Problem 30 - Arihant AIEEE Physics

Question

In case of molecules of an ideal gas, which of the following, average velocities cannot be zero? (a) \(\langle\bar{v}\rangle\) (b) \(\left.<\bar{v}^{3}\right\rangle\) (c) \(\left\langle\bar{v}^{4}\right\rangle\) (d) \(\left\langle\bar{v}^{5}\right\rangle\)

Step-by-Step Solution

Answer

(c) \(\left\langle\bar{v}^{4}\right\rangle\) cannot be zero; it is always positive.

Step 1: Understanding the Terms

We need to differentiate between different average velocity equations. In an ideal gas, the velocities are distributed equally in all directions. Important statistical measures to consider include the average velocity \(\langle\bar{v}\rangle\), the mean cube velocity \(\left.\langle\bar{v}^{3}\right\rangle\), and higher power velocities like \(\left.\langle\bar{v}^{4}\right\rangle\), \(\left.\langle\bar{v}^{5}\right\rangle\).

Step 2: Zero Average Velocity Analysis

For symmetric distributions, like the Maxwell-Boltzmann distribution of an ideal gas, the average value \(\langle\bar{v}\rangle\) over all directions results in zero, since positive and negative velocities will cancel out. Therefore, \(\langle\bar{v}\rangle = 0\).

Step 3: Cube and Odd-Power Velocities

Further analysis reveals that any odd-powered average velocity, such as \(\left.\langle\bar{v}^{3}\right\rangle\) and \(\left\langle\bar{v}^{5}\right\rangle\), can also result in zero due to symmetry; the integral over all possible states will be symmetric and cancel out.

Step 4: Even-Power Velocities

However, for even powers like \(\left\langle\bar{v}^{4}\right\rangle\), no cancellations occur since velocities are squared or taken to an even power, making all results non-negative and ultimately leading to a non-zero average.

Key Concepts

Average Velocity

The average velocity in a sample of ideal gas molecules is a key concept to understand. In simple terms, when we consider all the random velocities of gas particles moving in different directions, their average can often turn out to be zero. This happens because velocities in one direction might be canceled by equal and opposite velocities in the other direction. This characteristic is particularly true in an ideal gas scenario because of its symmetrical distribution of velocities.

The formula for calculating average velocity is crucial here. We denote it as \( \langle \bar{v} \rangle \). This is a statistical average derived over large numbers of particle velocities, considering motion in all directions. For gases under the Maxwell-Boltzmann distribution, since particle motion is assumed symmetrical around zero, the net average velocity can often sum to zero.

Maxwell-Boltzmann Distribution

The Maxwell-Boltzmann distribution is fundamental when talking about the velocities of gas molecules. This statistical distribution provides insight into how speeds of particles in a gas are distributed at a given temperature. Here's why it's vital:

• It shows most particles have speeds around the most probable speed.
• Lesser particles have either very high or very low speeds.
• The distribution shape is symmetrical about the mean velocity.

Due to this symmetry, a significant outcome arises that for a gas under this distribution, the average velocity across many particles tends to cancel out, as we discussed earlier. This makes the Maxwell-Boltzmann distribution integral to predicting behaviors such as average velocity or when considering constraints like symmetry in velocities.

Odd-power Velocities

Odd-power velocities refer to calculations like cube or fifth powers of velocities. Analyzing whether these can be zero hinges on the nature of symmetry. Let's explore two such cases:

• Mean cube velocity \( \langle \bar{v}^{3} \rangle \)
• Mean fifth power velocity \( \langle \bar{v}^{5} \rangle \)

When velocities are raised to an odd power, if the velocity distribution around zero is even, each positive velocity has a corresponding negative one. This balance results in a cancellation effect — this is exactly why in some instances these average calculations also end up as zero. Practically, the symmetrical qualities of these distributions in an ideal gas context ensure that odd-power velocities can also average to zero.

Even-power Velocities

Unlike their odd counterparts, even-power velocities do not benefit from the symmetry cancellation. For these calculations:

• Let’s take the mean fourth power velocity \( \langle \bar{v}^{4} \rangle \)

Even-power calculations involve exponentiating the velocities to powers like two or four, which makes all velocities in individual calculations non-negative. This results in them summing up instead of canceling out.

Thus, calculations like \( \langle \bar{v}^{4} \rangle \) in gases are always non-zero because they reflect the sum of positive-only values. The symmetries that apply to odd-powered situations are irrelevant here, ensuring that even-power velocity averages are never zero, providing insights into the kinetic energies that particles carry in a gas.