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. $\frac{15}{11}$
• B. $\frac{5}{13}$
• C. $\frac{5}{11}$
• D. $\frac{15}{13}$

Answer 1

Answer: (A)

For $\gamma =\frac{7}{5}\text{,}$ $C_{\text{v}}=\frac{R}{\gamma -1}=\frac{R}{\frac{7}{5}-1}=\frac{5R}{2}$
$C_{\text{p}}=\frac{\gamma R}{\gamma -1}=\frac{\left(7/5\right)R}{\frac{7}{5}-1}=\frac{7R}{2}$
For $\gamma =\frac{4}{3}\text{, }{\text{C}}_{\text{v}}=3\text{R}\text{, }{\text{C}}_{\text{p}}=4R$
$\therefore {\gamma }_{\text{mix}}=\frac{\frac{7}{2}+4}{\frac{5}{2}+3}=\frac{15}{11}$