Question:
The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \mathrm{~m}$ and $600 \mathrm{~m} / \mathrm{s}$, respectively. Find the frequency of its collisions.
The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \mathrm{~m}$ and $600 \mathrm{~m} / \mathrm{s}$, respectively. Find the frequency of its collisions.
Show Hint
The frequency of collisions is given by the average speed divided by the mean free path.
Updated On: Jan 14, 2026
- $2 \times 10^{10} / \mathrm{s}$
- $9 \times 10^{9} / \mathrm{s}$
- $2 \times 10^{9} / \mathrm{s}$
- $5 \times 10^{8} / \mathrm{s}$
Show Solution
Verified By Collegedunia
The Correct Option is C
Approach Solution - 1
The problem asks us to find the frequency of collisions of oxygen molecules, given their mean free path and average speed at 300 K and 1 atm.
Concept Used:
The frequency of molecular collisions \( f \) is related to the mean free path \( \lambda \) and the average speed \( \bar{c} \) by the formula:
\[ f = \frac{\bar{c}}{\lambda} \]where:
- \( f \) = frequency of collisions (in s\(^{-1}\))
- \( \bar{c} \) = average speed of molecules (in m/s)
- \( \lambda \) = mean free path (in m)
Step-by-Step Solution:
Step 1: Write down the given data.
\[ \lambda = 3 \times 10^{-7} \, \mathrm{m}, \quad \bar{c} = 600 \, \mathrm{m/s} \]Step 2: Use the formula for collision frequency:
\[ f = \frac{\bar{c}}{\lambda} \]Step 3: Substitute the given values into the formula.
\[ f = \frac{600}{3 \times 10^{-7}} \]Step 4: Simplify the expression.
\[ f = 2 \times 10^{9} \, \mathrm{s^{-1}} \]Final Computation & Result:
Therefore, the frequency of collisions of oxygen molecules is:
\[ \boxed{f = 2 \times 10^{9} \, \mathrm{s^{-1}}} \]Final Answer: The frequency of collisions is \(2 \times 10^{9} \, \mathrm{s^{-1}}\).
Was this answer helpful?
0
0
Show Solution
Verified By Collegedunia
Approach Solution -2
1. Given:
- Mean free path, $\lambda = 3 \times 10^{-7} \mathrm{~m}$
- Average speed, $v_{\text{avg}} = 600 \mathrm{~m/s}$
2. Frequency of collisions: \[ \text{Frequency} = \frac{v_{\text{avg}}}{\lambda} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9} \mathrm{~s^{-1}} \] Therefore, the correct answer is (3) $2 \times 10^{9} / \mathrm{s}$.
2. Frequency of collisions: \[ \text{Frequency} = \frac{v_{\text{avg}}}{\lambda} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9} \mathrm{~s^{-1}} \] Therefore, the correct answer is (3) $2 \times 10^{9} / \mathrm{s}$.
Was this answer helpful?
0
0
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
View More Questions
Top JEE Main The Kinetic Theory of Gases Questions
- The temperature of a gas having \( 2.0 \times 10^{25} \) molecules per cubic meter at 1.38 atm (Given, \( k = 1.38 \times 10^{-23} \, \text{JK}^{-1} \)) is:
- The average kinetic energy of a monatomic molecule is 0.414 eV at temperature: (Use \( k_B = 1.38 \times 10^{-23} \, \text{J/mol-K} \))
- The parameter that remains the same for molecules of all gases at a given temperature is :
- The temperature of a gas is \(-78^\circ\text{C}\) and the average translational kinetic energy of its molecules is \( K \). The temperature at which the average translational kinetic energy of the molecules of the same gas becomes \( 2K \) is:
- The speed of sound in oxygen at S.T.P. will be approximately: Given, \( R = 8.3 \, \text{J K}^{-1} \), \( \gamma = 1.4 \))
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions