Question

One box containing `1` mole of `He` is `7//3 T_0` and other box containing `1` mole of a polyatomic gas `(gamma =1.33)` at `T_0` are placed together to attain thermal equilibrium. The final temperature `T_f`. Then :
A. `T_(f)=9/13 T_(0)`
B. `T_(f)=13/9 T_(0)`
C. `T_(f)=(T_(0))/2`
D. `T_(f)=3/2T_(0)`

Answer 1

Correct Answer - B
Heat given by `He`=heat taken by polyatomic gas
`C_(v) for He=3/2R`
`C_(v)for gas=3R`
`:. 1xx3/2R[7/3T_(0)-T_(f)]=1xx3Rxx(T_(f)-T_(0))`
`:. T_(f)=13/9T_(0)`