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Chapter 18: Problem 30

A flask contains a mixture of neon (Ne), krypton (Kr), and radon (Rn) gases. Compare (a) the average kinetic energies of the three types of atoms and; (b) the root-mean-square speeds. (Hint: The periodic table in Appendix D shows the molar mass (in g/mol) of each element under the chemical symbol for that element)

Short Answer

Expert verified
(a) The average kinetic energies are equal for Ne, Kr, and Rn. (b) Ne has the highest rms speed, followed by Kr, and then Rn.

Step by step solution

01

Understand Molecular Kinetic Energy

The average kinetic energy of gas particles is given by the equation \( KE = \frac{3}{2} k T \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. This shows that average kinetic energy depends only on the temperature, not the type of gas.
02

Compare Average Kinetic Energies

Since the average kinetic energy depends only on temperature (and temperature is the same for all gases in the flask), the average kinetic energy is the same for Ne, Kr, and Rn.
03

Understand Root-Mean-Square Speed Formula

Root-mean-square speed is given by the equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( m \) is the molar mass of the gas. This means it depends on both the temperature and the molar mass of the gas.
04

Calculate Root-Mean-Square Speeds

The root-mean-square speed inversely depends on the square root of the molar mass of each gas. Using the molar masses from the periodic table (Ne: 20.18 g/mol, Kr: 83.80 g/mol, Rn: 222 g/mol), calculate \( v_{rms} \) for each: Ne will have the highest \( v_{rms} \), followed by Kr, and Rn will have the lowest.

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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 an important concept in understanding gas behavior. It provides us with an average measure of the speed at which gas particles move in a system. It is derived from the formula: \[ v_{rms} = \sqrt{\frac{3kT}{m}} \] where \( k \)is the Boltzmann constant, \( T \)represents the temperature in Kelvin, and \( m \)is the molar mass of the gas.
  • The term \( \sqrt{3kT} \)reflects the influence of temperature. Higher temperature means faster particle speeds.
  • The denominator expresses the impact of the molar mass, indicating that lighter gases have higher root-mean-square speeds.
To practically assess this in a gas mixture, one would compare the given molar masses of the gases and their effect on \( v_{rms} \). In the example provided, neon (Ne), with the smallest molar mass, moves fastest. Radon (Rn), being the heaviest, has the slowest \( v_{rms} \).
Molar Mass
Molar mass is a key factor when dealing with gas calculations, influencing properties such as root-mean-square speed. Molar mass, characterized by the symbol \( m \), is the mass of one mole of a substance, commonly expressed in grams per mole (g/mol).
  • It essentially tells us how heavy or light a particular molecule is and thus plays a significant role in kinetic computations like \( v_{rms} \).
  • Heavier particles, due to their higher molar mass, tend to move slower compared to lighter ones at the same temperature.
For the gases in our flask鈥攏eon, krypton, and radon鈥攖he molar masses are 20.18 g/mol, 83.80 g/mol, and 222 g/mol respectively. These values guide us in predicting physical behaviors like speed, where neon, being the lightest, moves quickest while radon trails behind due to its substantial mass.
Boltzmann Constant
The Boltzmann constant (\( k \)) is a fundamental constant utilized in statistical mechanics and thermodynamics. Its value, approximately \( 1.38 \times 10^{-23} \text{ J/K} \), connects the macroscopic and microscopic worlds, converting temperature energy into kinetic energy.
  • It allows us to translate the temperature of a system into the kinetic energy of particles, a central concept in gas behavior models.
  • In the formula for average kinetic energy \( KE = \frac{3}{2} k T \), it clarifies how temperature affects the movement of particles.
The presence of \( k \) in both kinetic energy and root-mean-square speed equations underscores its pivotal role in determining the dynamic properties of gases, independent of their chemical identity.

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Most popular questions from this chapter

(a) Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 \(\mathrm{K}\) . (b) Calculate the moment of inertia of an oxygen molecule \(\left(\mathrm{O}_{2}\right)\) for rotation about either the \(y\) - or \(z\) -axis shown in Fig. 18.18 . Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance of \(1.21 \times 10^{-10} \mathrm{m}\) . The molar mass of oxygen atoms is \(16.0 \mathrm{g} / \mathrm{mol} .\) (c) Find the rms angular velocity of rotation of an oxygen molecule about either the \(y\) - or \(z\) -axis shown in Fig. 18.15. How does your answer compare to the angular velocity of a typical piece of rapidly rotating machinery \((10,000 \mathrm{rev} / \mathrm{min}) ?\)

Puffy cumulus clouds, which are made of water droplets, occur at lower altitudes in the atmosphere. Wispy cirrus clouds, which are made of ice crystals, occur only at higher altitudes. Find the altitude \(y\) (measured from sea level) above which only cirrus clouds can occur. On a typical day and at altitudes less than 11 \(\mathrm{km}\), the temperature at an altitude \(y\) is given by \(T=T_{0}-\alpha y,\) where \(T_{0}=15.0^{\circ} \mathrm{C}\) and \(\alpha=6.0 \mathrm{C}^{\circ} / 1000 \mathrm{m} .\)

A diver observes a bubble of air rising from the bottom of a lake (where the absolute pressure is 3.50 atm) to the surface (where the pressure is 1.00 atm). The temperature at the bottom is \(4.0^{\circ} \mathrm{C},\) and the temperature at the surface is \(23.0^{\circ} \mathrm{C}\) , (a) What is the ratio of the volume of the bubble as it reaches the surface to its volume at the bottom? (b) Would it be safe for the diver to hold his breath while ascending from the bottom of the lake to the surface? Why or why not?

(a) A process called gaseous diffusion is often used to separate isotopes of uranium-that is, atoms of the elements that have different inasses, such as \(^{235} \mathrm{U}\) and \(^{238} \mathrm{U}\) . The only gaseous compound of uranium at ordinary temperatures is uranium hexafluoride, \(\mathrm{UF}_{6}\) . Speculate on how 235 \(\mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6}\) inolecules might be separated by diffusion. (b) The molar masses for \(^{235} \mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6}\) molecules are 0.349 \(\mathrm{kg} / \mathrm{mol}\) and \(0.352 \mathrm{kg} / \mathrm{nol},\) respectively. If uranium hexafluoride acts as an ideal gas, what is the ratio of the root-mean- square speed of \(^{235} \mathrm{UF}_{6}\) molecules to that of \(^{238} \mathrm{UF}_{6}\) molecules if the temperature is uniform?

Atmosphere of Titan. Titan, the largest satellite of Saturn, has a thick nitrogen atmosphere. At its surface, the pressure is 1.5 Earth-atmospheres and the temperature is 94 \(\mathrm{K}\) . (a) What is the surface temperature in \(^{\circ} \mathrm{C}\) ? (b) Calculate the surface density in Titar's atmosphere in molecules per cubic meter (c) Compare the density of Titan's surface atmosphere to the density of Earth's atmosphere at \(22^{\circ} \mathrm{C}\) . Which body has denser atmosphere?
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