Question

A gas mixture contains one mole of oxygen gas and one mole of helium gas. Find the ratio of specific heat at constant pressure to that at constant volume of the gaseous mixture

Answer 1

Helium (He) is a monoatomic gas and dioxygen (O2) is a diatomic gas.

For a monoatomic gas and a diatomic gas, value of Cv are (3/2)R and (5/2)R respectively.

For a gaseous mixture, (Cv)mix = [n1 (Cv)1 + n2.(Cv)2] / [n1 + n2].

Hence for the given mixture, (Cv)mix = [2.(3/2).R + 1.(5/2).R] / [1 + 2] =[3R + (5/2)R] / 3 = (11R/6).

Now since Cp- Cv = R hence Cp = Cv + R = (17) R / 6.

So we get, Cp / Cv = (17 / 11).

Also you can use, (Cp/Cv) - 1 = (R/Cv).

Therefore, (Cp/Cv) = 1 + (R/Cv) = 1 + (6/11) = 17/11.

Answer 2

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