Question

$3$ moles of mono-atomic gas ($\gamma =5/3$) is mixed with $1$ mole of a diatomic gas ($\gamma =7/3$). The value of $\gamma$ for the mixture will be

• A. $\frac{9}{11}$
• B. $\frac{11}{7}$
• C. $\frac{12}{7}$
• D. $\frac{15}{7}$

Answer 1

Answer: (B)

$\frac{11}{7}$

The number of degrees of freedom for the mixture is:
$f_{mix}=\frac{n_1{f}_1+n_2{f}_2}{n_1-n_2}$$=\frac{3\times 3+1\times 5}{4}=\frac{7}{2}$
$\therefore \gamma =1+\frac{2}{f}\Rightarrow \gamma =1+\frac{4}{7}=\frac{11}{7}$