One mole of an ideal monoatomic gas is mixed with one mole of an ideal diatomic gas. The molar specific heat of this mixture at constant volume is:
Correct answer: C. 2R
- A. R
- B. 3/2R
- C. 2R
- D. 2.5R
Explanation
To find the molar specific heat at constant volume (Cv) for the mixture, we take the average of the Cv values of the monoatomic and diatomic gases. For a monoatomic gas, Cv = 3/2R, and for a diatomic gas, Cv = 5/2R. The mixture contains one mole of each gas, so the average Cv is:\(\frac{1}{2}(\frac{3}{2}R + \frac{5}{2}R) = \frac{1}{2}(\frac{3R + 5R}{2}) = \frac{8R}{4} = 2R\)Thus, the correct answer is 2R. Option A (R) is incorrect because R is not the specific heat capacity for either gas. Option B (3/2R) is incorrect as it only represents the Cv for monoatomic gases. Option D (2.5R) is incorrect as it only represents the Cv for diatomic gases.
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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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