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NEET CHEMISTRYThermodynamicsMedium

Question

Equal volumes of monoatomic and diatomic gases at the same initial temperature and pressure are mixed. The ratio of specific heats of the mixture (Cp/CvC_p/C_v) will be [AFMC 2002]:

A

1

B

2

C

1.67

D

1.5

Step-by-Step Solution

To find the ratio of specific heats (γ\gamma ) for a mixture, we use the values of molar heat capacities for monoatomic and diatomic gases.

  1. Avogadro's Law: According to the sources, equal volumes of gases at the same temperature and pressure contain an equal number of moles (nn) . Thus, we assume n1n_1 (monoatomic) = n2n_2 (diatomic) = 1 mole.
  2. Molar Heat Capacities: From standard thermodynamic data :
  • Monoatomic gas: Cv1=32RC_{v1} = \frac{3}{2}R and Cp1=52RC_{p1} = \frac{5}{2}R.
  • Diatomic gas: Cv2=52RC_{v2} = \frac{5}{2}R and Cp2=72RC_{p2} = \frac{7}{2}R.
  1. Heat Capacities of the Mixture: Cv,mix=n1Cv1+n2Cv2n1+n2=1.5R+2.5R2=2RC_{v,mix} = \frac{n_1C_{v1} + n_2C_{v2}}{n_1 + n_2} = \frac{1.5R + 2.5R}{2} = 2R. Cp,mix=n1Cp1+n2Cp2n1+n2=2.5R+3.5R2=3RC_{p,mix} = \frac{n_1C_{p1} + n_2C_{p2}}{n_1 + n_2} = \frac{2.5R + 3.5R}{2} = 3R.
  2. Ratio (\gamma ): The ratio of specific heats of the mixture is γmix=Cp,mixCv,mix=3R2R=1.5\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{3R}{2R} = 1.5.

This calculation aligns with Option D.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsvolumesmonoatomicdiatomicinitialtemperature

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