Two moles of `O_(2)(gamma=7/5)` at temperature `T_(0)` and 3 moles of `CO_(2)(gamma=4/3)` at temperature `2T_(0)` are allowed to mix together in a rigid closed adiabatic vessel. The resulting mixture finally comes in thermal equilibrium. Then,

Options

A. final temperature of the mixture is `(23T_(0))/(14)`
B. final temperature of the mixture is `(31T_(0))/19`
C. adiabatic exponent of the mixture formed is 14/3
D. adiabatic exponent of the mixture formed is `19/14`

Answer

A, D

Explanation

(1)`n_(1)C_(v1) T_(1)+n_(2)C_(v2) T_(2)=(n_(1)+n_(2)) C_(v)T`
`2((5R)/2)T_(0)+3(3R)2T_(0)=5C_(v)T`
`23/5RT_(0)=C_(v)T`
(2) `C_(v)=(n_(1)C_(v1)+n_(2)C_(v))/(n_(1)+n_(2))=(2xx(5R)/2+3xx3R)/(2+3)=14/5R`
`C_(p)=19/5R`
`r=19/14,` Put in(1)`C_(v)=14/5R`
`T=23/14 T_(0)`