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Chapter 19: Problem 40

Two containers are at the same temperature. The first contains gas with pressure \(p_{1}\), molecular mass \(m_{1}\), and \(\mathrm{rms}\) speed \(v_{\mathrm{rms} 1}\). The second contains gas with pressure \(2.0 p_{1}\), molecular mass \(m_{2}\), and average speed \(v_{\text {avg2 }}=2.0 v_{\text {rms } 1} .\) Find the mass ratio \(m_{1} / m_{2}\).

Short Answer

Expert verified
The mass ratio \( \frac{m_1}{m_2} = \frac{3\pi}{2} \).

Step by step solution

01

Understand given parameters

You are given two gases at the same temperature: 1. Gas 1 with pressure \( p_1 \), molecular mass \( m_1 \), and \( v_{\mathrm{rms}1} \). 2. Gas 2 with pressure \( 2p_1 \), molecular mass \( m_2 \), and average speed \( v_{\text{avg2}} = 2v_{\mathrm{rms}1} \).
02

Define formulas for rms speed and average speed

For any gas, the root mean square (rms) speed \( v_{\mathrm{rms}} \) is given by:\[ v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} \]Where \( k \) is the Boltzmann constant, \( T \) is temperature, and \( m \) is molecular mass.The average speed \( v_{\text{avg}} \) is given by:\[ v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} \]
03

Relate speeds of the gases

Since the two gases are at the same temperature and we are given \( v_{\text{avg2}} = 2v_{\mathrm{rms}1} \),\[ \sqrt{\frac{8kT}{\pi m_2}} = 2\sqrt{\frac{3kT}{m_1}} \]
04

Equate to find ratio

Square both sides to eliminate the square roots:\[ \frac{8kT}{\pi m_2} = 4 \times \frac{3kT}{m_1} \]Cancel out \( kT \) from both sides:\[ \frac{8}{\pi m_2} = \frac{12}{m_1} \]
05

Solve for \( m_1/m_2 \)

Cross-multiply and rearrange the terms to find:\[ m_1 = \frac{12}{8/\pi} m_2 \]This simplifies to:\[ \frac{m_1}{m_2} = \frac{12\pi}{8} \]Thus,\( \frac{m_1}{m_2} = \frac{3\pi}{2} \).

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Key Concepts

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

rms speed
The root mean square (rms) speed of molecules in a gas provides insight into the energy and temperature of the gas. It's a type of average velocity based on the squares of the individual molecular speeds.
  • For any gas, the rms speed is given by the formula: \[ v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}} \]
  • In this equation, \( k \) represents the Boltzmann constant, \( T \) is the absolute temperature, and \( m \) is the molecular mass of the gas molecule.
  • It's important to note that the rms speed enables us to understand how fast, on average, molecules are moving in a given state of a gas. This speed increases with temperature because molecules gain kinetic energy.
Breaking down the formula, the \( 3kT \) part accounts for the kinetic energy of translational motion in three dimensions, assuming perfect gas behavior. The mass in the denominator inversely affects the rms speed, indicating heavier molecules move slower at the same energy level.
average speed
Average speed in the context of gases is another way to measure molecular motion, though it differs slightly from rms speed. The average speed of gas molecules is given by: \[ v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}} \]
  • Similar to rms speed, \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) indicates the molecular mass.
  • This equation highlights that average speed is also influenced by temperature and molecular mass; however, the factor \( 8/\pi \) highlights the differentiation between the overarching patterns observed in molecular behavior compared to rms speed.
  • The average speed is often slightly lower than the rms speed and provides another perspective on the spread and distribution of molecular velocities.
Like rms speed, average speed is crucial for deriving statistical properties and understanding physical phenomena related to gases. It simplifies calculations in thermodynamics, especially when dealing with large numbers of particles, like those in gases.
molecular mass
Molecular mass is essential in understanding gas behavior, as it influences both rms speed and average speed. Here鈥檚 why molecular mass matters:
  • Represented as \( m \) in the speed formulas, molecular mass defines how much a single molecule weighs.
  • In gases, as molecular mass increases, the speed of individual molecules, both rms and average speeds, decreases when comparing at the same temperature.
  • Heavier molecules have more inertia, meaning they move slower for the same given kinetic energy.
  • Understanding molecular mass is crucial in calculating gas density, pressure, and rate of diffusion.
In the context of gases, identifying molecular mass allows students to understand how molecular properties influence other gas laws, such as the Ideal Gas Law and Graham's Law of Effusion. Through these concepts, we link microscopic particle behavior to macroscopic measurable properties.
Boltzmann constant
The Boltzmann constant, denoted as \( k \), is a fundamental constant that connects macroscopic and microscopic ideas in thermodynamics. It is central to our understanding of kinetic theory:
  • Boltzmann constant is a proportionality factor that relates the average kinetic energy of particles in a gas with the temperature of the gas.
  • It is expressed as a constant value: \( k = 1.38 \times 10^{-23} \text{ J/K} \).
  • In both the rms speed and average speed formulas, \( k \) translates temperature from a macroscopic scale (Kelvin) to an energy scale (Joules), which works at the molecular level.
  • The constant also appears in the Boltzmann distribution law, which explains the distribution of speeds in a gas.
The Boltzmann constant is fundamental in statistical mechanics, allowing us to make sense of thermal motion and providing insights into the inherent "randomness" of particles at a given temperature. Understanding \( k \) helps link how gases behave at different temperatures, a crucial aspect of studying thermodynamic systems.

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Most popular questions from this chapter

At what frequency do molecules (diameter \(290 \mathrm{pm}\) ) collide in (an ideal) oxygen gas \(\left(\mathrm{O}_{2}\right)\) at temperature \(400 \mathrm{~K}\) and pressure \(2.00\) atm?

The mean free path of nitrogen molecules at \(0.0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\) is \(0.80 \times 10^{-5} \mathrm{~cm}\). At this temperature and pressure there are \(2.7 \times\) \(10^{19}\) molecules/cm \(^{3}\). What is the molecular diameter?

A quantity of ideal gas at \(10.0^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) occupies a volume of \(2.50 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? (b) If the pressure is now raised to \(300 \mathrm{kPa}\) and the temperature is raised to \(30.0^{\circ} \mathrm{C}\), how much volume does the gas occupy? Assume no leaks.

What is the internal energy of \(1.0\) mol of an ideal monatomic gas at \(273 \mathrm{~K}\) ?

What is the average translational kinetic energy of nitrogen molecules at \(1600 \mathrm{~K} ?\)
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