Problem 30 - University Physics with Modern Physics

Question

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. (a) On average, how many molecules of each type will there be in either half of the box? (b) What is the change in entropy of the system when the partition is punctured? (c) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured- that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

Step-by-Step Solution

Answer

(a) 250 nitrogen and 50 oxygen molecules in each half. (c) Probability is extremely low: \(\left(\frac{1}{2}\right)^{600}\).

Step 1: Determine Final Distribution of Molecules

Initially, we have 500 nitrogen molecules on the left and 100 oxygen molecules on the right. Once the partition is removed and equilibrium is reached, the molecules will distribute evenly due to diffusion. Since the total number of nitrogen molecules is 500, on average, 250 nitrogen molecules will be in each half. For oxygen, with 100 molecules, on average, 50 molecules will be in each half of the box.

Step 2: Calculate Change in Entropy

The change in entropy \(\Delta S\) for a given gas undergoing free expansion can be calculated using the formula: \[\Delta S = nR \ln(\frac{V_{final}}{V_{initial}})\]Here, since the gases expand to double their initial volume, \(\frac{V_{final}}{V_{initial}} = 2\).For nitrogen, \(n = 500\) moles, therefore the entropy change is:\[\Delta S_{N2} = 500R \ln(2)\].For oxygen, \(n = 100\) moles, similarly the entropy change is:\[\Delta S_{O2} = 100R \ln(2)\].Total change in entropy of the system is:\[\Delta S = \Delta S_{N2} + \Delta S_{O2} = 600R \ln(2)\].

Step 3: Calculate Probability of Initial Configuration

The probability \(P\) of returning to the initial distribution (500 nitrogen in the left and 100 oxygen in the right) is determined by the probability of each molecule staying on the side where it initially was. For nitrogen:\[P_{N2} = \left(\frac{1}{2}\right)^{500}\]And for oxygen:\[P_{O2} = \left(\frac{1}{2}\right)^{100}\]Therefore, the overall probability is the product:\[P = P_{N2} \times P_{O2} = \left(\frac{1}{2}\right)^{600}\].

Key Concepts

Diffusion

Diffusion is a process that describes the spread of particles from areas of high concentration to areas of lower concentration, resulting in an even distribution over time. It's a natural phenomenon that occurs in gases, liquids, and even in solids on a microscopic level. At the molecular level, when the partition in the box is punctured, nitrogen and oxygen molecules begin to move randomly, bouncing off each other and the walls of the box. This random movement leads to diffusion.

• Initially, nitrogen molecules are crowded on the left, and oxygen on the right.
• Diffusion drives the system towards an equilibrium, where on average, half the nitrogen molecules find themselves in each part of the box, and the same for oxygen molecules.

The driving force behind this process is the natural tendency for systems to increase in entropy, moving towards a state of disorder which, in this context, means an even spread of particles.

Probability

In statistical mechanics, probability plays a crucial role in understanding molecular distributions. When reflecting on the probability of molecules returning to their original configuration after the partition is removed, we must consider each molecule’s chance of staying on its initial side.

• The probability of finding a particular nitrogen atom on one side after diffusion is \( \frac{1}{2} \).
• Similarly, for oxygen, the chance is also \( \frac{1}{2} \).
• Hence, the likelihood of all 500 nitrogen molecules and 100 oxygen molecules staying in their initial halves is extremely low.

This is because each molecule acts independently, and the total probability is the product of individual probabilities. Hence, an outcome where particles return to the original configuration is so improbable that it is practically impossible in a real-world scenario.

Free Expansion

Free expansion refers to the process where a gas expands into a vacuum without being hindered by external pressure. In this particular exercise, each gas undergoes a free expansion once the partition is removed, doubling its available volume without any resistance or work done on the surroundings.

• During free expansion, gas molecules spread out due to their natural movement and occupy the entire box.
• Important to note is that during this expansion, the temperature remains constant because there is no work done on or by the system.

This type of expansion results in an increase in entropy because the gases have more space to occupy, leading to more possible arrangements of molecules — essentially increasing the system's disorder. The concept of free expansion is vital to understanding how gas molecules behave when they transition from a constrained state to one of complete freedom, maximizing disorder or entropy in a system.