Animated Solution for Physics - Thermodynamics: Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently, the mean collision time between the gas molecule changes from ${\tau }_1$ to ${\tau }_2$. If $\frac{C_p}{C_V}=\gamma$ for this gas, then a good estimate for $\frac{{\tau }_1}{{\tau }_2}$ is given by

What is Mean Collision Time? Imagine you are in a crowded room, walking in a straight line until you bump into someone. The average time you spend walking freely before a collision is your mean collision time ($\tau$). In a gas, this time depends on two critical factors: how far apart the molecules are (the mean free path, $\lambda$), and how fast they are moving (the average speed, $v_{\text{mean}}$).

Mathematically, this is expressed as:

When a gas expands, the volume $V$ increases. This means the molecules spread out, and the mean free path $\lambda$ increases directly with volume ($\lambda \propto V$). If the molecules have to travel further to hit each other, the collision time naturally goes up.

Lambda and Average Speed But wait, there's a second factor: speed. The average speed of a gas molecule is dictated by its temperature ($v_{\text{mean}}\propto \sqrt{T}$).

If we combine these two dependencies, we get a master proportionality for the mean collision time:

This equation is beautiful because it captures the tug-of-war between space and speed.

Eliminating Temperature Here is where the specific thermodynamic process comes into play. The problem states the expansion is adiabatic. In an adiabatic expansion, the gas does work on its surroundings without any heat entering the system. It pays for this work using its own internal energy, which causes the temperature to drop!

We know the adiabatic equation of state relating temperature and volume is:

Rearranging this, we find how temperature scales with volume:

Now, we substitute this temperature scaling back into our master proportionality for $\tau$:

Let's carefully simplify the exponents. The denominator is $V^{\frac{1-\gamma }{2}}$. When we bring it to the numerator, we subtract the exponents:

The problem asks for the ratio of the initial collision time to the final collision time, $\frac{{\tau }_1}{{\tau }_2}$.

Using our derived proportionality:

We are given that the volume doubles, meaning $V_2=2V_1$, or $\frac{V_1}{V_2}=\frac{1}{2}$. Substituting this into our ratio yields the final, elegant result:

This perfectly matches option (a). The physics tells us that because the gas expanded (increasing distance) AND cooled down (decreasing speed), the time between collisions increased significantly, making the ratio ${\tau }_1/{\tau }_2$ a fraction less than 1.