/*! 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 29 "The density of nitrogen gas in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 29

"The density of nitrogen gas in a container at \(300 \mathrm{K}\) and 1.0 bar pressure is \(1.25 \mathrm{g} \mathrm{dm}^{-3}\) (Section 8.5 ). (a) Calculate the rms speed of the molecules. (b) At what the temperature will the ms speed be twice as fast?

Short Answer

Expert verified
(a) rms speed is approximately 517 m/s. (b) Temperature required is 1200 K.

Step by step solution

01

Understand Root Mean Square (rms) Speed Formula

The root mean square speed, or rms speed (\(v_{rms}\)), of gas molecules can be determined using the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where \(R\) is the ideal gas constant (8.314 J/mol K), \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kg/mol. To find \(M\) for nitrogen (N\(_2\)), we need the atomic masses: N has an atomic mass of about 14 g/mol, so \(M = 28 \times 10^{-3} \text{ kg/mol}\).
02

Calculate rms Speed at 300 K

Using the rms formula with \(T = 300 \, \mathrm{K}\) and substituting the other known values: \[ v_{rms} = \sqrt{\frac{3 \times 8.314 \, \mathrm{J/mol \, K} \times 300 \, \mathrm{K}}{28 \times 10^{-3} \, \mathrm{kg/mol}}} \] Perform the calculation inside the square root: \[ = \sqrt{\frac{7483.2}{0.028}} = \sqrt{267257.14} \] So, \[ v_{rms} \approx 517 \text{ m/s} \].
03

Understand Relationship for ms Speed Doubling

To find the temperature at which the ms speed is double, recognize that the ms speed (mean speed \(v\)) is related to temperature in a similar way to the rms speed. Thus, if \(v' = 2v\), we use: \[ v'_{rms} = \sqrt{\frac{3RT'}{M}} \] And set up the equation based on doubling condition: \[ 2 \times \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3RT'}{M}} \]Squaring both sides: \[ 4 \times \frac{3RT}{M} = \frac{3RT'}{M} \] Simplify to: \[ 4T = T' \].
04

Calculate New Temperature

Given \(T = 300 \, \mathrm{K}\), solve for \(T'\): \[ T' = 4 \times 300 \, \mathrm{K} = 1200 \, \mathrm{K} \]. So, the new temperature at which the ms speed is twice as fast is \(1200 \, \mathrm{K}\).

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.

Ideal Gas Constant
The ideal gas constant, denoted as \( R \), is a key parameter in understanding the behavior of gases. It's a universal constant that appears in many fundamental equations in thermodynamics and physical chemistry. The value of \( R \) is 8.314 J/mol K. This constant plays a critical role in various equations, including the ideal gas law and the root mean square speed formula.
\( R \) serves as a bridge between different units of measurement, such as pressure, volume, and temperature. By using the ideal gas constant, it becomes easier to relate these variables when analyzing the behavior of gases under different conditions.
Molar Mass
Molar mass is the mass of one mole of a substance, generally represented in g/mol or kg/mol. For nitrogen gas (\( N_2 \)), it's crucial to know the molar mass to compute the root mean square speed of its molecules.
Since nitrogen is a diatomic molecule consisting of two nitrogen atoms, you need to consider the individual atomic masses. Each nitrogen atom has an atomic mass of approximately 14 g/mol, so for nitrogen gas, the molar mass becomes 28 g/mol, or 0.028 kg/mol when converted for the formula.
  • Understanding molar mass helps in calculations involving gas properties and behaviors.
  • It allows conversion of mass quantities into moles, providing a link between the microscopic and macroscopic world.
Temperature
Temperature is a fundamental variable in thermodynamics that quantifies the average kinetic energy of particles in a substance. In the context of gases, temperature is often measured in Kelvin (K).
The root mean square speed formula relates the temperature of gas molecules to their average speed. This relationship shows that increasing the temperature will increase the kinetic energy and thus the speed of the molecules.
  • Temperature provides a direct measure of energy in a gas.
  • Understanding temperature helps in predicting how changes in conditions will affect the behavior of gas molecules.
Temperature changes influence many properties such as pressure, volume, and speed in gas calculations.
Nitrogen Gas
Nitrogen gas, denoted as \( N_2 \), is one of the most abundant gases in Earth's atmosphere. It accounts for about 78% of the Earth's air.
Being a diatomic molecule, nitrogen gas consists of two nitrogen atoms bonded together. This composition plays a role in its physical properties. When calculating the root mean square speed of nitrogen gas molecules, it's important to consider its molar mass and behavior as an inert gas.
  • Inert by nature, nitrogen doesn't easily react with other elements, allowing it to serve as a common standard in experiments.
  • Its abundance makes it significant in studies related to atmospheric and environmental sciences.
Root Mean Square
The root mean square (rms) speed is a measure of the speed of particles in a gas. It gives a sense of the average kinetic speed of gas molecules.
The formula used to calculate rms speed is \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the ideal gas constant, \( T \) is the temperature, and \( M \) is the molar mass. The rms speed incorporates all these variables to provide an accurate depiction of how fast the molecules are moving.
  • The rms speed offers insights into the energy distribution among molecules at a given temperature.
  • Changes in rms speed indicate changes in kinetic energy and potentially the physical state of the gas.
  • It's particularly useful in comparing the behavior of different gases under the same conditions.
Understanding the rms speed is essential for grasping how gas molecules respond to changes in environmental conditions.

