Sigma Percentile
JEE Main 2021
LEVELJEE Main

Animated Solution for Physics - Thermodynamics: Calculate the value of mean free path () for oxygen molecules at temperature and pressure . Assume the molecular diameter and the gas is ideal. ()

Select Answer:

Visualized Solution

The Sigma Insight: Kinetic Theory of Gases

Solution Diagram

The Chaos of the Microscopic World

Imagine you are shrunk down to the size of an oxygen molecule, trapped inside a container. It is absolute chaos. You are zipping around at hundreds of meters per second, but you can't go very far before—BAM!—you collide with another molecule. The average distance you manage to travel in a straight line between these violent collisions is what physicists call the mean free path ().
In this problem, we are tasked with finding exactly how far an oxygen molecule travels between collisions at a pleasant room temperature of and standard atmospheric pressure.

The Master Equation

To find the mean free path, we rely on a beautiful result from the Kinetic Theory of Gases:
Let's break down why this formula makes perfect physical sense: - Temperature () is in the numerator: Higher temperature means molecules are moving faster and the gas tends to expand (if not rigidly confined), increasing the space between molecules. - Pressure () is in the denominator: Higher pressure means the molecules are packed more densely. More crowding equals more frequent collisions, hence a shorter path. - Diameter squared () is in the denominator: The larger the molecule, the bigger its "target area" (). A bigger target means it's much harder to avoid hitting others. - The factor: This accounts for the fact that all molecules are moving, not just the one we are tracking. It represents the relative velocity between colliding particles.

Setting Up the Raw Data

Before we plug numbers into our master equation, we must ensure absolute harmony in our units. Physics is unforgiving when it comes to mixed units!
- Pressure: (Already in standard SI units). - Temperature: . We must convert this to absolute temperature (Kelvin).
- Diameter: . We convert this to meters.
- Boltzmann Constant: .

Executing the Calculation

Now, we carefully substitute our pristine data into the formula:
Let's tackle the numerator first. It's straightforward:
Next, the denominator. We must square the diameter first:
Now, multiply the constants in the denominator (, ):
Finally, we divide the numerator by our simplified denominator:

The Final Verdict

To match our options, we convert the result back into nanometers ():
The mean free path is approximately . This means an oxygen molecule travels about 340 times its own diameter before crashing into a neighbor. The microscopic world is indeed a crowded, chaotic place!

Similar Questions

JEE Main 2021
LEVELJEE Main

Consider a sample of oxygen behaving like an ideal gas. At , the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be (Molecular weight of oxygen is ; )

(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Advanced

A volume cylinder is filled with of gas at room temperature (). The molecular diameter of and its root mean square speed are found to be and , respectively. What is the average collision rate (per second) for an molecule?

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELBoard

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature ? ( is Boltzmann constant)

(A)
(B)
(C)
(D)
LEVELJEE Main

The average translational energy and the rms speed of molecules in a sample of oxygen gas at K are J and m/s respectively. The corresponding values at K are nearly (assuming ideal gas behaviour)

(A)
J, m/s
(B)
J, m/s
(C)
J, m/s
(D)
J, m/s
LEVELJEE Main

From the following statements concerning ideal gas at any given temperature , select the correct one (s).

* Multiple Correct Options
(A)
The coefficient of volume expansion at constant pressure is the same for all ideal gases
(B)
The average translational kinetic energy per molecule of oxygen gas is , being Boltzmann constant
(C)
The mean-free path of molecules increases with decrease in the pressure
(D)
In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different
LEVELJEE Main

Let , and respectively denote the mean speed, root mean square speed and most probable speed of the molecules in an ideal monoatomic gas at absolute temperature . The mass of a molecule is . Then,

* Multiple Correct Options
(A)
no molecule can have a speed greater than
(B)
no molecule can have speed less than
(C)
(D)
the average kinetic energy of a molecule is
JEE Main 2019
LEVELJEE Main

For a given gas at pressure, rms speed of the molecules is at . At pressure and at , the rms speed of the molecules will be

(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main

A polyatomic ideal gas has 24 vibrational modes. What is the value of ?

(A)
1.03
(B)
1.30
(C)
1.37
(D)
10.3
LEVELJEE Advanced

One gram mole of oxygen at 27°C and one atmospheric pressure is enclosed in a vessel. (a) Assuming the molecules to be moving with , find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (b) The vessel is next thermally insulated and moved with a constant speed . It is then suddenly stopped. The process results in a rise of the temperature of the gas by 1°C. Calculate the speed .

JEE Main 2020
LEVELJEE Advanced

Number of molecules in a volume of of a perfect monoatomic gas at some temperature and at a pressure of of mercury is close to (Given, mean kinetic energy of a molecule at is , , density of mercury )

(A)
(B)
(C)
(D)