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Oxygen \(\left(\mathrm{O}_{2}\right)\) gas at \(273 \mathrm{~K}\) and \(1.0 \mathrm{~atm}\) is confined to a cubical container \(10 \mathrm{~cm}\) on a side. Calculate \(\Delta U_{g} / K_{\text {avg }}\), where \(\Delta U_{g}\) is the change in the gravitational potential energy of an oxygen molecule falling the height of the box and \(K_{\text {avg }}\) is the molecule's average translational kinetic energy

Short Answer

Expert verified
\( \frac{\Delta U_g}{K_{\text{avg}}} \approx 9.20 \times 10^{-7} \).

Step by step solution

01

Calculate the Height of the Box

Given that the container is cubical with each side being 10 cm, the height of the box is 10 cm. We need to convert this value to meters for use in standard SI units. So, the height, \( h = 0.1 \text{ m} \).
02

Determine Mass of an Oxygen Molecule

The molecular mass of an oxygen molecule (\( \text{O}_2 \)) is approximately \( 32 \text{ g/mol} \), which equals \( 32 \times 10^{-3} \text{ kg/mol} \). Using Avogadro's number, \( 6.022 \times 10^{23} \text{ molecules/mol} \), the mass of one oxygen molecule is \[ m = \frac{32 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} = 5.31 \times 10^{-26} \text{ kg} \].
03

Calculate Change in Gravitational Potential Energy (\( \Delta U_g \))

Gravitational potential energy change is calculated using the formula \( \Delta U_g = mgh \). Substituting the values, where \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity, we get \[ \Delta U_g = 5.31 \times 10^{-26} \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.1 \text{ m} \approx 5.20 \times 10^{-27} \text{ J} \].
04

Calculate Average Translational Kinetic Energy (\( K_{\text{avg}} \))

The average translational kinetic energy of a molecule is given by \( K_{\text{avg}} = \frac{3}{2} kT \). The Boltzmann constant \( k \) is \( 1.38 \times 10^{-23} \text{ J/K} \) and \( T = 273 \text{ K} \), so \[ K_{\text{avg}} = \frac{3}{2} \times 1.38 \times 10^{-23} \text{ J/K} \times 273 \text{ K} = 5.65 \times 10^{-21} \text{ J} \].
05

Calculate \( \frac{\Delta U_g}{K_{\text{avg}}} \)

Now, we find the ratio of change in potential energy to average kinetic energy: \[ \frac{\Delta U_g}{K_{\text{avg}}} = \frac{5.20 \times 10^{-27} \text{ J}}{5.65 \times 10^{-21} \text{ J}} \approx 9.20 \times 10^{-7} \].

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy that an object possesses due to its position in a gravitational field. It is an essential concept in physics because it helps us understand how energy is stored and transferred in systems under the influence of gravity.

For an object near the surface of the Earth, the gravitational potential energy (\( U_g \)) can be calculated using the formula:\[U_g = mgh\]where:
  • \( m \) is the mass of the object in kilograms,
  • \( g \)is the acceleration due to gravity (approximately 9.8 m/s² on Earth), and
  • \( h \)is the height in meters above a reference point.
In practical applications, changes in gravitational potential energy (\( \Delta U_g \)) are often more relevant than the absolute value. When an object either rises or falls in a gravitational field, its potential energy changes. For the oxygen molecule in the example, it falls over a distance of 0.1 meters. By using the gravitational potential energy formula, the change experienced is calculated as 5.20 x 10^-27 Joules.

It's important to know that gravitational potential energy depends not just on the height, but also the mass of the object and the gravitational force, making it a pivotal concept in understanding motion.
Translational Kinetic Energy
Translational kinetic energy refers to the energy possessed by an object due to its motion along a straight line. If an object is moving, it certainly has kinetic energy. When a molecule, like oxygen, moves in a straight path, it exhibits translational kinetic energy.

The translational kinetic energy of a molecule is mathematically expressed as:\[K_{\text{avg}} = \frac{3}{2} kT\]where
  • \( k \) is the Boltzmann constant (1.38 x 10^-23 J/K),
  • \( T \)is the absolute temperature of the gas in Kelvin.
In this formula, the average translational kinetic energy (\( K_{\text{avg}} \)) of each molecule is directly related to the temperature of the gas. Thus, higher temperatures imply greater kinetic energy. For instance, at a temperature of 273 K, the average kinetic energy of an oxygen molecule is found to be 5.65 x 10^-21 Joules.

