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At what temperature is the root-mean-square speed of nitrogen molecules equal to the root-mean-square speed of hydrogen molecules at \(20.0^{\circ} \mathrm{C} ?\) (Hint The periodic table in Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element. The molar mass of \(\mathrm{H}_{2}\) is twice the molar mass of hydrogen atoms, and similarly for \(\mathrm{N}_{2}\) .)

Short Answer

Expert verified
The temperature for nitrogen is approximately 4067.08 K.

Step by step solution

01

Understand the Root-Mean-Square Speed Formula

The root-mean-square 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 molecule.
02

Relate Speeds of Nitrogen and Hydrogen

Since we want the root-mean-square speeds to be equal, set the equations for nitrogen \( v_{rms,N_2} \) and hydrogen \( v_{rms,H_2} \) equal: \( \sqrt{\frac{3kT_N}{m_N}} = \sqrt{\frac{3kT_H}{m_H}} \). This simplifies to \( \frac{T_N}{m_N} = \frac{T_H}{m_H} \).
03

Determine Mass of Nitrogen and Hydrogen Molecules

From the periodic table, the molar mass of hydrogen \( H \) is approximately \( 1.008 \, \text{g/mol} \) making \( H_2 \) \( 2.016 \, \text{g/mol} \). The molar mass of nitrogen \( N \) is approximately \( 14.007 \, \text{g/mol} \), making \( N_2 \) \( 28.014 \, \text{g/mol} \).
04

Solve for Temperature of Nitrogen

Given the temperature of hydrogen is \( 20^{\circ}\text{C} = 293.15 \text{K} \), use the relation \( \frac{T_N}{28.014} = \frac{293.15}{2.016} \). Solving for \( T_N \), multiply through: \( T_N = 28.014 \times \frac{293.15}{2.016} \approx 4067.08 \text{K} \).
05

Verify Solution

Double-check calculations for any arithmetic mistakes: \( T_N = 28.014 \times 145.312 \approx 4067.08 \text{K} \). Given the physical correctness and dimensions check out, this verifies the temperature result.

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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 (commonly abbreviated as \( v_{rms} \)) is a concept used in the kinetic theory of gases. It represents the average speed of particles within a gas. The formula to calculate \( v_{rms} \) is given by:\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]
This formula tells us that the root-mean-square speed depends on three primary factors:
  • \( k \): Boltzmann constant, which relates the average kinetic energy of particles in a gas with the temperature.
  • \( T \): Temperature of the gas in Kelvin. Temperature affects how fast the molecules are moving; higher temperatures mean faster molecules.
  • \( m \): Molecular mass, or the mass of a single molecule. Heavier molecules move slower at the same temperature.
Understanding \( v_{rms} \) helps us analyze how gases behave under different conditions, particularly in terms of their speed and energy.
Molecular Mass
When dealing with gases, knowing their molecular mass is crucial because it greatly influences the root-mean-square speed. Molecular masses are typically found using the periodic table.

Molecular mass of a molecule is the sum of the masses of its constituent atoms. For instance, the molecular mass of hydrogen \( H_2 \) is 2.016 g/mol because hydrogen atoms weigh around 1.008 g/mol each. Similarly, the molecular mass of nitrogen \( N_2 \) is 28.014 g/mol, as nitrogen atoms are about 14.007 g/mol.

In calculations, these masses are usually converted into kilograms for consistency in the International System of Units (SI). It’s important to accurately use these values, as they precisely determine the speed calculations of different gases.
Temperature Conversion
In physics, especially when working with the kinetic theory of gases, temperatures are commonly expressed in Kelvin. This is because Kelvin is an absolute temperature scale, translating more directly into energy and speed calculations.

To convert from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, a temperature of \( 20.0^{\circ} \text{C} \) converts to \( 293.15 \text{K} \).
This conversion is vital, as using Celsius directly in equations like that for \( v_{rms} \) would lead to incorrect results due to the need for an absolute scale where zero corresponds to absolute zero or no molecular movement.
Boltzmann Constant
The Boltzmann constant \( (k) \) is a fundamental constant in physics that provides a bridge between macroscopic and microscopic physical quantities. It relates the temperature of a gas to the average kinetic energy of its particles.

