Question

A mixture of ideal gases contains 7kg of $N_2$ and 11kg of $CO_2$. Then,
(Take $\gamma_{N_2} = 1.4$ and $\gamma_{CO_2} = 1.3$).
A.Equivalent molecular weight of the mixture is 36
B.Equivalent molecular weight of the mixture is 18
C.$\gamma$ for the mixture is $\dfrac{5}{2}$
D.$\gamma$ for the mixture is $\dfrac{47}{35}$

Answer 1

According to question, we will have to calculate both the equivalent molecular weight of the mixture as well as the heat capacity ratio of the mixture.
The equivalent molecular weight is given as the ratio of the total mass of the mixture (sum of nitrogen and carbon dioxide masses) and the total number of moles of gas present in the mixture, for which you will have to calculate the number of moles for both the gases.
Now, the $\gamma$ of the mixture can be found by taking the sum of specific heat $C_P$ or $C_V$ of individual gases. For this, use the relation between either specific heat at constant volume or constant pressure and $\gamma$ and plug in the given values and solve it. This should lead you to the value of $\gamma$ for the mixture.
Formula Used:
Molecular weight of a mixture of ideal gases: $MW_{mixture} = \dfrac{Total\;mass}{Total\;moles}$
The specific heat at constant volume for a mixture: $C_V = \dfrac{nR}{\gamma -1}$

Complete answer:
To verify the correctness of any choice we might make, we have to calculate both the equivalent molecular weight of the mixture as well as the heat capacity ratio of the mixture.
Let us first attempt to derive the molecular weight of the mixture.
We have mass of $N_2$ gas: $m_{N_2} = 7\;kg = 7 \times 10^{3}\;g$ and mass of $CO_2$ gas: $m_{CO_2} = 11\;kg = 11 \times 10^3\;g$
The molecular weight of $N_2$ is $M_{N_2} = 14 \times 2 =28\;gmol^{-1}$
The molecular weight of $CO_2$ is $M_{CO_2} = 12 + (16 \times 2) = 44\;gmol^{-1}$
The number of moles of $N_2$ contained in $7\;kg$ of it is given as: $n_{N_2} = \dfrac{m_{N_2}}{M_{N_2}} = \dfrac{7 \times 10^3}{28} = 250\;moles$
The number of moles of $CO_2$ contained in $11\;kg$ of it is given as: $n_{CO_2} = \dfrac{m_{CO_2}}{M_{CO_2}} = \dfrac{11 \times 10^3}{44} = 250\;moles$
Now, the effective molecular weight of the mixture is given as:
$MW_{mixture} = \dfrac{Total\;mass}{Total\;moles}$
$MW_{mixture} =\dfrac{m_{N_2}+m_{CO_2}}{n_{N_2}+n_{CO_2}}$
$\Rightarrow MW_{mixture} = \dfrac{(7+11)\times 10^3}{250+250} = \dfrac{18 \times 10^3}{500} = 0.036 \times 10^3 = 36\;g$
Now, let us find the heat capacity ratio for the mixture $\gamma_{mixture}$.
We know that $\gamma = \dfrac{C_P}{C_V}$, where $C_P$ and $C_V$ are the specific heats at constant volume and pressure respectively.
We have $C_V = \dfrac{nR}{\gamma -1}$ where, n is the number of moles, R is the universal gas constant and $\gamma$ is the specific heat ratio.
Now, $C_{V_{mixture}} = C_{V_{N_2}}+ C_{V_{CO_2}}$
$\Rightarrow \dfrac{n_{mixture}}{\gamma_{mixture}-1}= \dfrac{n_{N_2}}{\gamma_{N_2} -1} +\dfrac{n_{CO_2}}{\gamma_{CO_2}-1}$
Where $n_{mixture} = n_{N_2}+n_{CO_2} = 500\;moles$
$\Rightarrow \dfrac{500}{\gamma_{mixture}-1} = \dfrac{250}{1.4-1} +\dfrac{250}{1.3-1} \Rightarrow \dfrac{500}{\gamma_{mixture}-1} = 625+833.3 = 1458.3$
$\Rightarrow 500 =( \gamma_{mixture}-1)(1458.3) = 1458.3\gamma_{mixture} – 1458.3$
$\Rightarrow 1958.3 = 1458.3\gamma_{mixture} \Rightarrow \gamma_{mixture} = \dfrac{1958.3}{1458.3} = 1.3428$
However, $\dfrac{47}{35} = 1.3428 \Rightarrow \gamma_{mixture} = \dfrac{47}{35}$

Therefore, the correct option(s) would be A. Equivalent molecular weight of the mixture is 36 and D. $\gamma$ for the mixture is $\dfrac{47}{35}$.

Note:
An alternative way to go about finding the $\gamma$ of the mixture is by considering the number of degrees of freedom(f) of the individual gases. We have:
$f_{N_2} = \dfrac{2}{\gamma_{N_2}-1} = \dfrac{2}{1.4-1}=5$
$f_{CO_2} = \dfrac{2}{\gamma_{CO_2}-1}=\dfrac{2}{1.3-1}=\dfrac{20}{3}$
$f_{mixture} =\dfrac{n_{N_2}f_{N_2}+n_{CO_2}f_{CO_2}}{n_{N_2}+n_{CO_2}} = \dfrac{(250 \times 5)+(250 \times \dfrac{20}{3})}{250+250} \approx \dfrac{35}{6}$
Therefore, $f_{mixture} = \dfrac{2}{\gamma_{mixture}-1} \Rightarrow \dfrac{35}{6} = \dfrac{2}{\gamma_{mixture}-1} \Rightarrow 35(\gamma_{mixture}-1) = 12 \Rightarrow 35\gamma_{mixture} = 12+35 \Rightarrow \gamma_{mixture} = \dfrac{47}{35}$, which yields the same result.