(Take $\gamma$ for nitrogen and $CO_2$ as $1.4$ and $1.3$ respectively)
$f_{N_{2}}=\frac{2}{\gamma-1}=\frac{2}{1.4-1}=5$
$n_{\mathrm{CO}_{2}}=\frac{11 \mathrm{kg}}{44}=0.25 \mathrm{kmoles}$
$f_{C O_{2}}=\frac{2}{1.3-1}=\frac{2}{0.3}=\frac{20}{3}$
$\therefore$ Equivalent molecular weight
$=\frac{\mathrm{n}_{1} \mathrm{M}_{1}+\mathrm{n}_{2} \mathrm{M}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$
$=\frac{7+11}{0.25+0.25}=36 \mathrm{gm}$
$\gamma_{\operatorname{mix}} \neq \frac{n_{1} \gamma_{1}+n_{2} \gamma_{2}+\ldots . .}{n_{1}+n_{2}+\ldots . .}$
$f_{\operatorname{mix}}=\frac{n_{1} f_{1}+n_{2} f_{2}+\ldots .}{n_{1}+n_{2}}$
$=\frac{\left(0.25 \times 5+0.25 \times \frac{20}{3}\right)}{0.25+0.25 \times \frac{20}{3}}=\frac{35}{6}$
$\therefore \gamma_{\operatorname{mix}}=1+\frac{2}{f_{\operatorname{mix}}}$
$=1+\frac{2}{35 / 6}=\frac{47}{35}=1.34$
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