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JEE Main 2024 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

A closed container holds a mixture of 2 mol of helium and 1 mol of an ideal diatomic gas (rigid, no vibration).

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A closed container holds a mixture of 22 mol of helium and 11 mol of an ideal diatomic gas (rigid, no vibration). The ratio γ=Cp/Cv\gamma=C_p/C_v of the mixture is
  1. A. 32\dfrac{3}{2}
  2. B. 85\dfrac{8}{5}
  3. C. 1912\dfrac{19}{12}
  4. D. 1711\dfrac{17}{11}
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Answer: (D) 1711\dfrac{17}{11}

For a mixture, Cv=n1Cv1+n2Cv2n1+n2C_v=\dfrac{n_1C_{v1}+n_2C_{v2}}{n_1+n_2}. Helium: Cv=32RC_v=\tfrac32R; rigid diatomic: Cv=52RC_v=\tfrac52R. So Cv=2(1.5R)+1(2.5R)3=5.5R3=11R6C_v=\dfrac{2(1.5R)+1(2.5R)}{3}=\dfrac{5.5R}{3}=\dfrac{11R}{6}. Then Cp=Cv+R=17R6C_p=C_v+R=\dfrac{17R}{6} and γ=1711≈1.545\gamma=\dfrac{17}{11}\approx1.545. Averaging the two gammas (5/35/3 and 7/57/5) weighted by moles is a common error.
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