Question

A mixture of $n_{1}$ moles of monoatomic gas and $n_{2}$ moles of diatomic gas has $\frac{C_{p}}{C_{V}}=\gamma=1.5 .$ Then

• A. $ {{n}_{1}}={{n}_{2}} $
• B. $ 2{{n}_{1}}={{n}_{2}} $
• C. $ {{n}_{1}}=2{{n}_{2}} $
• D. $ 2{{n}_{1}}=3{{n}_{2}} $

Answer 1

Answer: (A)

$f_{a v}=\frac{\text { total number of degrees of freedom }}{\text { total number of molecules }}$
$=\frac{n_{1} N_{A} f_{1}+n_{2} N_{A} f_{2}}{n_{1} N_{A}+n_{2} N_{A}}$
$=\frac{n_{1} f_{1}+n_{2} f_{2}}{n_{1}+n_{2}}$
$=\frac{3 n_{1}+5 n_{2}}{n_{1}+n_{2}}$
Also $\gamma=1+\frac{2}{f_{a v}}=1.5$
Or $f_{a v}=4$
$\therefore \frac{3 n_{1}+5 n_{2}}{n_{1}+n_{2}}=4$
Or $n_{1}=n_{2}$