2 mole of an ideal monoatomic gas mix with 1 mole of a ideal diatomic gas. The Cp/Cv for the mixture is:
Correct answer: B. 17/11
- A. 15/11
- B. 17/11
- C. 13/11
- D. None
Explanation
To find the Cp/Cv (gamma, γ) for the mixture, consider the specific heat capacities for the individual gases:For a monoatomic gas, γ = 5/3, so Cv = 3R/2.For a diatomic gas, γ = 7/5, so Cv = 5R/2.Calculate the mean specific heat capacities for the mixture using the mole fractions:Cv(mixture) = (2*Cv(monoatomic) + 1*Cv(diatomic)) / 3 = (2*(3R/2) + 1*(5R/2)) / 3 = 11R/6Cp(mixture) = Cv(mixture) + R = 11R/6 + R = 17R/6Thus, γ(mixture) = Cp(mixture) / Cv(mixture) = (17R/6) / (11R/6) = 17/11.Therefore, the correct answer is 17/11. The other options are incorrect due to miscalculations in averaging the specific heat capacities.
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About Molar Specific Heat of a Gas
Molar specific heat is the heat required to raise the temperature of one mole of a gas by one kelvin, expressed through Q = nCΔT. Questions compare the constant volume value Cv with the constant pressure value Cp, apply the ideal gas relation Cp − Cv = R, and identify how the thermodynamic process affects heat capacity.
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