A mixture of a monoatomic gas and a diatomic gas has adiabatic exponent γ = 1.6 . Find the ratio n 2 n 1 of the number of moles of the monoatomic to the diatomic gas.

C v = γ − 1 R = 0.6 R = 3 5 R . So n 1 + n 2 2 3 n 1 + 2 5 n 2 = 3 5 , which gives 2 3 n 1 + 2 5 n 2 = 3 5 n 1 + 3 5 n 2 , so 6 5 n 2 = 6 1 n 1 and n 2 n 1 = 5 .

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PhysicsHeat & ThermoThermodynamics Laws & Specific Heats of GasesMedium2 minQB

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A mixture of a monoatomic gas and a diatomic gas has adiabatic exponent $γ = 1.6$ . Find the ratio $\frac{n _{1}}{n _{2}}$ of the number of moles of the monoatomic to the diatomic gas.

Answer:

Solution:

$C_{v} = \frac{R}{γ - 1} = \frac{R}{0.6} = \frac{5}{3} R$ . So $\frac{\frac{3}{2} n _{1} + \frac{5}{2} n _{2}}{n _{1} + n _{2}} = \frac{5}{3}$ , which gives $\frac{3}{2} n_{1} + \frac{5}{2} n_{2} = \frac{5}{3} n_{1} + \frac{5}{3} n_{2}$ , so $\frac{5}{6} n_{2} = \frac{1}{6} n_{1}$ and $\frac{n _{1}}{n _{2}} = 5$ .

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermodynamics Laws & Specific Heats of Gases

⏱ 2mℹ️ Source: QB

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