A gas mixture contain one mole O2 gas and one mole He gas.
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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.
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HOPE THIS HELPS YOU
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.
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HOPE THIS HELPS YOU
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