For gaseous state, if most probable speed is denoted by C*, average speed by C_bar and mean square s

Q: Why did we assume it's RMS speed instead of mean square speed as written in the question?

If we used the actual mean square speed ($3RT/M$), it wouldn't have a square root. The ratio would be $\sqrt{2} : \sqrt{8/\pi} : 3$, which doesn't match the magnitude or format of any of the given options. The options clearly indicate that all three terms have the same dimensions, implying a typo in the question.

Q: How do I easily remember the order of these three speeds?

You can use the acronym **RAM**: **R**oot mean square > **A**verage > **M**ost probable. Mathematically, $\sqrt{3} > \sqrt{8/\pi} > \sqrt{2}$.

Q: Why is the average speed not at the peak of the Maxwell-Boltzmann curve?

The Maxwell-Boltzmann distribution is not perfectly symmetrical like a bell curve; it is skewed to the right. The long tail of high-speed molecules pulls the mathematical average to the right of the highest peak (the most probable speed).

Q: What is the physical significance of the $\sqrt{RT/M}$ term?

It acts as a characteristic 'thermal speed scale' for a gas. It shows that molecular speeds increase with the square root of temperature ($T$) and decrease with the square root of molar mass ($M$).

Q: How can I quickly calculate $\sqrt{4/\pi}$ without a calculator?

Approximate $\pi$ as $3.14$. So, $4/3.14$ is roughly $1.27$. You know $1.1^2 = 1.21$ and $1.2^2 = 1.44$. Since $1.27$ is closer to $1.21$, the square root must be slightly more than $1.1$, which perfectly aligns with $1.128$ in the options.

Question

For gaseous state, if most probable speed is denoted by $C^*$, average speed by $\bar{C}$ and mean square speed by $C$, then for a large number of molecules, the ratios of these speeds are

Options:

(A) $C^* : \bar{C} : C = 1.225 : 1.128 : 1$

(B) $C^* : \bar{C} : C = 1.128 : 1.225 : 1$

(C) $C^* : \bar{C} : C = 1 : 1.128 : 1.225$

(D) $C^* : \bar{C} : C = 1 : 1.225 : 1.128$

Sigma Percentile · Accepted Answer

### The Maxwell-Boltzmann Symphony

Imagine you are looking at a bustling city from above. Not everyone is walking at the same speed. Some are strolling leisurely, some are jogging, and a few are sprinting to catch a train. A gas inside a container behaves exactly the same way. The molecules are in a state of constant, chaotic motion, colliding with each other and the walls of the container.

To make sense of this chaos, we use the **Maxwell-Boltzmann distribution curve**. This beautiful mathematical curve tells us exactly what fraction of molecules possesses a particular speed at a given temperature.

### The Three Speeds

When we analyze this distribution, three distinct statistical speeds emerge as our primary tools:

1. **Most Probable Speed ($C^*$):** 
If you look at the peak of the Maxwell-Boltzmann curve, the speed corresponding to this highest point is the most probable speed. It is the speed possessed by the maximum number of molecules in the gas. Mathematically, it is given by:

$$C^* = \sqrt{\frac{2RT}{M}}$$

2. **Average Speed ($\bar{C}$):** 
If you were to take the arithmetic mean of the speeds of all the individual molecules, you would get the average speed. Because the distribution curve is skewed to the right (it has a long tail towards higher speeds), the average speed is pulled slightly to the right of the peak. Its formula is:

$$\bar{C} = \sqrt{\frac{8RT}{\pi M}}$$

3. **Root Mean Square Speed ($C_{rms}$):** 
This is the square root of the average of the squares of the speeds. It is the most physically significant speed because it directly relates to the average kinetic energy of the gas. It is located even further to the right on the curve:

$$C_{rms} = \sqrt{\frac{3RT}{M}}$$

### The Typo Trap

Now, let's look closely at the question. It asks for the ratio involving the "mean square speed" denoted by $C$. However, there is a catch here! The mean square speed is simply $\frac{3RT}{M}$ (without the square root). If we were to use this, the units wouldn't match, and the ratio would be completely off from the given options.

This is a classic case of a misprint in the exam paper. By looking at the options, which are all in the neighborhood of $1$, it becomes glaringly obvious that the examiner intended to ask for the **Root Mean Square (RMS) speed**, not the mean square speed. We must proceed by assuming $C$ represents the RMS speed.

### The Grand Ratio

Let's set up the ratio of these three speeds exactly as requested:

$$C^* : \bar{C} : C = \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}}$$

Notice how the term $\sqrt{\frac{RT}{M}}$ is a common factor in all three expressions. This term represents the fundamental thermal velocity scale of the gas. We can elegantly cancel it out from the ratio:

$$C^* : \bar{C} : C = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3}$$

To match the format of the options, where the first term is exactly $1$, we need to normalize our ratio. We do this by dividing the entire ratio by $\sqrt{2}$:

$$C^* : \bar{C} : C = \frac{\sqrt{2}}{\sqrt{2}} : \frac{\sqrt{8/\pi}}{\sqrt{2}} : \frac{\sqrt{3}}{\sqrt{2}}$$

$$C^* : \bar{C} : C = 1 : \sqrt{\frac{4}{\pi}} : \sqrt{\frac{3}{2}}$$

### Final Calculation

Now, we just need to crunch the numbers.

For the middle term, we know $\pi \approx 3.1415$.

$$\frac{4}{\pi} \approx \frac{4}{3.1415} \approx 1.273$$

Taking the square root:

$$\sqrt{1.273} \approx 1.128$$

For the third term:

$$\sqrt{\frac{3}{2}} = \sqrt{1.5} \approx 1.225$$

Substituting these decimal values back into our normalized ratio, we get:

**$$C^* : \bar{C} : C = 1 : 1.128 : 1.225$$**

This perfectly matches option (c). The beauty of this problem lies not just in remembering the formulas, but in having the situational awareness to spot the typo and normalize the ratio to find the correct answer.

Animated Solution for Chemistry - States of Matter: For gaseous state, if most probable speed is denoted by $C^{*}$ , average speed by $\overset{ˉ}{C}$ and mean square speed by $C$ , then for a large number of molecules, the ratios of these speeds are

Select Answer:

Visualized Solution

$f (v)$

$C^{*}$

$\overset{ˉ}{C}$

$C$

$C^{*} : \overset{ˉ}{C} : C$

$2 : \frac{8}{π} : 3$

$1 : \frac{4}{π} : \frac{3}{2}$

$1 : 1.128 : 1.225$

$Option (c)$

The Sigma Insight: Gaseous State

The Maxwell-Boltzmann Symphony

The Three Speeds

The Typo Trap

The Grand Ratio

Final Calculation

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