Animated Solution for Physics - Thermodynamics: A $25\times 10^{-3}{\text{ m}}^3$ volume cylinder is filled with $1\text{ mole}$ of ${\text{O}}_2$ gas at room temperature ($300\text{ K}$). The molecular diameter of ${\text{O}}_2$ and its root mean square speed are found to be $0.3\text{ nm}$ and $200\text{ m/s}$, respectively. What is the average collision rate (per second) for an ${\text{O}}_2$ molecule?

Imagine you are standing inside a cylinder filled with oxygen gas. To the naked eye, it looks completely still. But if you could zoom in to the microscopic level, you would see absolute chaos. Billions of oxygen molecules are darting around at breakneck speeds, constantly crashing into the walls and, more importantly, into each other.

In the Kinetic Theory of Gases, one of the most fascinating questions we can ask is: How often does a single molecule experience a collision? This is known as the collision rate or collision frequency ($Z$). To solve this, we need to understand the geometry of a molecular collision.

Let's carefully lay out the parameters given in our problem:

Volume of the cylinder ($V$): $25\times 10^{-3}{\text{ m}}^3$ Amount of gas ($N$): $1\text{ mole}$, which means we have Avogadro's number of molecules, $N_A=6.022\times 10^{23}$. Molecular diameter ($d$): $0.3\text{ nm}=0.3\times 10^{-9}\text{ m}$ Root Mean Square speed ($v_{\text{rms}}$): $200\text{ m/s}$

To calculate the collision rate, we use a brilliant thought experiment. Imagine freezing all the molecules in the gas except for one. As this single molecule moves forward, it sweeps out an imaginary cylinder. If the center of any frozen molecule falls inside this cylinder, a collision occurs.

What is the radius of this swept cylinder? If two molecules, each with radius $r$, approach each other, they will touch if the distance between their centers is $2r$. Since $2r$ is exactly the diameter $d$, the effective "target area" or collision cross-section is $\pi d^2$.

This is a critical point where many students (and even official examiners!) make a mistake. The cross-section is $\pi d^2$, not $\pi r^2$.

The collision rate $Z$ is defined as the average speed of the molecule divided by the mean free path ($\lambda$), which is the average distance traveled between collisions:

The formula for the mean free path, accounting for the fact that all molecules are moving (which introduces a relative velocity factor of $\sqrt{2}$), is:

Where $n_V$ is the number density (number of molecules per unit volume). Substituting this back into our collision rate equation gives us our Master Equation:

Notice that our Master Equation requires the average speed ($v_{\text{av}}$), but the problem gave us the RMS speed ($v_{\text{rms}}$). These are not the same! From the Maxwell-Boltzmann distribution, we know the relationship between them:

Let's calculate this first:

Next, let's find the number density $n_V$:

Now, we bring all our pieces together into the Master Equation:

If you look at the options provided in the question—$10^{10},10^{11},10^{12},10^{13}$—none of them are close to our mathematically rigorous answer of $\sim 10^9$. Did we do something wrong?

No! This question is a classic example of a flawed problem in a competitive exam. The official solution key made a fundamental physics error: they used the radius squared ($r^2$) instead of the diameter squared ($d^2$) in their collision cross-section.

Because $r=d/2$, using $r^2$ introduces a factor of $1/4$. If you divide our correct answer by 4, you get:

This is exactly the calculation shown in the flawed official key! Even with their flawed calculation, the answer is $\sim 10^8$, which still doesn't match any of the options.

The ultimate takeaway? Trust your conceptual rigor. When you know the physics is solid, don't let incorrect options shake your confidence.