Multiple choice

A mixture of ideal gasses $N_{2}$ and $He$ are taken in the mass ratio of $14 : 1$ respectively. Molar heat capacity of the mixture at constant pressure is.

  1. $\dfrac{19 R}{6}$
  2. $\dfrac{6 R}{19}$
  3. $\dfrac{13 R}{6}$
  4. $\dfrac{6 R}{13}$
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A Correct answer
Explanation

Mass ratio N2:He = 14:1. Molar mass N2 = 28, He = 4. Moles ratio = (14/28) : (1/4) = 0.5 : 0.25 = 2:1. Mixture Cp = (n1*Cp1 + n2*Cp2) / (n1+n2). Cp(N2) = 7R/2, Cp(He) = 5R/2. Mixture Cp = (2 * 7R/2 + 1 * 5R/2) / 3 = (7R + 2.5R) / 3 = 9.5R / 3 = 19R/6.

AI explanation

Taking 14 g of N2 and 1 g of He gives 0.5 moles of N2 and 0.25 moles of He, for a total of 0.75 moles. The molar heat capacity at constant pressure for N2 is 7R/2 and for a monatomic gas like He is 5R/2. Using the mole fraction weighted average, Cp_mix = (0.5/0.75)(7R/2) + (0.25/0.75)(5R/2), which evaluates to 19R/6.

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