1 mol of an ideal monatomic gas `(gamma=5/3)` is mixed with 1 mol of an ideal diatomic gas `(gamma=7/5)`. Find the value of `gamma` for the mixture.

Explanation

For the monatomic gas , `C_v=3/2 R and C_p=5/2 R`,
For the diatomic gas, `C_v=5/2 R and C_p=7/2 R`
`C_v=1times3/2R+1times5/2R=4R`,
`C_p=1times5/2R+1times7/2R=6R`,
`therefore gamma C_p/C_v=(6R)/(4R)=3/2=1.5`