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Chapter 14: Problem 55

Interactive Solution \(\underline{14.55}\) at provides a model for problems of this type. The temperature near the surface of the earth is \(291 \mathrm{~K}\). A xenon atom (atomic mass \(=131.29 \mathrm{u}\) ) has a kinetic energy equal to the average translational kinetic energy and is moving straight up. If the atom does not collide with any other atoms or molecules, how high up would it go before coming to rest? Assume that the acceleration due to gravity is constant throughout the ascent.

Short Answer

Expert verified
The xenon atom would rise to approximately 28100 meters before coming to rest.

Step by step solution

01

Determine the Average Translational Kinetic Energy

The average translational kinetic energy of a gas particle can be given by the equation \( KE_{avg} = \frac{3}{2}kT \), where \( k \) is the Boltzmann constant \( 1.38 \times 10^{-23} \, \text{J/K} \) and \( T \) is the temperature in Kelvin. Here, \( T = 291 \, \text{K} \).Calculate the average translational kinetic energy:\[ KE_{avg} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 291 = 6.01 \times 10^{-21} \, \text{J} \]
02

Relate Kinetic Energy to Potential Energy

When the gas particle climbs to its highest point, all its initial kinetic energy will be converted into gravitational potential energy (GPE). The potential energy \( PE \) is given by \( PE = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity \( 9.8 \, \text{m/s}^2 \), and \( h \) is the height.Since \( KE_{avg} = PE \), we have:\[ 6.01 \times 10^{-21} = m \times 9.8 \times h \]
03

Calculate the Mass of a Xenon Atom

The mass \( m \) of a xenon atom can be calculated using its atomic mass \( 131.29 \mathrm{u} \) and converting it to kilograms. 1 atomic mass unit \( \mathrm{u} \) is equal to \( 1.66 \times 10^{-27} \, \text{kg} \).\[ m = 131.29 \times 1.66 \times 10^{-27} = 2.18 \times 10^{-25} \, \text{kg} \]
04

Solve for Height

Substitute the mass into the equation for potential energy:\[ 6.01 \times 10^{-21} = 2.18 \times 10^{-25} \times 9.8 \times h \]Solve for \( h \):\[ h = \frac{6.01 \times 10^{-21}}{2.18 \times 10^{-25} \times 9.8} = 2.81 \times 10^4 \, \text{m} \]
05

Conclusion

The final height that the xenon atom would reach before coming to a stop is approximately \( 2.81 \times 10^4 \, \text{m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
Potential energy is the energy possessed by an object due to its position or state. For example, an object at a height in a gravitational field has potential energy because gravity can act on it to move it to a lower height. In our scenario, when the xenon atom travels upward, it gains gravitational potential energy. This is because it moves against the gravitational pull of the Earth. As it moves higher, the potential energy increases, while its kinetic energy decreases until the atom comes to rest.
To calculate potential energy, use the formula:
  • \( PE = mgh \)
  • where \( m \) is the mass of the xenon atom, \( g \) is the acceleration due to gravity, and \( h \) is the height.
This formula shows that potential energy increases with height and mass.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. It depends on the mass of the object and its velocity. In this problem, the xenon atom has kinetic energy because it moves upwards through the air. Initially, its kinetic energy is at the maximum as it starts its journey upwards from the Earth's surface.
For gases, the average translational kinetic energy is related to temperature using the formula:
  • \( KE_{avg} = \frac{3}{2}kT \)
  • where \( k \) is the Boltzmann constant and \( T \) is the temperature.
As the xenon atom rises, its kinetic energy decreases, converting into potential energy until the atom halts at its highest point.
Gravitational Potential Energy
Gravitational potential energy is a specific type of potential energy related to an object's position in a gravitational field. Any object, including the xenon atom, that rises in this field gains gravitational potential energy because it moves further from the Earth's center.
In this exercise, as the xenon atom ascends without any collisions, its kinetic energy completely transforms into gravitational potential energy at its peak height. This energy change illustrates the conservation of energy principle 鈥� energy is never lost, only transformed from one type to another.
Gravitational potential energy can be described with the formula:
  • \( PE = mgh \), where
  • \( m \) is the mass of the object, \( g \) is the gravitational constant, and \( h \) is the height above the reference point.
The transformation from kinetic to gravitational potential energy is crucial for solving how high the atom will travel.

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Most popular questions from this chapter

If the translational rms speed of the water vapor molecules \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) in air is \(648 \mathrm{~m} / \mathrm{s},\) what is the translational rms speed of the carbon dioxide molecules \(\left(\mathrm{CO}_{2}\right)\) in the same air? Both gases are at the same temperature.

Hemoglobin has a molecular mass of \(64500 \mathrm{u}\). Find the mass (in \(\mathrm{kg}\) ) of one molecule of hemoglobin.

The drawing shows an ideal gas confined to a cylinder by a massless piston that is attached to an ideal spring. Outside the cylinder is a vacuum. The cross-sectional area of the piston is \(A\). The initial pressure, volume, and temperature of the gas are, respectively, \(P_{0}, V_{0}\), and \(T_{0}\), and the spring is initially stretched by an amount \(x_{0}\) with respect to its unstrained length. The gas is heated, so that its final pressure, volume, and temperature are \(P_{\mathrm{f}}, V_{\mathrm{f}},\) and \(T_{\mathrm{f}}\) and the spring is stretched by an amount \(x_{\mathrm{f}}\) with respect to its unstrained length. Assume that \(x_{0}\) and \(x_{\mathrm{f}}\) are positive variables. (a) What is the relation between the magnitude of the force required to stretch an ideal spring and the amount of the stretch with respect to the unstrained length of the spring? (b) What are the magnitudes, \(F_{0}\) and \(F_{\mathrm{F}}\), of the forces that the initial and final pressures apply to the piston and, hence, to the spring? Express your answers in terms of the pressures and the cross-sectional area of the piston. (c) According to the ideal gas law, how are the initial pressure, volume, and temperature related to the final pressure, volume, and temperature? (d) How is the final volume related to the initial volume, the cross-sectional area of the piston, and the initial and final amounts by which the spring is stretched? Account for your answer. Problem The initial temperature and volume of the gas described in the Concept Questions are \(273 \mathrm{~K}\) and \(6.00 \times 10^{-4} \mathrm{~m}^{3}\). The initial and final amounts by which the spring is stretched are, respectively, 0.0800 and \(0.1000 \mathrm{~m}\). The cross-sectional area of the piston is \(2.50 \times 10^{-3} \mathrm{~m}^{2}\). What is the final temperature of the gas?

In a diesel engine, the piston compresses air at \(305 \mathrm{~K}\) to a volume that is onesixteenth of the original volume and a pressure that is 48.5 times the original pressure. What is the temperature of the air after the compression?

A tank contains \(0.85 \mathrm{~mol}\) of molecular nitrogen \(\left(\mathrm{N}_{2}\right)\). Determine the mass (in grams) of nitrogen that must be removed from the tank in order to lower the pressure from 38 to 25 atm. Assume that the volume and temperature of the nitrogen in the tank do not change.
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