Question

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

• A. $\frac{1 5}{1 1}$
• B. $\frac{5}{1 3}$
• C. $\frac{5}{1 1}$
• D. $\frac{1 5}{1 3}$

Answer 1

Answer: (A)

For $\gamma = \frac{7}{5} \text{,}$ $C_{\text{v}}=\frac{R}{\gamma - 1}=\frac{R}{\frac{7}{5} - 1}=\frac{5 R}{2}$
$C_{\text{p}}=\frac{\gamma R}{\gamma - 1}=\frac{\left(7 / 5\right) R}{\frac{7}{5} - 1}=\frac{7 R}{2}$
For $\gamma =\frac{4}{3}\text{, }\textit{C}_{\text{v}}=3\textit{R}\text{, }\textit{C}_{\text{p}}=4R$
$\therefore \gamma _{\text{mix}} = \frac{\frac{7}{2} + 4}{\frac{5}{2} + 3} = \frac{1 5}{1 1}$