Question

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)`.

Answer 1

For two gases, we can wire
`(n_(1))/(gamma_(1) - 1) + (n_(2))/(gamma_(2) - 1) = (n_(1) + n_(2))/(gamma_(mean) - 1)`
Here, `n_(1) = 1` and `n_(2) = ?`
`gamma_(1) = 5//3` and `gamma_(2) = 7//5`
Substituting these value in the above equation, we get
`(1)/((5)/(3) -1) + (n_(2))/((7)/(5) -1) + (1 + n_(2))/((19)/(13) -1)`
or `(3)/(2) + (5 n_(2))/(2) = (13 (1 + n_(2)))/(6)`
or `3 (3 + 5 n_(2)) = 13 (1 + n_(2))`
`9 + 15 n_(2) = 13 + 13 n_(2)`
`n_(2) = 3 g mol`