One mole of gas having `gamma = 7//5` is mixed with 1 mole of a gas having `gamma = 4//3`. What will be `gamma` for the mixture ?

Options

A. `5/11`
B. `15/13`
C. `15/11`
D. `5/13`

Answer

C

Explanation

For `gamma=7/5 C_(V)=R/(gamma-1)=R/(7/5-1)=(5R)/(2)`
`C_(p)=(gammaR)/(gamma-1)=((7//5)R)/(7/5-1) =(7R)/2`
for `gamma = 4/3, C_(v)=3R, C_(p) =4R`
`gamma_(mix)=(7/2+4)/(5/2+3)= 15/11`.