Question

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature `T_0`, while Box contains one mole of helium at temperature `(7/3)T_0`. The boxes are then put into thermal contact with each other, and heat flows between them until the gasses reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gasses, `T_f` in terms of `T_0` is
A. `T_(f)=7/3 T_(0)`
B. `T_(f) = 3/2 T_(0)`
C. `T_(f) = 5/2 T_(0)`
D. `T_(f) = 3/7 T_(0)`

Answer 1

Correct Answer - B
When two gases are mixed together then
Heat lost by the Helium gas = Heat gained by the Nitrogen gas.
`n_(B)xx(c_(v))_(He) xx ((7/3)T_(0)-T_(f)) = n_(A) xx (c_(V))_N(2) xx (T_(f) - T_(0))`

`rArr (1 xx 3/2Rxx(7/3T_(0)-T_(f)) = 1 xx 5/2R xx (T_(f)-T_(0))`
By solving we get `T_(f) = 3/2T_(0)`