The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The

Q: Why do we use number density $N^*$ instead of the number of moles $n$?

The mean free path formula fundamentally depends on the number of target particles per unit volume, which is exactly what number density $N^*$ represents. Using moles $n$ would require an extra step to divide by Avogadro's number and volume.

Q: Where does the $T^{3/2}$ come from?

It comes from multiplying two temperature-dependent terms. The mean free path contributes a factor of $T$ (since volume expands with temperature at constant pressure), and the mean speed contributes a factor of $\sqrt{T}$ or $T^{1/2}$. Multiplying them gives $T^{1} \times T^{1/2} = T^{3/2}$.

Q: Does the nature of the gas, like its molar mass, affect this ratio?

No, not the *ratio*. While the absolute value of the diffusion coefficient depends on the molar mass (inversely proportional to $\sqrt{M}$), when we take the ratio for the *same* gas under different conditions, the mass terms cancel out completely.

Q: What is the difference between $R$ and $k$ in these formulas?

$R$ is the universal gas constant, used when dealing with macroscopic quantities like moles. $k$ is the Boltzmann constant, used for microscopic quantities like individual molecules. They are related by $R = N_A k$, where $N_A$ is Avogadro's number.

Q: Why is diffusion proportional to $\lambda \times U_{avg}$?

Diffusion is essentially a random walk process. The rate at which particles spread out depends on how far they can step between collisions ($\lambda$) and how frequently they take those steps, which is governed by their speed ($U_{avg}$).

Question

The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is

Sigma Percentile · Accepted Answer

### The Dance of Gas Molecules

Imagine a sealed container filled with an ideal gas. Inside, billions of molecules are engaged in a chaotic, never-ending dance.

They zip around at high speeds, constantly colliding with one another and the walls of the container.

This random motion is the fundamental mechanism behind **diffusion**, the process by which a gas spreads out to fill its available space.

The problem tells us that the diffusion coefficient, $D$, is proportional to two key microscopic properties.

First, the **mean free path** ($\lambda$), which is the average distance a molecule travels between collisions.

Second, the **mean speed** ($U_{avg}$), which dictates how fast the molecule covers that distance.

Mathematically, this is expressed as:

$$D \propto \lambda 	imes U_{avg}$$

### Decoding the Microscopic Variables

To understand how diffusion changes with macroscopic properties like temperature ($T$) and pressure ($P$), we need to break down $\lambda$ and $U_{avg}$.

The mean free path depends on how crowded the molecules are. If there are more molecules in a given volume, collisions happen more frequently, and the mean free path decreases.

This crowding is measured by the **number density** ($N^*$), giving us the relationship:

$$\lambda = \frac{1}{\sqrt{2}\pi\sigma^2 N^*}$$

Here, $\sigma$ is the collision diameter of the molecule. Notice that $\lambda$ is inversely proportional to $N^*$.

Next, we look at the mean speed. According to the kinetic theory of gases, the speed of molecules is driven entirely by thermal energy.

The formula for mean speed is:

$$U_{avg} = \sqrt{\frac{8RT}{\pi M}}$$

This tells us that the mean speed is directly proportional to the square root of the absolute temperature:

$$U_{avg} \propto \sqrt{T}$$

### The Ideal Gas Bridge

We have a slight problem. Our mean free path formula uses number density ($N^*$), but the question gives us changes in pressure ($P$) and temperature ($T$).

We need a bridge to connect the microscopic world to the macroscopic world.

This bridge is the **Ideal Gas Law**, expressed in terms of individual molecules:

$$P = N^* k T$$

By rearranging this equation, we can solve for the number density:

$$N^* = \frac{P}{kT}$$

Now, we substitute this expression for $N^*$ back into our mean free path proportionality.

Since $\lambda \propto \frac{1}{N^*}$, we get:

$$\lambda \propto \frac{T}{P}$$

This makes intuitive sense! Heating a gas at constant pressure causes it to expand, reducing crowding and increasing the mean free path. Increasing pressure compresses the gas, increasing crowding and reducing the mean free path.

### The Master Proportionality

We are now ready to construct our master equation for the diffusion coefficient.

We substitute our new proportionalities for $\lambda$ and $U_{avg}$ back into the original equation:

$$D \propto \left(\frac{T}{P}ight) 	imes \sqrt{T}$$

Combining the temperature terms ($T^1 	imes T^{1/2}$), we arrive at the crucial relationship:

$$D \propto \frac{T^{3/2}}{P}$$

### The Final Calculation

The problem states that the absolute temperature is increased by a factor of 4, and the pressure is increased by a factor of 2.

We can set up a ratio of the final diffusion coefficient to the initial diffusion coefficient:

$$\frac{D_{final}}{D_{initial}} = \frac{(4T)^{3/2} / (2P)}{T^{3/2} / P}$$

The initial $T$ and $P$ variables cancel out beautifully, leaving us with just the numerical factors:

$$\frac{D_{final}}{D_{initial}} = \frac{4^{3/2}}{2}$$

Let's evaluate the numerator carefully. Four to the power of three-halves is the same as taking the square root of four, and then cubing the result.

The square root of 4 is 2, and 2 cubed is 8.

$$\frac{D_{final}}{D_{initial}} = \frac{8}{2} = 4$$

The diffusion coefficient increases by a factor of 4. Therefore, our final answer is **$x = 4$**.

This elegant result shows how deeply interconnected the macroscopic properties of a gas are with the invisible, chaotic motion of its molecules!

Enter Numerical Value:

Visualized Solution

$D \propto λ \times U_{a v g}$

$Mean Free Path (λ)$

$Mean Speed (U_{a v g})$

$The Missing Link$

$Ideal Gas Law$

$λ in terms of P and T$

$Combining Proportionalities$

$The Master Equation$

$Setting up the Ratio$

$Substituting Values$

$Calculating x$

$Effect of Molar Mass$

The Sigma Insight: Gaseous State

The Dance of Gas Molecules

Decoding the Microscopic Variables

The Ideal Gas Bridge

The Master Proportionality

The Final Calculation

Similar Questions

The molecular velocity of any gas is

When $r$ , $p$ and $M$ represent rate of diffusion, pressure and molecular mass, respectively, then the ratio of the rates of diffusion $(r_{A} / r_{B})$ of two gases $A$ and $B$ , is given as

A certain gas obeys $p (V_{m} - b) = R T$ . The value of $(\frac{\partial Z}{\partial p})_{T}$ is $\frac{x b}{R T}$ . The value of $x$ is ............ . ( $Z =$ compressibility factor)

The volume of gas $A$ is twice than that of gas $B$ . The compressibility factor of gas $A$ is thrice than that of gas $B$ at same temperature. The pressures of the gases for equal number of moles are

For an ideal gas, number of moles per litre in terms of its pressure $p$ , temperature $T$ and gas constant $R$ is

Which one of the following graphs is not correct for ideal gas? $d = density$ , $p = pressure$ , $T = temperature$

Which one of the following statements is not true about the effect of an increase in temperature on the distribution of molecular speeds in a gas?

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

The compressibility factor for a real gas at high pressure is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D∝λ×Uavg​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mean Free Path (λ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mean Speed (Uavg​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Missing Link

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ideal Gas Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ in terms of P and T

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combining Proportionalities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Master Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Effect of Molar Mass