Question

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

• A. $9/11$
• B. $11/7$
• C. $12/7$
• D. $15/7$

Answer 1

Answer: (B)

$11/7$

Given : $n_1=3$ $n_2=1$

For monoatomic gas $\gamma =\frac{5}{3}$

Degrees of freedom in monoatmic gas $f_1=\frac{2}{\gamma -1}=\frac{2}{\frac{5}{3}-1}=3$

For diatomic gas $\gamma =\frac{7}{5}$

Degrees of freedom in diatomic gas $f_2=\frac{2}{\gamma -1}=\frac{2}{\frac{7}{5}-1}=5$

Degree of freedom of mixture $f_{eq}=\frac{n_1{f}_1+n_2{f}_2}{n_1+n_2}$

$⟹$ $f_{eq}=\frac{3\times 3+1\times 5}{3+1}=\frac{7}{2}$

Thus $\gamma$ of mixture ${\gamma }_{eq}=\frac{f_{eq}+2}{f_{eq}}$

$⟹$ ${\gamma }_{eq}=\frac{\frac{7}{2}+2}{\frac{7}{2}}=\frac{11}{7}$