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JEE Main 2019 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

A rigid container holds a mixture of 2 moles of helium and 1 mole of a diatomic gas (rigid molecules, no vibration) at ordinary temperature.

Single correct option

A rigid container holds a mixture of 22 moles of helium and 11 mole of a diatomic gas (rigid molecules, no vibration) at ordinary temperature. The ratio CP/CVC_P/C_V of the mixture is
  1. A. 7145\dfrac{71}{45}
  2. B. 32\dfrac32
  3. C. 85\dfrac85
  4. D. 1711\dfrac{17}{11}
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Answer: (D) 1711\dfrac{17}{11}

Heat capacities add for a mixture, so CVC_V must be averaged, not γ\gamma. Total CVC_V (per three moles) is 2(32R)+1(52R)=112R2\left(\tfrac32R\right)+1\left(\tfrac52R\right)=\tfrac{11}{2}R, so the mean molar value is CV=116RC_V = \tfrac{11}{6}R. Then CP=CV+R=176RC_P = C_V + R = \tfrac{17}{6}R and γ=CP/CV=17/11≈1.55\gamma = C_P/C_V = 17/11\approx 1.55. Averaging the individual γ\gamma values with mole weights, (2⋅53+75)/3=71/45(2\cdot\tfrac53+\tfrac75)/3 = 71/45, is the typical wrong shortcut.
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