1 text{ mole} of a gas with gamma = 7/5 is mi | vedclass

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Question

(C) The formula for the adiabatic exponent $\gamma_m$ of a mixture of two gases is given by:$\frac{n_1 + n_2}{\gamma_m - 1} = \frac{n_1}{\gamma_1 - 1}

Vedclass · Accepted Answer

$3/2$

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(C) The formula for the adiabatic exponent $\gamma_m$ of a mixture of two gases is given by:$\frac{n_1 + n_2}{\gamma_m - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}$Given $n_1 = 1, \gamma_1 = 7/5$ and $n_2 = 1, \gamma_2 = 5/3$:$\frac{1 + 1}{\gamma_m - 1} = \frac{1}{(7/5) - 1} + \frac{1}{(5/3) - 1}$$\frac{2}{\gamma_m - 1} = \frac{1}{2/5} + \frac{1}{2/3}$$\frac{2}{\gamma_m - 1} = \frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4$$\gamma_m - 1 = \frac{2}{4} = 0.5$$\gamma_m = 1 + 0.5 = 1.5 = 3/2$

$1 \text{ mole}$ of a gas with $\gamma = 7/5$ is mixed with $1 \text{ mole}$ of a gas with $\gamma = 5/3$. The value of $\gamma$ for the resulting mixture is:

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