An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time be

Q: Why do we use average speed instead of rms speed in the collision time formula?

For calculating collision frequency and mean collision time, we actually need the mean relative speed between molecules. The mean relative speed is proportional to the mean average speed ($v_{\text{avg}}$), not the rms speed. However, since both $v_{\text{avg}}$ and $v_{\text{rms}}$ are proportional to $\sqrt{T}$, the proportionality $t \propto \frac{\sqrt{T}}{p}$ remains exactly the same regardless of which speed you use in a ratio problem!

Q: Why does collision time decrease when temperature increases, shouldn't they be further apart?

If pressure is kept constant, increasing temperature does increase the volume, moving molecules further apart (increasing mean free path). However, the molecules are also moving much faster. The increase in speed outpaces the increase in distance, so they cover that longer distance in less time. Hence, collision time decreases.

Q: How does pressure affect the mean free path?

Higher pressure (at a constant temperature) means a higher number density of molecules—more molecules packed into the same space. With more obstacles around, a molecule can't travel as far before hitting another one, so the mean free path decreases.

Q: Can we use Celsius instead of Kelvin for the temperatures?

Absolutely not! All thermodynamic formulas derived from the kinetic theory of gases, including those for rms speed and mean free path, are based on absolute zero. You must always convert temperatures to Kelvin before taking ratios or square roots.

Q: What if the gas was diatomic instead of monoatomic? Would the formula change?

The formula for mean free path depends on the collision cross-section (diameter of the molecule), and the average speed depends on the molar mass. While the exact numerical values change for different gases, the functional dependence on Temperature ($T$) and Pressure ($p$) remains identical. The proportionality $t \propto \frac{\sqrt{T}}{p}$ holds true for any ideal gas.

Question

An ideal gas is enclosed in a cylinder at pressure of $2 	ext{ atm}$ and temperature, $300 	ext{ K}$. The mean time between two successive collisions is $6 	imes 10^{-8} 	ext{ s}$. If the pressure is doubled and temperature is increased to $500 	ext{ K}$, the mean time between two successive collisions will be close to

Options:

(A) $4 	imes 10^{-8} 	ext{ s}$

(B) $3 	imes 10^{-6} 	ext{ s}$

(C) $2 	imes 10^{-7} 	ext{ s}$

(D) $0.5 	imes 10^{-8} 	ext{ s}$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two States

Imagine a gas trapped inside a sturdy cylinder. In its initial state, the gas is at a pressure of $2 	ext{ atm}$ and a temperature of $300 	ext{ K}$. The molecules are in constant, chaotic motion, zipping around and crashing into each other. The average time a molecule travels before it hits another one is called the **mean collision time**, denoted by $t$. Initially, this time is $t_1 = 6 	imes 10^{-8} 	ext{ s}$.

Now, we change the conditions. We compress the gas, doubling the pressure to $p_2 = 4 	ext{ atm}$, and we heat it up to $T_2 = 500 	ext{ K}$. The question is: how does this affect the mean collision time? Will they collide more frequently or less frequently?

### The Master Equation: Mean Collision Time

To figure this out, we need to understand what governs the collision time. The mean time between collisions is simply the average distance a molecule travels between collisions (the **mean free path**, $\lambda$) divided by its average speed ($v_{	ext{avg}}$).

$$t = \frac{\lambda}{v_{	ext{avg}}}$$

Let's break down these two components. The mean free path $\lambda$ depends on the number density of the gas. According to the kinetic theory of gases, it can be expressed in terms of temperature and pressure:

$$\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 p}$$

This tells us that $\lambda$ is directly proportional to temperature $T$ and inversely proportional to pressure $p$.

Next, the average speed of the gas molecules is given by:

$$v_{	ext{avg}} = \sqrt{\frac{8 k_B T}{\pi M}}$$

This shows that the average speed is directly proportional to the square root of the temperature, $\sqrt{T}$.

### The Proportionality Trick

Instead of calculating everything from scratch, we can use a powerful trick: **proportionality**. By substituting the proportionalities of $\lambda$ and $v_{	ext{avg}}$ into our equation for $t$, we get:

$$t \propto \frac{T/p}{\sqrt{T}}$$

Simplifying this, we find a beautiful and elegant relationship:

$$t \propto \frac{\sqrt{T}}{p}$$

This means the collision time is directly proportional to the square root of the temperature and inversely proportional to the pressure.

### The Final Calculation

Since we are comparing two states of the same gas, we can set up a ratio:

$$\frac{t_2}{t_1} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \cdot \frac{p_1}{p_2}$$

Let's plug in our raw values: $T_1 = 300 	ext{ K}$, $T_2 = 500 	ext{ K}$, $p_1 = 2 	ext{ atm}$, and $p_2 = 4 	ext{ atm}$.

$$\frac{t_2}{t_1} = \sqrt{\frac{500}{300}} \cdot \frac{2}{4}$$

$$\frac{t_2}{t_1} = \sqrt{\frac{5}{3}} \cdot \frac{1}{2}$$

Now, we multiply this ratio by our initial time $t_1 = 6 	imes 10^{-8} 	ext{ s}$ to find $t_2$:

$$t_2 = \left(6 	imes 10^{-8}ight) \cdot \frac{1}{2} \cdot \sqrt{\frac{5}{3}}$$

$$t_2 = 3 	imes 10^{-8} \cdot \sqrt{1.666...}$$

$$t_2 \approx 3 	imes 10^{-8} \cdot 1.291$$

$$t_2 \approx 3.87 	imes 10^{-8} 	ext{ s}$$

Looking at our options, the closest value is **$4 	imes 10^{-8} 	ext{ s}$**.

Notice the physics here: even though the higher temperature made the molecules move faster (which would normally decrease collision time), the doubled pressure squeezed them so much closer together that the mean free path plummeted. The pressure effect dominated, leading to a shorter overall collision time.

Select Answer:

Visualized Solution

Visualizing the Two States

The Mean Collision Time Formula

Deriving the Proportionality

Setting Up the Ratio

Simplifying the Ratio

Final Calculation

Conclusion

The Sigma Insight: Kinetic Theory of Gases

The Setup

The Master Equation

The Proportionality Trick

The Final Calculation

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