Question

If α(alpha) moles of a monoatomic gas are mixed with β(beta) moles of a polyatomic gas and mixture behaves like diatomic gas, then [ neglect the vibrational mode of freedom

Answer 1

Answer on Question #48364-Physics-Molecular Physics-Thermodynamics

If $\alpha$ (alpha) moles of a monoatomic gas are mixed with $\beta$ (beta) moles of a polyatomic gas and mixture behaves like diatomic gas, then [neglect the vibrational mode of freedom]

Solution

Energy associated with $\alpha$ moles of a monoatomic gas is

$$U_1 = \alpha \frac{3}{2} RT.$$

Energy associated with $\beta$ moles of a polyatomic gas is

$$U_2 = \beta \frac{n}{2} RT.$$

Energy associated with $(\alpha + \beta)$ moles of a diatomic gas is

$$U_3 = (\alpha + \beta) \frac{5}{2} RT.$$

Then

$$U_1 + U_2 = U_3 \rightarrow \alpha \frac{3}{2} RT + \beta \frac{n}{2} RT = (\alpha + \beta) \frac{5}{2} RT.$$ $$n = \frac{5(\alpha + \beta) - 3\alpha}{\beta} = \frac{5\beta + 2\alpha}{\beta} = 5 + 2 \frac{\alpha}{\beta}.$$

Answer: $5 + 2 \frac{\alpha}{\beta}$

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