If \[7\,{\text{gm}}\] \[{{\text{N}}_{\text{2}}}\] is mixed with \[20\,{\text{gm}}\] \[{\text{Ar}}\]. What will be the \[{C_{\text{P}}}/{C_{\text{V}}}\] of the mixture?
A. \[\dfrac{{17}}{6}\]
B. \[\dfrac{{11}}{7}\]
C. \[\dfrac{{17}}{{11}}\]
D. \[\dfrac{{17}}{{13}}\]

Hint: For this, first of all, we will find the number of moles of each gas, which may differ as the proportion of each gas is different. It is important to note that degrees of freedom of a di-atomic gas is always more than that of a mono-atomic gas. Calculate \[{C_{{{\text{P}}_{{\text{mix}}}}}}\] and \[{C_{{{\text{V}}_{{\text{mix}}}}}}\] separately and find the ratio.

Formula used:
We apply the formula to calculate number of moles:
\[n = \dfrac{m}{M}\] …… (1)
Where,
\[n\] indicates the number of moles.
\[m\] indicates given mass.
\[M\] indicates molar or molecular mass.
\[{C_{{{\text{P}}_{{\text{mix}}}}}} = \dfrac{{{n_{{{\text{N}}_{\text{2}}}}} \times {C_{\text{P}}} + {n_{{\text{Ar}}}} \times {C_{\text{P}}}}}{{{n_{{{\text{N}}_{\text{2}}}}} + {n_{{\text{Ar}}}}}}\], and
\[{C_{{{\text{V}}_{{\text{mix}}}}}} = \dfrac{{{n_{{{\text{N}}_{\text{2}}}}} \times {C_{\text{V}}} + {n_{{\text{Ar}}}} \times {C_{\text{V}}}}}{{{n_{{{\text{N}}_{\text{2}}}}} + {n_{{\text{Ar}}}}}}\]

Complete step by step answer:
We know, for nitrogen gas \[\left( {{{\text{N}}_{\text{2}}}} \right)\], which is a di-atomic gas, the degrees of freedom i.e. \[f = 5\]. For argon gas \[\left( {{\text{Ar}}} \right)\], which is a mono-atomic gas, the degrees of freedom i.e. \[f = 3\].

Again, given, the mass of nitrogen gas taken is \[7\,{\text{gm}}\] and the mass of nitrogen gas taken is \[20\,{\text{gm}}\] . We need to convert these two masses into a number of moles to proceed the solution.

Molecular mass of \[{{\text{N}}_{\text{2}}}\] is: $2 \times 14\,{\text{g}} {\text{ = 28 g}} \\$
Molecular mass of \[{\text{Ar}}\] is \[40\,{\text{g}}\]. To calculate number of moles of nitrogen, substitute,
\[m = 7\,{\text{g}}\] and \[M = 28\,{\text{g/mol}}\] in equation (1):
${n_{{{\text{N}}_{\text{2}}}}} = \dfrac{m}{M} \\
\Rightarrow {n_{{{\text{N}}_{\text{2}}}}} = \dfrac{{7\,{\text{g}}}}{{28\,{\text{g/mol}}}} \\
\Rightarrow {n_{{{\text{N}}_{\text{2}}}}} = \dfrac{1}{4}\,{\text{mol}} \\$
To calculate number of moles of argon, substitute,
\[m = 20\,{\text{g}}\] and \[M = 40\,{\text{g/mol}}\] in equation (1):
${n_{{\text{Ar}}}} = \dfrac{m}{M} \\
\Rightarrow {n_{{\text{Ar}}}} = \dfrac{{20\,{\text{g}}}}{{40\,{\text{g/mol}}}} \\
\Rightarrow {n_{{\text{Ar}}}} = \dfrac{1}{2}\,{\text{mol}} \\$
For the nitrogen gas \[\left( {{{\text{N}}_{\text{2}}}} \right)\],
\[{C_{\text{V}}} = \dfrac{{5R}}{2}\] and \[{C_{\text{P}}} = \dfrac{{7R}}{2}\]

For the argon gas \[\left( {{\text{Ar}}} \right)\],
\[{C_{\text{V}}} = \dfrac{{3R}}{2}\] and \[{C_{\text{P}}} = \dfrac{{5R}}{2}\]

