Question

If one mole of a monatomic gas $\left( {\gamma = \dfrac{5}{3}} \right)$ is mixed with one mole of a diatomic gas $\left( {\gamma = \dfrac{7}{5}} \right)$, the value for the mixture is
A.1.4
B.1.5
C.1.53
D.3.07

Answer 1

For an ideal monatomic gas, ${C_v} = \dfrac{3}{2}RT$, and ${C_p} = \dfrac{5}{2}RT$ and for an ideal diatomic gas, ${C_v} = \dfrac{5}{2}RT$, and ${C_p} = \dfrac{7}{2}RT$. We can deduce the values of ${C_p}$ and ${C_v}$ for both the gases by using the equation ${C_p} = \dfrac{{\gamma nR}}{{\gamma - 1}}$ and ${C_v} = \dfrac{{nR}}{{\gamma - 1}}$. Once we have found the ${C_p}$ and ${C_v}$ values for both the gases, we will find $\gamma$ for the mixture from the equation $\dfrac{{{C_p}}}{{{C_v}}}$

Complete step by step answer:
For a monatomic gas,
${n_1} = 1$
For diatomic gas,
${n_2} = 1$
For one mole of monoatomic gas,
${C_{v1}} = \dfrac{3}{2}R$, and ${C_{p1}} = \dfrac{5}{2}R$ s
For one mole of diatomic gas,
${C_{v2}} = \dfrac{5}{2}RT$, and ${C_{p2}} = \dfrac{5}{2}R$
${C_p}$ denotes the amount of heat that is required to raise the temperature of 1 mole by ${1^\circ }C$ while maintaining constant pressure and ${C_v}$ denotes the amount of heat that is required to raise the temperature of 1 mole by ${1^\circ }C$ while maintaining constant volume.
Thus, for mixture of 1 mole each,
${C_v}$ for the mixture $= \dfrac{{{n_1}{C_{v1}}{n_2}{C_{v2}}}}{{{n_1} + {n_2}}}$
${C_v}$ for the mixture $3 = \dfrac{{\dfrac{3}{2}RT + \dfrac{5}{2}RT}}{2} = \dfrac{8}{4}RT = 2RT$
And, ${C_p} = \dfrac{{\dfrac{5}{2}RT + \dfrac{7}{2}RT}}{2} = \dfrac{{12}}{4}RT = 3RT$
So, the specific heat ratio for the mixture will be:
$\Rightarrow \gamma = \dfrac{{{C_p}}}{{{C_v}}} = \dfrac{{3RT}}{{2RT}} = \dfrac{3}{2} = 1.5$
$\gamma = 1.5$

So, the correct answer is option (B).

Note: The alternate method to solve this problem is,
For a monatomic gas,
${n_1} = 1;\gamma 1 = \dfrac{5}{3}$
For diatomic gas,
${n_2} = 1;\gamma 2 = \dfrac{7}{5}$
For mixture of gases,
From, $\dfrac{{nR}}{{\gamma - 1}} =$ constant
$\dfrac{{{n_1} + {n_2}}}{{{\gamma _m} - 1}} = \dfrac{{{n_1}}}{{{\gamma _1} - 1}} + \dfrac{{{n_1}}}{{{\gamma _2} - 1}}$
$\Rightarrow$ $\dfrac{{1 + 1}}{{{\gamma _m} - 1}} = \dfrac{1}{{\dfrac{5}{3} - 1}} + \dfrac{1}{{\dfrac{7}{5} - 1}}$
$\Rightarrow$ $\dfrac{2}{{{\gamma _m}}} - 1 = \dfrac{3}{2} + \dfrac{5}{2}$
$\Rightarrow$ $\dfrac{2}{{{\gamma _m}}} - 1 = 4$
$\Rightarrow$ ${\gamma _m} - 1 = 0.5$
$\Rightarrow$ ${\gamma _m} = 1.5$