Start Learning Free

Before You Begin

Question #82015

Difficulty: 5/10 | 🟢 Free

The Question

A rigid vessel contains a mixture of two non-interacting ideal gases, AA and BB, in thermal equilibrium at absolute temperature TT. The mass of a single molecule of gas AA is mm, and the mass of a single molecule of gas BB is 4m4m. The collision cross-section area between a molecule of AA and a molecule of BB is σ\sigma. If the number density of gas BB is nBn_B and the mean speed of gas AA molecules is v0=8kTπmv_0 = \sqrt{\frac{8kT}{\pi m}} (where kk is the Boltzmann constant), what is the collision frequency ZABZ_{AB} of a single molecule of gas AA with the molecules of gas BB?

Options

  • A.

    52nBσv0\frac{\sqrt{5}}{2} n_B \sigma v_0

  • B.

    32nBσv0\frac{3}{2} n_B \sigma v_0

  • C.

    2nBσv0\sqrt{2} n_B \sigma v_0

  • D.

    12nBσv0\frac{1}{2} n_B \sigma v_0

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Mean Free Path|Frame of Reference|Vector Vector Direction

Why This Question Is Tricky

Common Mistakes

Students often memorize the collision frequency formula Z=2nσvavgZ = \sqrt{2} n \sigma v_{avg} for a single gas without understanding its statistical mechanics origin.

The 2\sqrt{2} factor arises because the mean relative speed ⟨vrel⟩=2vavg\langle v_{rel} \rangle = \sqrt{2} v_{avg} is a specific case of the general relative velocity distribution using reduced mass μ=m/2\mu = m/2 for identical particles.

When faced with a gas mixture, students fall into traps: they either blindly apply the 2\sqrt{2} factor (Option C), or incorrectly assume scalar addition/subtraction of mean speeds like v0±v0/2v_0 \pm v_0/2 (Options B and D).

This question tests the fundamental understanding that relative velocity in a 3D Maxwellian gas must be treated as the difference of two random vectors, requiring the reduced mass μ=mAmBmA+mB\mu = \frac{m_A m_B}{m_A + m_B} to correctly compute the mean relative speed magnitude.

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterKinetic Theory - Class 11 Part 2
ConceptMean Free Path
Difficulty5
Cognitive Load6
Target Solve Time140 sec
AccessFree