4 mol of an ideal gas having `(gamma = 1.67)` are mixed with 2 mol of another ideal gas having `(gamma =1.4). Find the equivalent value of `gamma ` for the mixture.

Explanation

Let,
`(C'_v)`= molar heat capacity of the first gas,
C''_v = molar heat capacity of the second gas,
C_v= molar heat capacity of the mixture
and similar symbols for other quantities. Then,
`(gamma = C''_p / C''_v = 1.67)`
C''_p = C''_v + R.
This gives `C''_v = 3 / 2 R and C'' _v = 5/2 R.
Similarly, `gamma = 1.4 gives C''_v = 5/2 R and C''_p 7/2 R.
Suppose the temperature of the mixture is increased by
dT. The increase in the internal enerngy of the first gas
`= n_1 C''_vdT. The increase in the internal enerngy of the second The increase in the internal enerngy of the gas `= n_2 C''_vdT` and The increase in the internal enerngy of the
mixture `= (n_1 + n_2) C_vdT. Thus, ` (n_1 + n_2) C_vdT = n_1 C''_vdT + n_2 C''_vdT`
`C_v = (n_1 C''_v + n_2 C''_v ) / (n_1 + n_2)`
`C_p = C_v + R = (n_1 C''_v + n_2 C''_v ) / n_1 + n_2`
= `(n_1C''p + n_2C''_p ) / n_1 + n_2 `
From (i) and (ii), `(gamma` = (C_p) / (C_v)= (n_1 C''_p) /(n_1 C''_v) + (n_2 C''_p ) / + n_2 C''_v))`
`(4 xx 5/2 R +2 xx 7/2 R )/ (4 xx 3/2 R + 2 xx 5/2 R) = 1.54