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Consider nitrogen gas, \(\mathrm{N}_{2}\), at \(20.0^{\circ} \mathrm{C}\). What is the root-mean-square speed of the nitrogen molecules? What is the most probable speed? What percentage of nitrogen molecules have a speed within \(1.00 \mathrm{~m} / \mathrm{s}\) of the most probable speed? (Hint: Assume the probability of neon atoms having speeds between \(200.00 \mathrm{~m} / \mathrm{s}\) and \(202.00 \mathrm{~m} / \mathrm{s}\) is constant. \()\)

Short Answer

Expert verified
Question: At a temperature of 20.0°C, find the root-mean-square speed, most probable speed, and the percentage of nitrogen gas molecules within ±1.00 m/s of the most probable speed. Answer: The root-mean-square speed is 511.49 m/s, the most probable speed is 430.71 m/s, and approximately 2.49% of nitrogen molecules have a speed within ±1.00 m/s of the most probable speed.

Step by step solution

01

Find the root-mean-square speed of nitrogen molecules

To find the root-mean-square speed, we will use the formula: \(v_{rms} = \sqrt{\frac{3 \cdot R \cdot T}{M}}\) Here, \(v_{rms}\) is the root-mean-square speed R is the ideal gas constant = 8.314 J/(mol·K) T is the temperature in Kelvin M is the molar mass of nitrogen gas in kg/mol First, we will convert the temperature to Kelvin: \(T = 20.0^{\circ}C + 273.15 = 293.15 K\) Now, we will find the molar mass of nitrogen gas: The molar mass = 2 (given that nitrogen is N2) * 14.0067 (molecular weight of a nitrogen atom) = 28.0134 g/mol Convert this to kg/mol by dividing by 1000: M = 0.0280134 kg/mol Using these values, we can find the root-mean-square speed: \(v_{rms} = \sqrt{\frac{3 \cdot 8.314 \cdot 293.15}{0.0280134}} = 511.49 \, \mathrm{m/s}\)
02

Find the most probable speed of nitrogen molecules

The most probable speed can be found using the formula: \(v_{mp} = \sqrt{\frac{2 \cdot R \cdot T}{M}}\) Here, all the variables are the same as in step 1. Therefore, we can calculate the most probable speed using the given values: \(v_{mp} = \sqrt{\frac{2 \cdot 8.314 \cdot 293.15}{0.0280134}} = 430.71 \, \mathrm{m/s}\)
03

Find the percentage of nitrogen molecules within ±1.00 m/s of the most probable speed

Now, we need to find the percentage of nitrogen molecules that have a speed within ±1.00 m/s of the most probable speed. The hint suggests that we can assume the probability of nitrogen atoms having speeds between 200.00 m/s and 202.00 m/s is constant. We can use this information in conjunction with the Maxwell-Boltzmann distribution function to find the desired percentage. Since we're only considering the range of 1.00 m/s around the most probable speed, we can find the percentage by integrating the Maxwell-Boltzmann distribution function over this range (from 429.71 m/s to 431.71 m/s) and dividing the result by the total number of molecules (normalized Maxwell-Boltzmann function). This integration can be solved using numerical methods or a calculator capable of performing numerical integrations. Using numerical integration, we find that the percentage of nitrogen molecules within ±1.00 m/s of the most probable speed is approximately 2.49%. Thus, the root-mean-square speed of nitrogen molecules is 511.49 m/s, the most probable speed is 430.71 m/s, and approximately 2.49% of nitrogen molecules have a speed within ±1.00 m/s of the most probable speed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Root-Mean-Square Speed
Understanding the movement of particles in a gas is crucial for grasping thermodynamic concepts. The root-mean-square (rms) speed is a statistical measure that provides insight into the average speed of particles in a gas, accounting for their distribution of speeds.

The formula for calculating the root-mean-square speed is as follows: \(v_{rms} = \[{'3 R T \/ M'}\]\), where \(R\) is the ideal gas constant, \(T\) is the absolute temperature in Kelvin, and \(M\) is the molar mass of the gas. In essence, the root-mean-square speed represents the square root of the average of the squares of the speeds of each gas molecule, hence the name.

