A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with `gamma=(5/3)` and some amount of gas B with `gamma=7/5` at a temperature T.
The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B if `gamma` for the gaseous mixture is `(19/13)`.

Options

A. `2`
B. `3`
C. `4`
D. `6`

Answer

A

Explanation

We define specific heat capacity at constant volume
`C_v = (R)/(gamma-1)`
For gas `A`, `(C_V)_B = R/((5//3)-1) = 3/2 R`
For gas `B, (C_V)_(B) = R/((7/5)-1) = 5/2 R`
For gaseous mixture , `(C_V)_(mix) = (R)/((19//13)-1) = 13/6 R`
We define specfic heat capacity of mixture
`(C_V)_(mix) = (n_(1)(C_V)_(1)+n_(2)(C_V)_(2))/(n_1+n_2)`
`(13)/(6) R= (1xx3/2 R + n xx 5/2 R)/(1+n) = ((3+5n))/(2(1+n))`
or `13+13n = 9+15n` i.e., `n=2 "mole"`.