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Q.

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature To, while box B contains one mole of helium at temperature (7/3) T0.The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then the final temperature of the gases, Tf in terms of To is

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a

Tf=37T0

b

Tf=73T0

c

Tf=32T0

d

Tf=52T0

answer is C.

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Detailed Solution

Here, the change in internal energy of the system is zero. The increase in the internal energy of one box will be equal to the decrease in internal energy of the other box. The boxes are shown in fig.The change in internal energy is given bydUA=1×5R2Tf−T0dUB=1×3R2Tf−73T0Now,  dUA+dUB=0∴ 5R2Tf−T0+3R2Tf−73T0=0or   5Tf−5T0+3Tf−7T0=0or  8Tf=12T0  or  Tf=128T0∴ Tf=32T0
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Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature To, while box B contains one mole of helium at temperature (7/3) T0.The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then the final temperature of the gases, Tf in terms of To is