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Kinetic Theory of Gases appeared 26 times across 3 years — 3% of Physics. This question is from Specific Heat Capacities of Gases.

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γA is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. γB is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If (γA)/(γB) = (1 + (1)/(n)) then the value of n is

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

Solution & Explanation

Related Formula
γ = 1 + (2)/(f)
Core Logic

Let's find the specific heat ratio for each gas based on degrees of freedom:

  • Monoatomic gas A:
  • Degrees of freedom, fA = 3 (translational only)
γA = 1 + (2)/(3) = (5)/(3)
  • Polyatomic gas B:
  • Translational degrees of freedom = 3
  • Rotational degrees of freedom = 3
  • Vibrational modes = 1.

    • Note: Each active vibrational mode has 2 degrees of freedom (kinetic + potential energy terms). This contributes 2 × 1 = 2 degrees of freedom.

  • Therefore, the total active degrees of freedom is:
fB = 3 + 3 + 2 = 8

The specific heat ratio of B is:

γB = 1 + (2)/(fB) = 1 + (2)/(8) = 1 + (1)/(4) = (5)/(4)

Now, find the ratio of specific heat capacities:

(γA)/(γB) = (5/3)/(5/4) = (4)/(3)

We are given:

(γA)/(γB) = 1 + (1)/(n) (4)/(3) = 1 + (1)/(n) (1)/(n) = (1)/(3) n = 3
Step 1: Final Conclusion

The value of n is 3.

Pattern Recognition

Always remember that each vibrational mode contributes exactly 2 degrees of freedom because it holds both kinetic and potential energy components (fvib = 2 × modes).

Chapter Mix

Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics

Reference Study Guides

Kinetic Theory Of Gases

More Kinetic Theory of Gases Previous-Year Questions — Page 6

Q31 jee_main_2024_31_jan_morning Kinetic Energy Of Gas Molecules
The parameter that remains the same for molecules of all gases at a given temperature is :
  • A. kinetic energy
  • B. momentum
  • C. mass
  • D. speed

Solution

Related Formula
KE = (f)/(2)kT
Core Logic

The average translational kinetic energy of any gas molecule depends only on the absolute temperature of the gas and is independent of the nature or mass of the gas.

For 1 mole of any ideal gas, the average translational kinetic energy is (3)/(2)RT. Therefore, at a given temperature, the kinetic energy parameter is uniform across all ideal gases.

Pattern Recognition

Temperature is directly proportional to average translational kinetic energy. If T is constant, KE is constant for all gases regardless of mass.

Chapter Mix

Class 11 Physics: Kinetic Theory

More Kinetic Theory of Gases Questions — jee_main_2025_02_april_morning

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)