Question

Two moles of an ideal gas with $\dfrac{{{C_P}}}{{{C_V}}} = \dfrac{5}{3}$ are mixed with $3$ moles of another ideal gas with $\dfrac{{{C_P}}}{{{C_V}}} = \dfrac{4}{3}$. The value of $\dfrac{{{C_P}}}{{{C_V}}}$ for the mixture is:
A.$1.47$
B.$1.45$
C.$1.42$
D.$1.50$

Answer 1

Using the data given in the question, we need to find out the values of ${C_{v1}}$, ${C_{v2}}$ and ${C_{p1}}$,${C_{p2}}$ respectively. This can be done by rearranging the given ratios. These individual ${C_p}$ and ${C_v}$ can be used to obtain the required value of the mixture.

Complete step by step answer:
We know that ${C_v}$ represents molar heat capacity at constant volume and ${C_p}$ represents molar heat capacity at constant pressure.
The ratio $\dfrac{{{C_p}}}{{{C_v}}}$ is called heat capacity ratio and is sometimes also known as an adiabatic index.
Let us consider ${n_1}$ to be the number of moles in gas $1$ and ${n_2}$ be the number of moles in gas $2$.
In the first case:
$\dfrac{{{C_{P1}}}}{{{C_{V1}}}} = \dfrac{5}{3}$
On rearranging this equation, we can write:
$3{C_p} = 5{C_v}$
Also, we know in thermodynamics, ${C_p} = {C_v} + R.....(i)$
Here, $R$ is the universal gas constant.
Now, putting ${C_{p1}} = \dfrac{5}{3}{C_{v1}}$ in equation$(i)$, we get:
$\dfrac{5}{3}{C_{v1}} = {C_{v1}} + R$
$\Rightarrow \dfrac{5}{3}{C_{v1}} = R$,
$\Rightarrow {C_{v1}} = \dfrac{3}{5}R$
Solving similarly, we also obtain ${C_{p1}} = \dfrac{5}{2}R$

Now, in the second case, we have:
$\dfrac{{{C_{P2}}}}{{{C_{V2}}}} = \dfrac{4}{3}$
This gives us, ${C_{P2}} = \dfrac{4}{3}{C_{V2}}$
Putting this value in the equation$(i)$, we get:
$\dfrac{4}{3}{C_{V2}} = {C_{V2}} + R$
On solving this, we get:
${C_{V2}} = 3R$
Thus, we obtain:
${C_{P2}} = 3R + R = 4R$
Therefore, when the two mixtures are mixed, in the resultant mixture:
${C_{Vmix}} = \dfrac{{{n_1} \times {C_{V1}} + {n_2} \times {C_{V2}}}}{{{n_1} + {n_2}}}$
On putting the values, as obtained above. We get:
${C_{Vmix}} = \dfrac{{12R}}{5}$
Similarity,
For ${C_{pmix}} = \dfrac{{{n_1} \times {C_{p1}} + {n_2} \times {C_{p2}}}}{{{n_1} + {n_2}}}$
Again, on solving, we get:
${C_{pmix}} = \dfrac{{17R}}{5}$
Therefore, the heat capacity ratio is:
$\dfrac{{{C_{pmix}}}}{{{C_{vmix}}}} = \dfrac{{17}}{{12}} = 1.42$
This is the required solution.
Therefore, option (C) is correct.

Note:
As evident from the equation $(i)$,${C_p}$ is larger than ${C_v}$. This happens because when a substance is added at constant pressure, work is done in the process as a result of which expansion occurs. This requires more heat to be supplied to get the work done. Thus ${C_v}$ can never be greater than ${C_p}$.