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Question: If \[7\,{\text{gm}}\] \[{{\text{N}}_{\text{2}}}\] is mixed with \[20\,{\text{gm}}\] \[{\text{Ar}}\]....

If 7 gm7\,{\text{gm}} N2{{\text{N}}_{\text{2}}} is mixed with 20 gm20\,{\text{gm}} Ar{\text{Ar}}. What will be the CP/CV{C_{\text{P}}}/{C_{\text{V}}} of the mixture?
A. 176\dfrac{{17}}{6}
B. 117\dfrac{{11}}{7}
C. 1711\dfrac{{17}}{{11}}
D. 1713\dfrac{{17}}{{13}}

Explanation

Solution

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 CPmix{C_{{{\text{P}}_{{\text{mix}}}}}} and CVmix{C_{{{\text{V}}_{{\text{mix}}}}}} separately and find the ratio.

Formula used:
We apply the formula to calculate number of moles:
n=mMn = \dfrac{m}{M} …… (1)
Where,
nn indicates the number of moles.
mm indicates given mass.
MM indicates molar or molecular mass.
CPmix=nN2×CP+nAr×CPnN2+nAr{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
CVmix=nN2×CV+nAr×CVnN2+nAr{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 (N2)\left( {{{\text{N}}_{\text{2}}}} \right), which is a di-atomic gas, the degrees of freedom i.e. f=5f = 5. For argon gas (Ar)\left( {{\text{Ar}}} \right), which is a mono-atomic gas, the degrees of freedom i.e. f=3f = 3.

Again, given, the mass of nitrogen gas taken is 7 gm7\,{\text{gm}} and the mass of nitrogen gas taken is 20 gm20\,{\text{gm}} . We need to convert these two masses into a number of moles to proceed the solution.

Molecular mass of N2{{\text{N}}_{\text{2}}} is: 2 \times 14\,{\text{g}} {\text{ = 28 g}} \\\
Molecular mass of Ar{\text{Ar}} is 40 g40\,{\text{g}}. To calculate number of moles of nitrogen, substitute,
m=7 gm = 7\,{\text{g}} and M=28 g/molM = 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 gm = 20\,{\text{g}} and M=40 g/molM = 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 (N2)\left( {{{\text{N}}_{\text{2}}}} \right),
CV=5R2{C_{\text{V}}} = \dfrac{{5R}}{2} and CP=7R2{C_{\text{P}}} = \dfrac{{7R}}{2}

For the argon gas (Ar)\left( {{\text{Ar}}} \right),
CV=3R2{C_{\text{V}}} = \dfrac{{3R}}{2} and CP=5R2{C_{\text{P}}} = \dfrac{{5R}}{2}

To calculate Cp{C_p} for the mixture, we apply the formula:
CPmix=nN2×CP+nAr×CPnN2+nAr{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 CV{C_{\text{V}}} for the mixture, we apply the formula:
CVmix=nN2×CV+nAr×CVnN2+nAr{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 CP{C_{\text{P}}} and CV{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 1711\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,
CPmix/CVmix=1+2fmix{C_{{{\text{P}}_{{\text{mix}}}}}}/{C_{{{\text{V}}_{{\text{mix}}}}}} = 1 + \dfrac{2}{{{f_{{\text{mix}}}}}}.