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In a sample of hydrogen sulfide \((M=34.1 \mathrm{g} / \mathrm{mol})\) at a temperature of \(3.00 \times 10^{2} \mathrm{K},\) estimate the ratio of the number of molecules that have speeds very close to \(v_{\mathrm{rms}}\) to the number that have speeds very close to \(2 v_{\mathrm{rms}}\)

Short Answer

Expert verified
The ratio of the number of hydrogen sulfide molecules that have speeds very close to \(v_{rms}\) to those that have speeds very close to \(2v_{rms}\) is approximately 0.223.

Step by step solution

01

Understand the Maxwell-Boltzmann Distribution

The Maxwell-Boltzmann distribution function tells us how the speeds (or energies) of molecules in a gas are distributed at a given temperature. The main equation we will be using is \[f(v) = \sqrt{ \frac{2}{\pi} } \times \frac{v^2}{v_{rms}^3} \times e^{ -v^2/2v_{rms}^2 }\] where the root mean square speed (\(v_{rms}\)) is given by the formula \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] and \(m\) is the molar mass, \(k\) is Boltzmann's constant and \(T\) is the absolute temperature.
02

Calculate the \(v_{rms}\)

The first thing to calculate is the root mean square speed of the molecules in the gas, using the formula mentioned above. For hydrogen sulfide, we have the molar mass \(M=34.1 \, g/mol = 34.1 \times 10^-3 \, kg/mol\). Boltzmann's constant \(k = 1.38 \times 10^{-23} \, J/K\), and the temperature \(T = 3.00 \times 10^{2} \, K\). Substituting all these values in, we get \(v_{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 3.00 \times 10^2}{34.1 \times 10^-3}} = 317.4 \, m/s\)
03

Estimate the Ratio

The question asks for the ratio of the number of molecules with speeds close to \(v_{rms}\)to the number with speeds close to \(2v_{rms}\). This ratio is equivalent to the ratio of the Maxwell-Boltzmann distribution function values at these two speeds. Let's denote by \(f1\) the value of the function at \(v_{rms}\) and by \(f2\) its value at \(2v_{rms}\). We then have the ratio \(f1 / f2 = \frac{v_{rms}^2}{v_{rms}^3} \times e^{ -v_{rms}^2/2v_{rms}^2 } \, / \, \frac{(2v_{rms})^2}{v_{rms}^3} \times e^{ -(2v_{rms})^2/2v_{rms}^2 } = 0.25e^{ -3/2 } \) After calculating this expression, the ratio comes out to be approximately \(0.223\)

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Key Concepts

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

Root Mean Square Speed
The root mean square speed, abbreviated as \(v_{rms}\), is a significant concept in understanding gas molecules' motion. It’s mathematically defined as the square root of the average of the squares of speeds. To visualize this concept, imagine squaring each molecule’s speed, finding an average of these squares, and then taking the square root of that average. This gives you the \(v_{rms}\).
\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
In this equation, \(k\) stands for Boltzmann’s constant, \(T\) is the temperature in Kelvin, and \(m\) represents the molar mass of the gas. The \(v_{rms}\) provides crucial insight into the speed distribution within a gas, linking thermal energy to molecular motion.
For example, in our exercise involving hydrogen sulfide (H\(_2\)S), the molar mass is given as 34.1 g/mol. Students find this practical calculation useful because it connects theoretical understanding with real-world applications, like how gases behave under different temperatures.
Molecular Speed Distribution
The concept of molecular speed distribution is central to understanding how gas molecules move at a given temperature. The Maxwell-Boltzmann distribution describes this concept, defining the probability of molecules having certain speeds in a gas. It provides a spread, showing how many molecules are moving at slow, average, and fast speeds.
The formula used to express this distribution function is:
\[ f(v) = \sqrt{ \frac{2}{\pi} } \times \frac{v^2}{v_{rms}^3} \times e^{ -v^2/2v_{rms}^2 } \]
This formula tells us that at any given speed \(v\), the probability distribution function \(f(v)\) gives a peak, often around the most probable speed, indicating most molecules are not moving extremely slow or fast, but somewhere in between. When considering speeds close to \(v_{rms}\) like in our exercise, comparing molecules with different speeds becomes possible, enhancing the understanding of real gas behavior under normal conditions.
Kinetic Theory of Gases
The kinetic theory of gases posits that gases consist of small particles in constant, random motion. This theory bridges the molecular world with macroscopic observations, like pressure and temperature. A critical element of this framework is that the temperature of a gas impacts the average kinetic energy of its particles.
Using this theory helps explain why and how molecules distribute their speeds in a certain way, as captured by the Maxwell-Boltzmann distribution. The kinetic theory provides a basis for predicting how gases react under various conditions. For example, higher temperatures increase particles' kinetic energy, leading to faster molecular speeds.
Key points of the kinetic theory include:
  • Gas particles are in constant, random motion.
  • Collisions between gas particles are elastic, meaning no energy is lost.
  • The total volume of the gas particles is negligible compared to the volume of the container.
This theoretical basis, combined with real-life calculations like those of \(v_{rms}\), allows students to comprehend why gases behave the way they do at different temperatures, pressures, and volumes.

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Most popular questions from this chapter

Unreasonable results. (a) Find the temperature of \(0.360 \mathrm{kg}\) of water, modeled as an ideal gas, at a pressure of \(1.01 \times 10^{5} \mathrm{Pa}\) if it has a volume of \(0.615 \mathrm{m}^{3} .\) (b) What is unreasonable about this answer? How could you get a better answer?

A company advertises that it delivers helium at a gauge pressure of \(1.72 \times 10^{7} \mathrm{Pa}\) in a cylinder of volume 43.8 L. How many balloons can be inflated to a volume of 4.00 L with that amount of helium? Assume the pressure inside the balloons is \(1.01 \times 10^{5} \mathrm{Pa}\) and the temperature in the cylinder and the balloons is \(25.0^{\circ} \mathrm{C}\).

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People buying food in sealed bags at high elevations often notice that the bags are puffed up because the air inside has expanded. A bag of pretzels was packed at a pressure of 1.00 atm and a temperature of \(22.0^{\circ} \mathrm{C}\). When opened at a summer picnic in Santa Fe, New Mexico, at a temperature of \(32.0^{\circ} \mathrm{C},\) the volume of the air in the bag is 1.38 times its original volume. What is the pressure of the air?

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