e . dQ = {nC}_{p} {dT}_{1} = {nC}_{v} {dT}_{1} + {nRdT}_{1} P i s t o n B, t h e h e a t i s a d d e d a t c o n s \tan t v o l u m e dQ = dU = {nC}_{V} {dT}_{2} G i v e n t h a t s a m e a m o u n t o f h e a t i s s u p p l i e d . \frac{3 R}{2} {dT}_{1} + R {dT}_{1} = \frac{3 R}{2} {dT}_{2} {dT}_{2} = \frac{5 {dT}_{1}}{3} = \frac{5}{3} (42) = 70 K$