Understanding this concept is crucial because kinetic energy explains how energy transitions through motion - a key principle in thermodynamics and kinetic theory. This knowledge is applicable when calculating energy transformations and interactions in gases like oxygen.
Molecular Mass
Molecular mass refers to the sum of the atomic masses of all atoms in a molecule. It represents how heavy a molecule is based on the constituent elements. Learning about molecular mass is critical to understand how individual molecules behave and interact.

For oxygen (Oâ‚‚), the molecular mass is approximately 32 grams per mole. This value is obtained from the periodic table, where the atomic mass of a single oxygen atom is roughly 16 atomic mass units (AMU). With two atoms in each molecule of oxygen, the molecular mass doubles.

When converting molecular mass for physics exercises, it often needs to be expressed in kilograms for consistency with SI units. Therefore, the mass of a single Oâ‚‚ molecule is determined using Avogadro's number: \[m = \frac{32 \times 10^{-3} \text{ kg/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} = 5.31 \times 10^{-26} \text{ kg}\]This conversion plays an important role when discussing gases and their physical properties. Understanding molecular mass helps in calculations of motion, energy, and many other chemical and physical properties connected to molecules.

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Most popular questions from this chapter

A quantity of ideal gas at \(10.0^{\circ} \mathrm{C}\) and \(100 \mathrm{kPa}\) occupies a volume of \(2.50 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? (b) If the pressure is now raised to \(300 \mathrm{kPa}\) and the temperature is raised to \(30.0^{\circ} \mathrm{C}\), how much volume does the gas occupy? Assume no leaks.

When the U. S. submarine Squalus became disabled at a depth of \(80 \mathrm{~m}\), a cylindrical chamber was lowered from a ship to rescue the crew. The chamber had a radius of \(1.00 \mathrm{~m}\) and a height of \(4.00 \mathrm{~m}\), was open at the bottom, and held two rescuers. It slid along a guide cable that a diver had attached to a hatch on the submarine. Once the chamber reached the hatch and clamped to the hull, the crew could escape into the chamber. During the descent, air was released from tanks to prevent water from flooding the chamber. Assume that the interior air pressure matched the water pressure at depth \(h\) as given by \(p_{0}+\rho g h\), where \(p_{0}=1.000 \mathrm{~atm}\) is the surface pressure and \(\rho=\) \(1024 \mathrm{~kg} / \mathrm{m}^{3}\) is the density of seawater. Assume a surface temperature of \(20.0^{\circ} \mathrm{C}\) and a submerged water temperature of \(-30.0^{\circ} \mathrm{C}\). (a) What is the air volume in the chamber at the surface? (b) If air had not been released from the tanks, what would have been the air volume in the chamber at depth \(h=80.0 \mathrm{~m} ?\) (c) How many moles of air were needed to be released to maintain the original air volume in the chamber?

An automobile tire has a volume of \(1.64 \times 10^{-2} \mathrm{~m}^{3}\) and contains air at a gauge pressure (pressure above atmospheric pressure) of \(165 \mathrm{kPa}\) when the temperature is \(0.00^{\circ} \mathrm{C}\). What is the gauge pressure of the air in the tires when its temperature rises to \(27.0^{\circ} \mathrm{C}\) and its volume increases to \(1.67 \times 10^{-2} \mathrm{~m}^{3} ?\) Assume atmospheric pressure is \(1.01 \times 10^{5} \mathrm{~Pa}\).

It is found that the most probable speed of molecules in a gas when it has (uniform) temperature \(T_{2}\) is the same as the rms speed of the molecules in this gas when it has (uniform) temperature \(T_{1} .\) Calculate \(T_{2} / T_{1}\).

Air that initially occupies \(0.140 \mathrm{~m}^{3}\) at a gauge pressure of \(103.0 \mathrm{kPa}\) is expanded isothermally to a pressure of \(101.3 \mathrm{kPa}\) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)

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