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
Here, `n_(1) "mole" , n_(2) = 1 "mole"`
For nitrogen, `C_(p_1)=7/2 R`
for helium, `C_(p_2) = 5/2 R`
`T_(1)=T_(0), T_(2) = 7/3 T_(0).T_(f)=?`
when gases are put into thermal contact, heat is exchanged between them till final temperature `T_(f)` is reached.
Heat gained by nitrogen = Heat lost by He
`n_(1)C_(P_1)(T_(f)-T_(1)) = n_(2)C_(P_2) (T_(2)-T_(f))`
`(n_(1)C_(P_1)+n_(2)C_(p_2)) T_(f) = n_(2)C_(p_2) T_(2) n_(1)C_(p_1)T_(1)`
`(1 xx 7.2 R + 5/2 R) T_(f) = 1 xx 5/2 R xx 7/3 T_(0) + 1 xx 7/2 RT_(0)`
`6RT_(f) = (35/6 + 7/2)RT_(0)`
`T_(f) = (56)/(6 xx 6) T_(0)~~3/2 T_(0)`.