Question

One mole of an ideal monatomic gas $\left(\gamma=\dfrac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma = \dfrac{7}{5}\right)$. What is the $\gamma$ for the mixture? ($\gamma$ denotes the ratio of specific heat at constant pressure, to that at constant volume).
A. $\dfrac{3}{2}$
B. $\dfrac{23}{15}$
C. $\dfrac{35}{23}$
D. $\dfrac{4}{3}$

Answer 1

Begin by deducing $C_P$ and $C_V$ for both the gases by using the relation between the specific heat capacities, specific heat ratio, the gas constant and the number of moles. Once you have found the $C_P$ and $C_V$ values for both the gases, find the effective $C_P$ and $C_V$ by using the rule of mixtures. Then by using the definition of $\gamma$ given in the question, find the $\gamma$ for the mixture.

Formula used: Specific heat capacity at constant pressure $C_P = \dfrac{\gamma n R}{\gamma -1}$
Specific heat capacity at volume $C_V = \dfrac{nR}{\gamma -1}$, where $\gamma$ is the specific heat ratio, n is the number of moles and R is the gas constant.
General formula for the resultant heat capacity of a gas mixture: $C = \dfrac{n_1 C_1 + n_2 C_2}{n_1 +n_2}$, where $n_1$ and $n_2$ are the number of moles of gases 1 and 2, and $C_1$ and $C_2$ are the specific heat capacity (at constant pressure or volume) of gases 1 and 2 respectively.

Complete step by step answer:
Let us begin by first establishing an understanding of what $\gamma$ is.
As given in the question, $\gamma$ is the ratio of specific heat at constant pressure $C_P$ to the specific heat at constant volume $C_V$.
$\gamma = \dfrac{C_P}{C_V}$
Now let us understand what $C_P$ and $C_V$ mean.
$C_P$ denotes the amount of heat that is required to raise the temperature of 1 mol of a gas by $1^{\circ}C$ while maintaining a constant pressure.
$C_V$ denotes the amount of heat that is required to raise the temperature of 1 mol of a gas by $1^{\circ}C$ while maintaining a constant volume.
Now, the specific heat ratio $\gamma$ is related to the number of degrees of freedom $f$ as follows:
$\gamma = 1 +\dfrac{2}{f}$
A monatomic gas has three degrees of freedom: $\gamma = 1 + \dfrac{2}{3} = \dfrac{5}{3}$
A diatomic gas has five degrees of freedom: $\gamma = 1 + \dfrac{2}{5} = \dfrac{7}{5}$ .
Now, the specific heats $C_P$ and $C_V$ can be expressed in terms of the specific heat ratio $\gamma$, number of moles $n$ and the gas constant $R$ as follows:
$C_P = \dfrac{\gamma n R}{\gamma -1}$ and $C_V = \dfrac{nR}{\gamma -1}$
Therefore, for 1 mol of monatomic gas:
$C_{P_1} = \dfrac{\dfrac{5}{3}R}{\dfrac{5}{3}-1} = \dfrac{5R}{2}$
$C_{V_1} = \dfrac{R}{\dfrac{5}{3}-1} = \dfrac{3R}{2}$
And for 1 mol of diatomic gas:
$C_{P_2} = \dfrac{\dfrac{7}{5}R}{\dfrac{7}{5}-1} = \dfrac{7R}{2}$
$C_{V_2} = \dfrac{R}{\dfrac{7}{5}-1} = \dfrac{5R}{2}$
For a mixture having $n_1$ moles of gas 1 and $n_2$ moles of gas 2, the resultant heat capacity of the mixture is given as:
$C = \dfrac{n_1 C_1 + n_2 C_2}{n_1 +n_2}$
At this point let us revisit the question.
We are mixing 1 mole of monoatomic and 1 mole of diatomic gas together. Therefore, $n_1 = n_2 =1$
Thus, the specific heat capacities of the mixture will be an average of their individual specific heats, i.e.,
$C_{P_{mix}} = \dfrac{C_{P_1}+C_{P_2}}{2} = \dfrac{\dfrac{5R}{2}+\dfrac{7R}{2}}{2} =\dfrac{12R}{4} = 3R$
$C_{V_{mix}} = \dfrac{C_{V_1}+C_{V_2}}{2} = \dfrac{\dfrac{3R}{2}+\dfrac{5R}{2}}{2} = 2R$
Therefore, the specific heat ratio for the mixture will be:
$\gamma = \dfrac{ C_{P_{mix}} }{ C_{V_{mix}}} = \dfrac{3R}{2R} = \dfrac{3}{2}$

So, the correct answer is “Option A”.

Note: An alternative understanding of $C_P$ and $C_V$ is that $C_P$ is applicable when work in done on the system, or work is done by the system, whereas, $C_V$ only applies when the work done, or $P\;dV$ is zero.
There is also a way to calculate $C_P$ from $C_V$ or vice versa in terms of the gas constant $R$. This is known as the Mayer’s relation and is given by:
$C_V = C_P - nR$, where n is the number of moles.