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Chapter 18: Problem 11

Show that for a mixture of two gases at the same temperature, the ratio of their rms speeds is equal to the inverse ratio of the square roots of their molecular masses.

Short Answer

Expert verified
The rms speed ratio is the inverse ratio of the square roots of their molecular masses.

Step by step solution

01

Understanding the rms speed

The root mean square (rms) speed of a gas is given by the formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molecular mass of the gas. This formula helps us understand how fast the molecules in a gas move, on average, at a given temperature.
02

Set up the ratio of rms speeds

Let's consider two different gases, Gas 1 and Gas 2. If \( v_{rms,1} \) and \( v_{rms,2} \) are the rms speeds for Gas 1 and Gas 2 respectively, we want to find the ratio \( \frac{v_{rms,1}}{v_{rms,2}} \). Using the formula for rms speed, we have:\[ v_{rms,1} = \sqrt{\frac{3kT}{m_1}} \] \[ v_{rms,2} = \sqrt{\frac{3kT}{m_2}} \] Thus, the ratio is:\[ \frac{v_{rms,1}}{v_{rms,2}} = \frac{\sqrt{\frac{3kT}{m_1}}}{\sqrt{\frac{3kT}{m_2}}} \]
03

Simplify the expression

Notice that the \( 3kT \) term appears in both the numerator and the denominator. Since they are at the same temperature, these terms cancel each other out. This simplifies our expression to:\[ \frac{v_{rms,1}}{v_{rms,2}} = \frac{\sqrt{m_2}}{\sqrt{m_1}} \] which is the inverse ratio of the square roots of the molecular masses. This shows that the rms speed of a gas is inversely proportional to the square root of its molecular mass.

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Key Concepts

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

Molecular Mass
Molecular mass is fundamentally important in understanding the behavior of gases. It represents the mass of a single molecule, measured in atomic mass units (amu). Each gas molecule possesses a distinct molecular mass depending on the types and numbers of atoms it contains.

For instance, oxygen, which is Oeon:21, has a higher molecular mass compared to hydrogen, which is Heon:2. This variation plays a critical role when calculating the root mean square (rms) speed of gases. In the formula for rms speed, \( v_{rms} = \sqrt{\frac{3kT}{m}} \), the molecular mass \( m \) is in the denominator under a square root.

This arrangement means that the rms speed is inversely proportional to the square root of the molecular mass. So, lighter molecules, like hydrogen, move faster compared to heavier molecules like oxygen at the same temperature.
Boltzmann Constant
The Boltzmann constant \( k \) is a vital link between the microscopic and macroscopic worlds of physics. It effectively acts as a bridge between the temperature kelvin and the energy scale.

Its value is approximately \( 1.38 \times 10^{-23} \) J/K, which is quite small! This smallness reflects how tiny energy changes are in the molecular world.

In the context of gases, the Boltzmann constant features prominently in the rms speed formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where temperature \( T \) is converted into kinetic energy per molecule. This allows for an understanding of how energetic molecules of a gas will be at a given temperature, revealing that more energy equates to faster-moving molecules.
Temperature
The concept of temperature is a measure of the average kinetic energy of the molecules in a substance. In the context of gases, it signifies how warm or cold they are.

Gases at higher temperatures have molecules that move faster compared to those at lower temperatures. When looking at the rms speed formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), temperature \( T \) is directly proportional to the square of the rms speed. Thus, as temperature increases, the rms speed increases as well.

Because temperature is usually measured in Kelvin \( K \) for gas-related calculations, it ensures that the values are non-negative and that they have a direct physical significance. Many physical properties such as pressure and volume are influenced by the temperature of a gas, making it a critical factor in thermodynamics.

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Most popular questions from this chapter

The second postulate of kinetic theory is that the molecules are, on the average, far apart from one another. That is, their average separation is much greater than the diameter of each molecule. Is this assumption reasonable? To check, calculate the average distance between molecules of a gas at STP, and compare it to the diameter of a typical gas molecule, about \(0.3 \mathrm{nm}\). If the molecules were the diameter of ping-pong balls, say \(4 \mathrm{~cm},\) how far away would the next ping-pong ball be on average?

(III) Starting from the Maxwell distribution of speeds, Eq. \(6,\) show \((a) \int_{0}^{\infty} f(v) d v=N,\) and \((b)\) $$\begin{array}{l}{\int_{0}^{\infty} v^{2} f(v) d v / N=3 k T / m} \\ {f(v)=4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{-\frac{1}{2} \frac{m v^{2}}{k T}}}\end{array}$$

(III) A 0.5-mol sample of \(\mathrm{O}_{2}\) gas is in a large cylinder with a movable piston on one end so it can be compressed. The initial volume is large enough that there is not a significant difference between the pressure given by the ideal gas law and that given by the van der Waals equation. As the gas is slowly compressed at constant temperature (use \(300 \mathrm{~K}\) ), at what volume does the van der Waals equation give a pressure that is \(5 \%\) different than the ideal gas law pressure? Let \(a=0.14 \mathrm{~N} \cdot \mathrm{m}^{4} / \mathrm{mol}^{2}\) and \(b=3.2 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{mol}\).

(II) Below a certain threshold pressure, the air molecules (0.3-nm diameter) within a research vacuum chamber are in the "collision-free regime," meaning that a particular air molecule is as likely to cross the container and collide first with the opposite wall, as it is to collide with another air molecule. Estimate the threshold pressure for a vacuum chamber of side \(1.0 \mathrm{~m}\) at \(20^{\circ} \mathrm{C}\).

(II) For oxygen gas the van der Waals constants are \(a=0.14 \mathrm{N} \cdot \mathrm{m}^{4} / \mathrm{mol}^{2}\) and \(b=3.2 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) Using these values, graph six curves of pressure vs. volume between \(V=2 \times 10^{-5} \mathrm{m}^{3}\) to \(2.0 \times 10^{-4} \mathrm{m}^{3},\) for 1 \(\mathrm{mol}\) of oxygen gas at temperatures of \(80 \mathrm{K}, 100 \mathrm{K}, 120 \mathrm{K}, 130 \mathrm{K}\) , \(150 \mathrm{K},\) and 170 \(\mathrm{K}\) . From the graphs determine approximately the critical temperature for oxygen.
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