Question

A gaseous mixture enclosed in a vessel consists of one gram mole of a gas A with $\gamma = \dfrac{5}{3}$ and some amount of gas B with $\gamma = \dfrac{7}{5}$ at a temperature T. The gases A and B do not react with each other and are assumed to be ideal. Find the number of gram moles of the gas B, if $\gamma$ for the gaseous mixture is $\left( {\dfrac{{19}}{{13}}} \right)$.
A. ${\mu _2} = 2gmol$
B. ${\mu _2} = 2.2gmol$
C. ${\mu _2} = 2.1gmol$
D. ${\mu _2} = 2.6gmol$

Answer 1

In this question, we need to determine the number of gram moles of gas B such that the heat capacity for the gaseous mixture is $\left( {\dfrac{{19}}{{13}}} \right)$. For this, we will use the relation between the heat capacity and volume of the gas. Moreover, a relation between the internal heat change, number of moles, volume and change in temperature will be used.

Complete step by step answer:
Heat capacity of gas 1 $\gamma = \dfrac{5}{3}$
Heat capacity of gas 2$\gamma = \dfrac{7}{5}$
Also the heat capacity of the mixture $\gamma = \dfrac{{19}}{{13}}$
Since the gaseous mixture is in an enclosed vessel hence the volume of the gas is constant so we can say the mixture is going through the isochoric process which is given by the formula
${C_v} = \dfrac{R}{{\gamma - 1}} - - (i)$
So we can say ${C_v}$ for gas 1 will be

$${\left( {{C_v}} \right)_A} = \dfrac{R}{{\gamma - 1}} \\\ = \dfrac{R}{{\left( {\dfrac{5}{3} - 1} \right)}} \\\ = \dfrac{R}{{\left( {\dfrac{{5 - 3}}{3}} \right)}} \\\ = \dfrac{{3R}}{2} \\\$$

Also the${C_v}$for gas 2

$${\left( {{C_v}} \right)_B} = \dfrac{R}{{\gamma - 1}} \\\ = \dfrac{R}{{\left( {\dfrac{7}{5} - 1} \right)}} \\\ = \dfrac{R}{{\left( {\dfrac{{7 - 5}}{5}} \right)}} \\\ = \dfrac{{5R}}{2} \\\$$

${C_v}$for the mixture of gas 1 and 2

$${\left( {{C_v}} \right)_{mix}} = \dfrac{R}{{\gamma - 1}} \\\ = \dfrac{R}{{\left( {\dfrac{{19}}{{13}} - 1} \right)}} \\\ = \dfrac{R}{{\left( {\dfrac{{19 - 13}}{{13}}} \right)}} \\\ = \dfrac{{13R}}{6} \\\$$

Now since the two gases, 1 and 2 do not react with each other so the total internal energy of the gases will be conserved; hence we can write
${\left( {\Delta U} \right)_{mix}} = {\left( {\Delta U} \right)_1} + {\left( {\Delta U} \right)_2} - - (ii)$
Where internal energy of the gases at constant volume under changing temperature is given as
$\Delta U = \mu {C_V}\Delta T$
So the internal energy of the mixture from equation (ii) can be written as

$${\left( {\Delta U} \right)_{mix}} = {\left( {\Delta U} \right)_1} + {\left( {\Delta U} \right)_2} \\\ \left( {{\mu _1} + {\mu _2}} \right)\left( {\dfrac{{13R}}{6}} \right)\Delta T = {\mu _1}\left( {\dfrac{{3R}}{2}} \right)\Delta T + {\mu _2}\left( {\dfrac{{5R}}{2}} \right)\Delta T \\\$$

Hence by solving this equation, we get

$$\left( {{\mu _1} + {\mu _2}} \right)\left( {\dfrac{{13R}}{6}} \right)\Delta T = {\mu _1}\left( {\dfrac{{3R}}{2}} \right)\Delta T + {\mu _2}\left( {\dfrac{{5R}}{2}} \right)\Delta T \\\ \implies \dfrac{{13}}{6}\left( {{\mu _1} + {\mu _2}} \right) = \dfrac{3}{2}{\mu _1} + \dfrac{5}{2}{\mu _2} \\\ \implies \dfrac{{13}}{6}{\mu _1} + \dfrac{{13}}{6}{\mu _2} = \dfrac{3}{2}{\mu _1} + \dfrac{5}{2}{\mu _2} \\\$$

Since the number of gram mole of gas A is one so, we can write${\mu _1} = 1$, hence the number of gram moles of the gas B

$$\dfrac{{13}}{6} \times 1 + \dfrac{{13}}{6}{\mu _2} = \dfrac{3}{2} \times 1 + \dfrac{5}{2}{\mu _2} \\\ \implies \dfrac{{13}}{6} - \dfrac{3}{2} = \dfrac{5}{2}{\mu _2} - \dfrac{{13}}{6}{\mu _2} \\\ \implies \dfrac{{13 - 9}}{6} = \dfrac{{15 - 13}}{6}{\mu _2} \\\ \implies \dfrac{2}{6}{\mu _2} = \dfrac{4}{6} \\\ \therefore {\mu _2} = 2 \\\$$

Hence, the number of gram moles of the gas ${\mu _2} = 2$

So, the correct answer is “Option A”.

Note:
The process which does not involve any change in the volume of the material involved in the process is known as the Isochoric process. Moreover, the enthalpy is the sum of the internal energy and the product of the pressure and the volume of the given material.