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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 T1 , in terms of T0 is

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By Expert Faculty of Sri Chaitanya
a
Tf=73T0
b
Tf=32T0
c
Tf=52T0
d
Tf=37T0

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detailed solution

Correct option is B

When two gases are mixed together, thenHeat lost by the helium gas = Heat gained by the nitrogen gasnB×cvHe×73T0−Tf=nA×cvN2×Tf−T0       Box A      1 mole N2  Temperature = T0              Box B      1 mole He  Temperature = 73T0 ⇒ 1×32R×73T0−Tf=1×52R×Tf−T0By solving, we getTf=32T0
ctaimg

ctaimg

Similar Questions

Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300 K. Piston A is free to move and piston of B is fixed. Same amount of heat is given to the gases in the two cylinders. Temperature of the gas in cylinder A increases by 30 K, then increase in temperature of the gas in the cylinder B is ( γ = 14 for diatomic gas) [EAMCET 2013] 

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