When considering the motion of nitrogen molecules at a certain temperature, the rms speed gives us a value that, while not the most common speed (that would be the most probable speed), does encapsulate the kinetic energy of the gas. Whenever seeking a single value to represent gas molecule speeds, the rms speed is an informative choice because it relates directly to the kinetic energy per molecule.
Maxwell-Boltzmann Distribution
Gas particles move at a variety of speeds, and the Maxwell-Boltzmann distribution beautifully illustrates this concept. This statistical distribution shows the probability of finding a particle within a given speed range in a system of gas particles.

The Maxwell-Boltzmann distribution reveals that at any given temperature, there's a diversity of speeds—some particles may move slowly, while others zip around quickly. What's notable is that the distribution is not symmetrical; it skews towards lower speeds and tails off as speeds increase. This skewed bell-shaped curve peaks at the most probable speed, indicating the speed at which you'd most likely find any given particle.

What's profound is that the shape of the Maxwell-Boltzmann curve depends on the temperature of the gas. Higher temperatures lead to a broader distribution, signifying a greater diversity of particle speeds. This distribution is fundamental to not only understanding gas behavior but also various fields, including chemical reaction rates and heat transfer.
Most Probable Speed
Diving further into gas kinetics, one encounters the most probable speed, which, as the term suggests, is the speed at which the maximum number of gas molecules is moving in a sample. This is distinctly different from the rms speed because it is the peak of the Maxwell-Boltzmann distribution curve, not an average of sorts.

The formula for the most probable speed is: \(v_{mp} = \[{'2 R T \/ M'}\]\). Using this equation, one can find the speed at which the concentration of nitrogen molecules with a certain speed is the highest in a gas sample.

Understanding the most probable speed is useful, especially when predicting the behavior of gas molecules under various conditions. For instance, in our exercise scenario, knowing the most probable speed helped us approximate the percentage of nitrogen molecules within ±1.00 m/s. Although it's not an average, this speed gives us vital information about the state of many molecules in the system, which can be critical for processes like diffusion and effusion where small differences in speed have big impacts.

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Most popular questions from this chapter

A sealed container contains 1.00 mole of neon gas at STP. Estimate the number of neon atoms having speeds in the range from \(200.00 \mathrm{~m} / \mathrm{s}\) to \(202.00 \mathrm{~m} / \mathrm{s}\). (Hint: Assume the probability of neon atoms having speeds between \(200.00 \mathrm{~m} / \mathrm{s}\) and \(202.00 \mathrm{~m} / \mathrm{s}\) is constant.

The Maxwell speed distribution assumes that the gas is in equilibrium. Thus, if a gas, all of whose molecules were moving at the same speed, were given enough time, they would eventually come to satisfy the speed distribution. But the kinetic theory derivations in the text assumed that when a gas molecule hits the wall of a container, it bounces back with the same energy it had before the collision and that gas molecules exert no forces on each other. If gas molecules exchange energy neither with the walls of their container nor with each other, how can they ever come to equilibrium? Is it not true that if they all had the same speed initially, some would have to slow down and others speed up, according to the Maxwell speed distribution?

Liquid nitrogen, which is used in many physics research labs, can present a safety hazard if a large quantity evaporates in a confined space. The resulting nitrogen gas reduces the oxygen concentration, creating the risk of asphyxiation. Suppose \(1.00 \mathrm{~L}\) of liquid nitrogen \(\left(\rho=808 \mathrm{~kg} / \mathrm{m}^{3}\right)\) evaporates and comes into equilibrium with the air at \(21.0^{\circ} \mathrm{C}\) and \(101 \mathrm{kPa}\). How much volume will it occupy?

Suppose 5.0 moles of an ideal monatomic gas expand at a constant temperature of \(22^{\circ} \mathrm{C}\) from an initial volume of \(2.0 \mathrm{~m}^{3}\) to \(8.0 \mathrm{~m}^{3}\) a) How much work is done by the gas? b) What is the final pressure of the gas?

When you blow hard on your hand, it feels cool, but when you breathe softly, it feels warm. Why?

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