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Heat and Thermodynamics question

2025 · 4 Apr · Shift 1 · Q67
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  5. /2025 · 4 Apr · Shift 1 · Q67

Heat and Thermodynamics question

2025 · 4 Apr · Shift 1 · Q67

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are 3×10−7 m3 \times 10^{-7} \mathrm{~m}3×10−7 m and 600 m/s600 \mathrm{~m} / \mathrm{s}600 m/s, respectively. Find the frequency of its collisions.
  1. A
    5×108/s5 \times 10^8 / \mathrm{s}5×108/s
  2. B
    9×105/s9 \times 10^5 / \mathrm{s}9×105/s
  3. C
    2×1010/s2 \times 10^{10} / \mathrm{s}2×1010/s
  4. D
    2×109/s2 \times 10^9 / \mathrm{s}2×109/s
View written solutionFree

Correct answer: D

  1. Use the relation between mean free path and collision frequency

For a gas molecule,

collision frequency=average speedmean free path\text{collision frequency} = \frac{\text{average speed}}{\text{mean free path}}collision frequency=mean free pathaverage speed​

because the molecule travels, on average, one mean free path between two successive collisions.

So,

z=vavgλz = \frac{v_{\text{avg}}}{\lambda}z=λvavg​​

where:

  • vavg=600 m/sv_{\text{avg}} = 600\ \text{m/s}vavg​=600 m/s
  • λ=3×10−7 m\lambda = 3 \times 10^{-7}\ \text{m}λ=3×10−7 m
  1. Substitute the values
z=6003×10−7z = \frac{600}{3 \times 10^{-7}}z=3×10−7600​
  1. Simplify
6003=200\frac{600}{3} = 2003600​=200

Thus,

z=200×107=2×109 s−1z = 200 \times 10^{7} = 2 \times 10^{9}\ \text{s}^{-1}z=200×107=2×109 s−1
  1. Match with the options
2×109 /s2 \times 10^{9}\ /\text{s}2×109 /s

So the correct option is D.

  1. Verification with stored answer

Stored correct answer = D

Derived answer = D

Hence, they agree.

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