/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 If the temperature of a gas is d... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 25: Problem 3

If the temperature of a gas is doubled, by how much is the root-mean-square speed of the molecules increased?

Short Answer

Expert verified
The RMS speed increases by a factor of \( \sqrt{2} \) when the temperature is doubled.

Step by step solution

01

Understand the relationship between RMS speed and temperature

The root-mean-square (RMS) speed of gas molecules 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 mass of a gas molecule. This shows that \( v_{rms} \) is directly proportional to the square root of the temperature.
02

Set up the relationship for initial and final RMS speeds

Let the initial temperature be \( T_1 \) and the corresponding RMS speed be \( v_{rms,1} = \sqrt{\frac{3kT_1}{m}} \). When the temperature is doubled, \( T_2 = 2T_1 \), the new RMS speed is \( v_{rms,2} = \sqrt{\frac{3k \cdot 2T_1}{m}} \).
03

Simplify the expression for the new RMS speed

Substitute \( T_2 = 2T_1 \) into the RMS speed formula for \( v_{rms,2} \), resulting in \[ v_{rms,2} = \sqrt{\frac{6kT_1}{m}} = \sqrt{2} \cdot \sqrt{\frac{3kT_1}{m}} = \sqrt{2} \cdot v_{rms,1}. \]
04

Analyze the increase factor

The factor by which the RMS speed increases is \( \sqrt{2} \). This implies that if the temperature of the gas is doubled, the RMS speed increases by a factor of \( \sqrt{2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 (\( v_{rms} \)) is a crucial concept in the kinetic theory of gases. It represents the speed of gas molecules in a given system, providing an average speed indicator influenced by the kinetic energy of the molecules.
To understand this better, the RMS speed formula is \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where:
  • \( k \) is the Boltzmann constant,
  • \( T \) is the temperature in Kelvin, and
  • \( m \) is the mass of a single gas molecule.

This equation shows that RMS speed depends on both the temperature of the gas and the mass of its molecules. A higher temperature or lower molecular mass will result in a higher RMS speed. Thus, RMS speed is directly proportional to the square root of the temperature, highlighting the effect temperature changes have on gas molecules' movements.
Boltzmann Constant
The Boltzmann constant (\( k \)) plays a fundamental role in the kinetic theory of gases. It serves as a bridge between macroscopic and microscopic physical quantities. This constant enables us to connect the thermal energy at the particle level to the temperature observed at the macro level.
The Boltzmann constant is defined as \( k = 1.38 \times 10^{-23} \) Joules per Kelvin and is instrumental in formulas involving energy, temperature, and molecular speed. Its presence in the RMS speed formula, \( v_{rms} = \sqrt{\frac{3kT}{m}} \), ensures that we can accurately relate temperature effects to the speed of particles.
In essence, the Boltzmann constant allows us to comprehend how microscopic processes impact observable phenomena and explains why molecules move faster as they gain thermal energy.
Temperature Dependence of Gas Speed
The temperature of a gas significantly influences the speed of its molecules. According to the kinetic theory of gases, as described by the RMS speed formula \( v_{rms} = \sqrt{\frac{3kT}{m}} \), there's a clear relationship between temperature and molecular speed.
At higher temperatures, gas molecules possess greater kinetic energy, resulting in increased speeds. For example, if the temperature of a gas is doubled, the RMS speed increases by a factor of \( \sqrt{2} \). This direct relationship highlights that the speed of gas molecules is proportional to the square root of its temperature.
Understanding the temperature dependence of gas speed is crucial because it helps predict how gases will behave under different thermal conditions. Additionally, it reveals how energy input into a gas affects particle speed, informing various applications ranging from engine dynamics to environmental modeling.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the root-mean-square speed of a nitrogen molecule at \(200 \mathrm{K}, 300 \mathrm{K}, 500 \mathrm{K}\) and \(1000 \mathrm{K}\).

We can use Equation 25.33 to calculate the probability that the \(x\) -component of the velocity of a molecule lies within some range. For example, show that the probability that \(-u_{x 0} \leq u_{x} \leq u_{x 0}\) is given by \\[ \begin{aligned} \operatorname{Prob}\left\\{-u_{x 0} \leq u_{x} \leq u_{x 0}\right\\} &=\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{1 / 2} \int_{-u_{x 0}}^{u_{x 0}} e^{-m u_{x}^{2} / 2 k_{\mathrm{B}} T} d u_{x} \\ &=2\left(\frac{m}{2 \pi k_{\mathrm{B}} T}\right)^{1 / 2} \int_{0}^{u_{x 0}} e^{-m u_{x}^{2} / 2 k_{\mathrm{B}} T} d u_{x} \end{aligned} \\] Now let \(m u_{x}^{2} / 2 k_{\mathrm{B}} T=w^{2}\) to get the cleaner-looking expression \\[ \operatorname{Prob}\left\\{-u_{x 0} \leq u_{x} \leq u_{x 0}\right\\}=\frac{2}{\pi^{1 / 2}} \int_{0}^{w_{0}} e^{-w^{2}} d w \\] where \(w_{0}=\left(m / 2 k_{\mathrm{B}} T\right)^{1 / 2} u_{x 0}\) It so happens that the above integral cannot be evaluated in terms of any function that we have encountered up to now. It is customary to express the integral in terms of a new function called the error function, which is defined by \\[ \operatorname{erf}(z)=\frac{2}{\pi^{1 / 2}} \int_{0}^{z} e^{-x^{2}} d x \\] The error function can be evaluated as a function of \(z\) by evaluating its defining integral numerically. Some values of erf(z) are $$\begin{array}{cccc} z & \operatorname{erf}(z) & z & \operatorname{erf}(z) \\ \hline 0.20 & 0.22270 & 1.20 & 0.91031 \\ 0.40 & 0.42839 & 1.40 & 0.95229 \\ 0.60 & 0.60386 & 1.60 & 0.97635 \\ 0.80 & 0.74210 & 1.80 & 0.98909 \\ 1.00 & 0.84270 & 2.00 & 0.99532 \end{array}$$ Now show that \\[ \operatorname{Prob}\left\\{-u_{x 0} \leq u_{x} \leq u_{x 0}\right\\}=\operatorname{erf}\left(w_{0}\right) \\] Calculate the probability that \(-\left(2 k_{\mathrm{B}} T / m\right)^{1 / 2} \leq u_{x} \leq\left(2 k_{\mathrm{B}} T / m\right)^{1 / 2} ?\)

Calculate the pressures at which the mean free path of a hydrogen molecule will be \(100 \mu \mathrm{m}, 1.00 \mathrm{mm},\) and \(1.00 \mathrm{m}\) at \(20^{\circ} \mathrm{C}\).

Arrange the following gases in order of increasing root-mean-square speed at the same temperature: \(\mathrm{O}_{2}, \mathrm{N}_{2}, \mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}, \mathrm{NO}_{2},^{235} \mathrm{UF}_{6},\) and \(^{238} \mathrm{UF}_{6}\).

Consider a mixture of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g})\). Calculate the ratio of the root-mean-square speed of \(\mathrm{H}_{2}(\mathrm{g})\) and \(\mathrm{I}_{2}(\mathrm{g})\) molecules in the reaction mixture.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.