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, $C_V={\displaystyle \frac{3}{2}}RT$ and $C_P={\displaystyle \frac{5}{2}}RT$ and for diatomic gas, $C_V={\displaystyle \frac{5}{2}}RT$ and $C_P={\displaystyle \frac{7}{2}}RT$.

Complete step-by-step answer:

Formula used –

1) $C_V={\displaystyle \frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}}$

2) $C_P={\displaystyle \frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1+n_2}}$

3) $\gamma ={\displaystyle \frac{C_P}{C_V}}$

Given, for monatomic gas is $\gamma =5/3$
For diatomic gas, $\gamma =7/5$
We know that, for monatomic gas, $C_V={\displaystyle \frac{3}{2}}RT$ and $C_P={\displaystyle \frac{5}{2}}RT$ and for diatomic gas, $C_V={\displaystyle \frac{5}{2}}RT$ and $C_P={\displaystyle \frac{7}{2}}RT$
No. Monatomic gas moles=1
No. diatomic gas moles = 1
For mixture, $C_V={\displaystyle \frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}}$
$$C_V={\displaystyle \frac{{\displaystyle \frac{3}{2}}RT+{\displaystyle \frac{5}{2}}RT}{2}}={\displaystyle \frac{8}{4}}RT=2RT$$
Similarly, $$C_P={\displaystyle \frac{{\displaystyle \frac{5}{2}}RT+{\displaystyle \frac{7}{2}}RT}{2}}={\displaystyle \frac{12}{4}}RT=3RT$$
$\gamma ={\displaystyle \frac{C_P}{C_V}}$$$={\displaystyle \frac{3RT}{2RT}}=1.5$$
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.
$C_P={\displaystyle \frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1+n_2}}$ and $C_V={\displaystyle \frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}}$.