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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 C shows the molar mass (in \(\mathrm{g} / \mathrm{mol}\) ) 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 is 4102 K.

Step by step solution

01

Understand the Problem

We need to find the temperature at which the root-mean-square (rms) speed of nitrogen ( 2) molecules is equal to the rms speed of hydrogen ( h2) molecules at 20.0°C.
02

RMS Speed Formula

The formula for the root-mean-square speed is given by \(v_{rms} = \sqrt{\frac{3kT}{m}}\) or \(v_{rms} = \sqrt{\frac{3RT}{M}}\), where \(k\) is Boltzmann's constant, \(R\) is the gas constant, \(T\) is the temperature, \(m\) is the molecular mass, and \(M\) is the molar mass.
03

Given Conditions

At T = 20.0°C (or 293 K), we need to equate the rms speed of \(N_2\) with the rms speed of \(H_2\). Using the formula \(v_{rms} = \sqrt{\frac{3RT}{M}}\), we need to set the two expressions equal: \(\sqrt{\frac{3RT_{N_2}}{M_{N_2}}} = \sqrt{\frac{3RT_{H_2}}{M_{H_2}}}\).
04

Simplify the Equation

Square both sides to get \(\frac{3RT_{N_2}}{M_{N_2}} = \frac{3R(293)}{M_{H_2}}\). The \(3R\) terms cancel out, leaving us with \(\frac{T_{N_2}}{M_{N_2}} = \frac{293}{M_{H_2}}\).
05

Determine Molar Masses

The molar mass of \(H_2\) is 2 g/mol, and the molar mass of \(N_2\) is 28 g/mol. Substitute these values into the equation obtained: \(\frac{T_{N_2}}{28} = \frac{293}{2}\).
06

Solve for the Temperature of Nitrogen

Solve the equation \(\frac{T_{N_2}}{28} = \frac{293}{2}\) by multiplying both sides by 28: \(T_{N_2} = \frac{293 \times 28}{2} = 4102\) K.

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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 (\(v_{rms}\)) is a measure of the speed of particles in a gas. It provides an idea of how fast molecules are moving within a given sample.
The concept of root-mean-square speed is vital in the kinetic theory of gases, which explains the behavior of gases in terms of particle motion. To calculate it, we use the formula:\[v_{rms} = \sqrt{\frac{3RT}{M}}\]where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas.
This formula highlights that the speed of gas molecules is dependent on both the temperature and the mass. Lighter molecules and higher temperatures lead to higher speeds. This is because lighter molecules require less energy to increase their speed. Similarly, higher temperatures suggest more kinetic energy available to the molecules.
Molar Mass
Molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It is usually expressed in units of \(g/mol\).
In the context of gases, knowing the molar mass is crucial for computing the root-mean-square speed. When comparing two gases, as in the exercise with nitrogen (\(N_2\)) and hydrogen (\(H_2\)), the molar mass helps determine at what conditions similar speeds are achieved. For instance, the molar mass of hydrogen gas (\(H_2\)) is approximately 2 \(g/mol\), as hydrogen has an atomic mass of 1 for each atom, and there are two atoms in a hydrogen molecule.
Similarly, nitrogen molecules (\(N_2\)) have a molar mass of around 28 \(g/mol\) because nitrogen has an atomic mass of 14, and the nitrogen molecule consists of two atoms.
Understanding the molar mass is essential for calculating how temperature variations affect the speeds of different gases, allowing for the comparative analysis of molecular motion.
Temperature Conversion
Temperature conversion is an important step when dealing with gas laws and kinetic molecular theory. Different scientific contexts require temperatures in different units, typically Celsius or Kelvin.
For any kinetic calculations involving gases, temperature must be in Kelvin. This is because Kelvin is an absolute scale where 0 K represents absolute zero—the point at which molecular motion theoretically stops.
The conversion from Celsius (\(^{\circ}C\)) to Kelvin (\(K\)) is simple:
  • Add 273.15 to the Celsius temperature.
For the given problem, the hydrogen molecules are initially at 20°C, which is converted to:\[20 + 273.15 = 293.15\,\]approximated as 293 K for ease of calculation.
Accurate temperature conversion ensures that computations of gas properties, like the root-mean-square speed, are based on consistent and correct units.
Molecular Mass
Molecular mass refers to the sum of the atomic masses of all atoms in a molecule. Unlike molar mass, which is expressed in grams per mole, molecular mass is measured in atomic mass units (amu).
To find the molecular mass of a gas, one simply adds up the atomic masses of all the atoms in the molecule, as listed in the periodic table. For instance, in a nitrogen molecule (\(N_2\)), each nitrogen atom has an atomic mass of approximately 14 amu, resulting in a total molecular mass of about 28 amu for the molecule.
Similarly, a hydrogen molecule (\(H_2\)) has 2 atoms of hydrogen, each with an atomic mass of about 1 amu, totaling approximately 2 amu for the molecule. Although molecular mass and molar mass are related, it is important to distinguish between the two.
  • Molecular mass gives an idea of the individual molecule's mass.
  • Molar mass provides information on the mass of a mole of these molecules and assists in converting between microscopic and macroscopic quantities.
Understanding molecular mass is fundamental when working with equations and concepts in gas theory, ensuring clear and accurate molecular characterizations.

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Most popular questions from this chapter

(a) How much heat does it take to increase the temperature of 2.50 moles of an ideal monatomic gas from \(25.0^{\circ} \mathrm{C}\) to \(55.0^{\circ} \mathrm{C}\) if the gas is held at constant volume? (b) How much heat is needed if the gas is diatomic rather than monatomic? (c) Sketch a \(p V\) diagram for these processes.

A \(20.0 \mathrm{~L}\) tank contains \(0.225 \mathrm{~kg}\) of helium at \(18.0^{\circ} \mathrm{C}\). The molar mass of helium is \(4.00 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of helium are in the tank? (b) What is the pressure in the tank, in pascals and in atmospheres?

If deep-sea divers rise to the surface too quickly, nitrogen bubbles in their blood can expand and prove fatal. This phenomenon is known as the bends. If a scuba diver rises quickly from a depth of \(25 \mathrm{~m}\) in Lake Michigan (which is freshwater), what will be the volume at the surface of an \(\mathrm{N}_{2}\) bubble that occupied \(1.0 \mathrm{~mm}^{3}\) in his blood at the lower depth? Does it seem that this difference is large enough to be a problem? (Assume that the pressure difference is due only to the changing water pressure, not to any temperature difference, an assumption that is reasonable, since we are warm-blooded creatures.)

We have two equal-size boxes, \(A\) and \(B\). Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at a temperature of \(50^{\circ} \mathrm{C}\) while the gas in box \(B\) is at \(10^{\circ} \mathrm{C}\). This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? (a) The pressure in \(A\) is higher than in \(B\). (b) There are more molecules in \(A\) than in \(B\). (c) \(A\) and \(B\) cannot contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\). (e) The molecules in \(A\) are moving faster than those in \(B\). Explain the reasoning behind your answers.

The gas inside a balloon will always have a pressure nearly equal to atmospheric pressure, since that is the pressure applied to the outside of the balloon. You fill a balloon with helium (a nearly ideal gas) to a volume of \(0.600 \mathrm{~L}\) at a temperature of \(19.0^{\circ} \mathrm{C}\). What is the volume of the balloon if you cool it to the boiling point of liquid nitrogen \((77.3 \mathrm{~K}) ?\)

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