To calculate \[{C_p}\] for the mixture, we apply the formula:
\[{C_{{{\text{P}}_{{\text{mix}}}}}} = \dfrac{{{n_{{{\text{N}}_{\text{2}}}}} \times {C_{\text{P}}} + {n_{{\text{Ar}}}} \times {C_{\text{P}}}}}{{{n_{{{\text{N}}_{\text{2}}}}} + {n_{{\text{Ar}}}}}}\] …… (2)
Substitute the values of the respective gases in the equation (2), we get:
${C_{{{\text{P}}_{{\text{mix}}}}}} = \dfrac{{\dfrac{1}{4} \times \dfrac{{7R}}{2} + \dfrac{1}{2} \times \dfrac{{5R}}{2}}}{{\dfrac{1}{4} + \dfrac{1}{2}}} \\
\Rightarrow {C_{{{\text{P}}_{{\text{mix}}}}}} = \dfrac{{\dfrac{{7R}}{8} + \dfrac{{5R}}{4}}}{{\dfrac{3}{4}}} \\
\Rightarrow {C_{{{\text{P}}_{{\text{mix}}}}}} = \dfrac{{\dfrac{{17R}}{8}}}{{\dfrac{3}{4}}} \\
\Rightarrow {C_{{{\text{P}}_{{\text{mix}}}}}} = \dfrac{{17R}}{6} \\$
To calculate \[{C_{\text{V}}}\] for the mixture, we apply the formula:
\[{C_{{{\text{V}}_{{\text{mix}}}}}} = \dfrac{{{n_{{{\text{N}}_{\text{2}}}}} \times {C_{\text{V}}} + {n_{{\text{Ar}}}} \times {C_{\text{V}}}}}{{{n_{{{\text{N}}_{\text{2}}}}} + {n_{{\text{Ar}}}}}}\] …… (3)

Substitute the values of the respective gases in the equation (3), we get:
${C_{{{\text{V}}_{{\text{mix}}}}}} = \dfrac{{\dfrac{1}{4} \times \dfrac{{5R}}{2} + \dfrac{1}{2} \times \dfrac{{3R}}{2}}}{{\dfrac{1}{4} + \dfrac{1}{2}}} \\
\Rightarrow {C_{{{\text{V}}_{{\text{mix}}}}}} = \dfrac{{\dfrac{{5R}}{8} + \dfrac{{3R}}{4}}}{{\dfrac{3}{4}}} \\
\Rightarrow {C_{{{\text{V}}_{{\text{mix}}}}}} = \dfrac{{\dfrac{{11R}}{8}}}{{\dfrac{3}{4}}} \\
\Rightarrow {C_{{{\text{V}}_{{\text{mix}}}}}} = \dfrac{{11R}}{6} \\$
Now, the ratio of \[{C_{\text{P}}}\] and \[{C_{\text{V}}}\] of the gas mixture is:
$\dfrac{{{C_{{{\text{P}}_{{\text{mix}}}}}}}}{{{C_{{{\text{V}}_{{\text{mix}}}}}}}}=\dfrac{{\dfrac{{17R}}{6}}}{{\dfrac{{11R}}{6}}} \\
\Rightarrow\dfrac{{{C_{{{\text{P}}_{{\text{mix}}}}}}}}{{{C_{{{\text{V}}_{{\text{mix}}}}}}}}= \dfrac{{17R}}{6} \times \dfrac{6}{{11R}} \\
\therefore\dfrac{{{C_{{{\text{P}}_{{\text{mix}}}}}}}}{{{C_{{{\text{V}}_{{\text{mix}}}}}}}}= \dfrac{{17}}{{11}} \\$

Hence, the ratio is \[\dfrac{{17}}{{11}}\].So, option C is correct.

Note: It is important to remember that degrees of freedom of di-atomic gas is more than that of mono-atomic gases. Most of the students tend to make mistakes at this point. We can also solve the problem by finding the degree of freedom of the gas mixture and then find the ratio. By using the formula,
\[{C_{{{\text{P}}_{{\text{mix}}}}}}/{C_{{{\text{V}}_{{\text{mix}}}}}} = 1 + \dfrac{2}{{{f_{{\text{mix}}}}}}\].