The value of the Boltzmann constant is approximately \( 1.38 \times 10^{-23} \text{J/K} \). This means for every degree increase in temperature, each molecule's kinetic energy increases by this tiny amount times the temperature in Kelvin.
  • Enables calculations concerning thermal energy at the particle level.
  • Helps determine the speed and behavior of gas molecules as they relate to temperature.
Understanding and employing \( k \) allows us to apply kinetic theory accurately to describe and predict the behavior of gases in various environments.

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Most popular questions from this chapter

The speed of propagation of a sound wave in air at \(27^{\circ} \mathrm{C}\) is about 350 \(\mathrm{m} / \mathrm{s}\) . Calculate, for comparison, (a) \(v_{\mathrm{ms}}\) for nitrogen molecules and (b) the rms value of \(v_{x}\) at this temperature. The molar mass of nitrogen \(\left(\mathrm{N}_{2}\right)\) is \(28.0 \mathrm{g} / \mathrm{mol} .\)

The surface of the sun has a temperature of about 5800 \(\mathrm{K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is 1.67 \(\times 10^{-27} \mathrm{kg} . )\) (b) The escape speed for a particle to leave the gravitational influence of the sun is given by \((2 G M / R)^{1 / 2}\) , where \(M\) is the sun's mass, \(R\) its radius, and \(G\) the gravitational constant (see Example 12.5 of Section \(12.3 ) .\) Use the data in Appendix \(\mathrm{F}\) to calculate this escape speed. (c) Can appreciable quantities of hydrogen escape from the sun? Can any hydrogen escape? Explain.

(a) Compute the specific heat capacity at constant volume of nitrogen \(\left(\mathrm{N}_{2}\right)\) gas, and compare with the specific heat capacity of liquid water. The molar mass of \(\mathrm{N}_{2}\) is 28.0 \(\mathrm{g} / \mathrm{mol}\) . (b) You warm 1.00 \(\mathrm{kg}\) of water at a constant volume of 1.00 \(\mathrm{L}\) from \(20.0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C}\) in a kettle. For the same amount of beat, how many kilograms of \(20.0^{\circ} \mathrm{C}\) air would you be able to warm to \(30.0^{\circ} \mathrm{C} ?\) What volume (in liters) would this air occupy at \(20.0^{\circ} \mathrm{C}\) and a pressure of 1.00 atm? Make the simplifying assumption that air is 100\(\% \mathrm{N}_{2}\) .

It is possible to make crystalline solids that are only one layer of atoms thick. Such "two-dimensional" crystals can be created by depositing atoms on a very flat surface. (a) If the atoms in such a two-dimensional crystal can move only within the plane of the crystal, what will be its molar heat capacity near room temperature? Give your answer as a multiple of \(R\) and in \(\mathrm{J} / \mathrm{nol} \cdot \mathrm{K}\) . (b) At very low temperatures, will the molar heat capacity of a two- dimensional crystal be greater than, less than, or equal to the result you found in part (a)? Explain why.

A balloon whose volume is 750 \(\mathrm{m}^{3}\) is to be filled with hydrogen at atmospheric pressure \(\left(1.01 \times 10^{5} \mathrm{Pa}\right) .\) (a) If the hydrogen is stored in cylinders with volumes of 1.90 \(\mathrm{m}^{3}\) at a gauge pressure of \(1.20 \times 10^{6} \mathrm{Pa},\) how many cylinders are required? Assume that the temperature of the hydrogen remains constant. (b) What is the total weight (in addition to the weight of the gas) that can be supported by the balloon if the gas in the balloon and the surrounding air are both at \(15.0^{\circ} \mathrm{C}\) ? The molar mass of hydro\(\operatorname{gen}\left(\mathrm{H}_{2}\right)\) is 2.02 \(\mathrm{g} / \mathrm{mol}\) . The density of air at \(15.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.23 \(\mathrm{kg} / \mathrm{m}^{3}\) . See Chapter 14 for a discussion of buoyancy. (c) What weight could be supported if the balloon were filled with helium (molar mass 4.00 \(\mathrm{g} / \mathrm{mol}\) ) instead of hydrogen, again at \(15.0^{\circ} \mathrm{C} ?\)

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