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

A \(10 \mathrm{dm}^{3}\) SCUBA cylinder is filled with air to a pressure of \(\left.300 \text { atm at a temperature of } 20^{\circ} \mathrm{C}(293 \mathrm{K}) . \text { (Section } 8.6\right)\) (a) Calculate the amount in moles of gas in the cylinder, assuming the air behaves as an ideal gas. (b) When the diver jumps into cold water at \(278 \mathrm{K}\), the pressure gauge shows an alarming drop in pressure. Explain the reason why and calculate the new pressure inside the cylinder. (c) In fact, the compressed gases do not behave as ideal gases. Explain why. Use the van der Waals equation (for air, \(a=0.137 \mathrm{Pam}^{6} \mathrm{mol}^{-2}\) and \(b=3.7 \times 10^{-5} \mathrm{m}^{3} \mathrm{mol}^{-1}\) ) to show that the amount of air in the cylinder is 115 mol. In view of your answer to part (a) above, what are the implications of this for divers?

A sample of gas has a volume of \(346 \mathrm{cm}^{3}\) at \(25^{\circ} \mathrm{C}\) when the pressure is 1.00 atm. What volume will it occupy if the conditions are changed to \(35^{\circ} \mathrm{C}\) and 1.25 atm? (Section 8.2)

A mixture of nitrogen and carbon dioxide contains \(38.4 \% \mathrm{N}_{2}\) by mass. What is the mole fraction of nitrogen in the mixture? If the total pressure is 1.2 atm, what is the partial pressure of each gas in Pa? (Section 8.3)

Divers' "bends' are caused by the formation of bubbles of nitrogen in blood as the solubility reduces when the diver returns to the surface. The solubility of nitrogen in water at 1.00 atm pressure is \(13.0 \mathrm{mg} \mathrm{kg}^{-1}\) at body temperature of \(37^{\circ} \mathrm{C}\) and increases linearly with pressure. In water, the pressure increases at the rate of 1.00 atm per \(10 \mathrm{m}\) depth. Estimate the volume of gas that comes out of solution when a diver who has \(4.5 \mathrm{kg}\) of blood rapidly ascends from a depth of \(50 \mathrm{m}\) of water to the surface. Assume the solubility of nitrogen in blood is the same as in water. (Section 8.2)

A vessel of volume \(50.0 \mathrm{dm}^{3}\) contains \(2.50 \mathrm{mol}\) of argon and \(1.20 \mathrm{mol}\) of nitrogen at \(273.15 \mathrm{K}\) (i) Calculate the partial pressure in bar of each gas. (ii) Calculate the total pressure in bar. (iii) How many additional moles of nitrogen must be pumped into the vessel in order to raise the pressure to 5 bar? (Sections \(8.2 \text { and } 8.3)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic 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.