KVPY Sample Paper KVPY Stream-SX Model Paper-22

  • 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

    Correct Answer: B

    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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