• # question_answer A gaseous mixture enclosed in a vessel consists of 1 gram mole of gas $A$with ($\gamma$= 5/3) and another $B$ with (g = 7/5) at a temperature$T$. The gases$A$ and $B$do not react with each other and assume to be idea. Then the number of gram moles of the gas $B$is (if $\gamma$ for the gaseous mixture is 19/13). A) 3 B) 2 C) 4 D) 6

Solution :

 As for ideal gas ${{C}_{P}}-{{C}_{V}}=R$ and $\gamma =({{C}_{P}}/{{C}_{V}}),$ $\gamma -1=\frac{R}{{{C}_{V}}}$ or ${{C}_{V}}=\frac{R}{\left( \gamma -1 \right)}$${{\left( {{C}_{V}} \right)}_{1}}=\frac{R}{\left( 5/3 \right)-1}=\frac{3}{2}R;$ ${{\left( {{C}_{V}} \right)}_{2}}=\frac{R}{\left( 7/2 \right)-1}=\frac{5}{2}R$ and ${{\left( {{C}_{V}} \right)}_{\operatorname{mix}}}=\frac{R}{\left( 19/13 \right)-1}=\frac{13}{6}R$
 now from the conservation of energy, $i.e.,$$\Delta U=\Delta {{U}_{1}}+\Delta {{U}_{2}},$we get $({{\mu }_{1}}+{{\mu }_{2}}){{({{C}_{V}})}_{\operatorname{mix}}}\Delta T=[{{\mu }_{1}}{{({{C}_{V}})}_{1}}+{{\mu }_{2}}{{({{C}_{V}})}_{2}}]\Delta T$ $\Rightarrow$${{({{C}_{V}})}_{\operatorname{mix}}}=\frac{{{\mu }_{1}}{{({{C}_{V}})}_{1}}+{{\mu }_{2}}{{({{C}_{V}})}_{2}}}{{{\mu }_{1}}+{{\mu }_{2}}}$ $\frac{13}{6}R=\frac{1\times \frac{3}{2}R+n\times \frac{5}{2}R}{1+n}=\frac{(3+5n)}{2(1+n)}$ or $13+13n=9+15n,$ i.e., number of gram mole, $n=2$

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