Based on kinetic theory of gases following laws can be proved
(a) Boyle's law
(b) Charles' law
(c) A

Q: Why do we use mean square speed ($\overline{v}^2$) instead of just average speed?

Pressure is caused by the change in momentum of molecules colliding with the walls. Momentum depends on velocity ($v$), and the rate of collisions also depends on velocity ($v$). When you multiply these effects, you get a $v^2$ term. Averaging this over all molecules gives the mean square speed.

Q: Is the Boltzmann constant ($k$) the same as the universal gas constant ($R$)?

They are closely related but not the same. The Boltzmann constant $k$ is the gas constant per *molecule*, while $R$ is the gas constant per *mole*. The relationship is $R = N_A k$, where $N_A$ is Avogadro's number.

Q: Can kinetic theory prove Graham's law of diffusion too?

Yes! From the kinetic energy equation $\frac{1}{2} m \overline{v}^2 = \frac{3}{2} k T$, we can see that the root mean square speed $v_{rms}$ is proportional to $1/\sqrt{m}$. Since the rate of diffusion depends on molecular speed, Graham's law ($r \propto 1/\sqrt{M}$) naturally follows.

Q: What assumptions are we making when we say 'all these laws can be proved'?

We are assuming the gas behaves ideally. This means we assume the gas molecules are point masses with negligible volume and that there are no intermolecular forces of attraction or repulsion between them.

Q: Where does the factor of $1/3$ come from in the equation $pV = \frac{1}{3} N m \overline{v}^2$?

The molecules move randomly in all three spatial dimensions ($x, y, z$). On average, the velocity squared in any one direction is one-third of the total mean square speed ($\overline{v}_x^2 = \frac{1}{3} \overline{v}^2$). The pressure on a specific wall only depends on the motion perpendicular to that wall, hence the $1/3$ factor.

Question

Based on kinetic theory of gases following laws can be proved

Options:

(A) Boyle's law

(B) Charles' law

(C) Avogadro's law

(D) All of these

Sigma Percentile · Accepted Answer

### The Microscopic World of Gases

Imagine a closed container filled with millions of tiny gas molecules, constantly darting around, colliding with each other and the walls of the container. This chaotic, microscopic dance is the foundation of the **Kinetic Theory of Gases**. But how does this invisible motion relate to the macroscopic properties we can measure, like pressure, volume, and temperature?

Let's start with the fundamental equation derived from the kinetic theory. The pressure $p$ exerted by an ideal gas in a volume $V$ is given by:

$$pV = \frac{1}{3} N m \overline{v}^2$$

Here, $N$ is the total number of molecules, $m$ is the mass of a single molecule, and $\overline{v}^2$ is the mean square speed of the molecules. This equation beautifully links the macroscopic world ($p$ and $V$) to the microscopic world ($m$ and $\overline{v}^2$).

### The Bridge to Temperature

Now, we need to bring temperature into the picture. According to the kinetic theory, the absolute temperature $T$ of a gas is a direct measure of the average translational kinetic energy of its molecules. Mathematically, this is expressed as:

$$\frac{1}{2} m \overline{v}^2 = \frac{3}{2} k T$$

where $k$ is the Boltzmann constant. This is a profound realization: temperature is simply the macroscopic manifestation of microscopic kinetic energy!

### The Master Equation

Let's cleverly manipulate our first equation to incorporate this kinetic energy term. We can multiply and divide the right side by 2:

$$pV = \frac{2}{3} N \left( \frac{1}{2} m \overline{v}^2 ight)$$

Notice how we have isolated the kinetic energy term $\left( \frac{1}{2} m \overline{v}^2 ight)$ inside the bracket. Now, we can substitute our temperature relation into this equation:

$$pV = \frac{2}{3} N \left( \frac{3}{2} k T ight)$$

The factors of 2 and 3 cancel out perfectly, leaving us with an incredibly elegant result:

$$pV = N k T$$

This is the **Ideal Gas Equation**! (Note: Since $N = n N_A$ and $R = N_A k$, this is equivalent to the familiar $pV = nRT$).

### Deducing the Empirical Laws

From this single master equation, $pV = N k T$, we can effortlessly prove all the classical empirical gas laws:

1. **Boyle's Law:** If the temperature $T$ and the number of molecules $N$ are kept constant, the right side of the equation is a constant. Therefore, $pV = 	ext{constant}$, which means $V \propto \frac{1}{p}$.

2. **Charles' Law:** If the pressure $p$ and the number of molecules $N$ are kept constant, we can rearrange the equation to $V = \left(\frac{Nk}{p}ight) T$. Since the term in the bracket is constant, $V \propto T$.

3. **Avogadro's Law:** If the pressure $p$ and temperature $T$ are kept constant, we get $V = \left(\frac{kT}{p}ight) N$. This shows that $V \propto N$, meaning equal volumes of gases at the same temperature and pressure contain an equal number of molecules.

Since all these fundamental laws naturally emerge from the kinetic theory derivation, the correct answer is that **all of these** laws can be proved.

Animated Solution for Chemistry - States of Matter: Based on kinetic theory of gases following laws can be proved

Select Answer:

Visualized Solution

$p V = \frac{1}{3} N m \overline{v}^{2}$

$\frac{1}{2} m \overline{v}^{2} = \frac{3}{2} k T$

$p V = \frac{2}{3} N (\frac{1}{2} m \overline{v}^{2})$

$p V = N k T$

\text{Deducing Gas Laws}

\text{The Way Forward}

The Sigma Insight: Gaseous State

The Microscopic World of Gases

The Bridge to Temperature

The Master Equation

Deducing the Empirical Laws

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