Question

If 1 mole of a monatomic gas ($\gamma = 5/3$) is mixed with 1 mole of a diatomic gas ($\gamma = 7/5$), the value of $\gamma$ for the mixture is:
A. 1.40
B. 1.50
C. 1.53
D. 3.07

Answer 1

Hint – We know that for monatomic gas, ${C_V} = \dfrac{3}{2}RT$ and ${C_P} = \dfrac{5}{2}RT$ and for diatomic gas, ${C_V} = \dfrac{5}{2}RT$ and ${C_P} = \dfrac{7}{2}RT$.

Complete step-by-step answer:

Formula used –

1. ${C_V} = \dfrac{{{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}}{{{n_1} + {n_2}}}$

2. ${C_P} = \dfrac{{{n_1}{C_{{P_1}}} + {n_2}{C_{{P_2}}}}}{{{n_1} + {n_2}}}$

3. $\gamma = \dfrac{{{C_P}}}{{{C_V}}}$

Given, for monatomic gas is $\gamma = 5/3$
For diatomic gas, $\gamma = 7/5$
We know that, for monatomic gas, ${C_V} = \dfrac{3}{2}RT$ and ${C_P} = \dfrac{5}{2}RT$ and for diatomic gas, ${C_V} = \dfrac{5}{2}RT$ and ${C_P} = \dfrac{7}{2}RT$
No. Monatomic gas moles=1
No. diatomic gas moles = 1
For mixture, ${C_V} = \dfrac{{{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}}{{{n_1} + {n_2}}}$
${C_V} = \dfrac{{\dfrac{3}{2}RT + \dfrac{5}{2}RT}}{2} = \dfrac{8}{4}RT = 2RT$
Similarly, ${C_{P}} = \dfrac{{\dfrac{5}{2}RT + \dfrac{7}{2}RT}}{2} = \dfrac{{12}}{4}RT = 3RT$
$\gamma = \dfrac{{{C_P}}}{{{C_V}}}$$$= \dfrac{{3RT}}{{2RT}} = 1.5$$
Hence, the correct answer is 1.5.
Therefore, the correct option is B.

Note – In these types of questions, we should remember the basic concepts and the formulae associated with monatomic, diatomic and triatomic gases and formula for mixture of gases i.e.
${C_P} = \dfrac{{{n_1}{C_{{P_1}}} + {n_2}{C_{{P_2}}}}}{{{n_1} + {n_2}}}$ and ${C_V} = \dfrac{{{n_1}{C_{{V_1}}} + {n_2}{C_{{V_2}}}}}{{{n_1} + {n_2}}}$.