/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 (I) (a) What is the average tran... [FREE SOLUTION] | 91影视

91影视

(I) (a) What is the average translational kinetic energy of an oxygen molecule at STP? (b) What is the total translational kinetic energy of \(1.0 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) molecules at \(25^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) 5.65 x 10鈦宦猜 J; (b) 3714 J

Step by step solution

01

Identify the formula for translational kinetic energy

The average translational kinetic energy per molecule of gas is given by the formula \( \frac{3}{2} kT \), where \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \text{ J K}^{-1} \)) and \( T \) is the temperature in Kelvin.
02

Calculate average kinetic energy at STP

STP (Standard Temperature and Pressure) is defined as 273.15 K. The average kinetic energy is \( \frac{3}{2} kT = \frac{3}{2} \times 1.38 \times 10^{-23} \times 273.15 \approx 5.65 \times 10^{-21} \text{ J} \).
03

Calculate temperature in Kelvin at 25掳C

To convert Celsius to Kelvin, add 273.15. Therefore, \( 25^{\circ} \text{C} = 25 + 273.15 = 298.15 \text{ K} \).
04

Calculate total translational kinetic energy for 1.0 mol of \( \mathrm{O}_2 \)

Use the formula \( KE = \frac{3}{2} nRT \), where \( n \) is the number of moles, \( R \) is the ideal gas constant (8.31 J/mol K), and \( T \) is the temperature in Kelvin. Here, \( n = 1.0 \text{ mol} \), \( T = 298.15 \text{ K} \), so,\[ KE = \frac{3}{2} \times 1.0 \times 8.31 \times 298.15 \approx 3714 \text{ J} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental relation in physics and chemistry that describes the behavior of an ideal gas. It combines several other gas laws and can be expressed as \( PV = nRT \). Here, \( P \) represents the pressure of the gas, \( V \) is the volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

This equation allows us to understand how changes in one property will affect others, assuming the gas behaves ideally. The constant \( R \), often valued at \( 8.31 \, \text{J/mol K} \), is crucial as it bridges the macroscopic and molecular scales. By knowing any three of the variables, you can solve for the fourth, making it immensely useful in a wide range of calculations involving gases.
  • Pressure (\( P \)) is often measured in pascals or atmospheres.
  • Volume (\( V \)) is typically measured in liters or cubic meters.
  • Temperature (\( T \)) must always be in Kelvin.
This law is essential for understanding various phenomena, such as why a balloon inflates when heated or why a tire might burst when overfilled.
Boltzmann Constant
The Boltzmann Constant \( k \) plays a key role in statistical mechanics, linking the macroscopic and microscopic worlds. Its value, \( 1.38 \times 10^{-23} \, \text{J K}^{-1} \), allows calculations involving individual molecules.

When dealing with gases, each molecule works as a tiny unit of energy and movement. The Boltzmann Constant helps quantify this energy, especially in terms of average kinetic energy per molecule at a given temperature.

The formula \( \frac{3}{2} kT \) gives the average translational kinetic energy per molecule, showcasing the critical relationship between molecular movement and temperature. As temperature rises, kinetic energy increases, leading to faster molecular movement. This connection supports our understanding of temperature as a measure of energy stored in molecular motion.
Utilizing the Boltzmann Constant makes it possible to calculate the exact movement in individual gas molecules, offering a detailed insight into their behavior. This knowledge forms the foundation for deeper explorations in thermodynamics and thermal physics.
Translational Motion
Translational Motion refers to the straight-line movement of molecules and is a principal component of kinetic energy in gases. It's this motion that contributes to the pressure exerted by gases on their containers.

In physics, translational kinetic energy is the energy due to the motion of a particle moving in space. The average translational kinetic energy for a molecule in an ideal gas is expressed by \( \frac{3}{2} kT \), where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.
Translational motion is essential because:
  • It helps explain why gases mix and spread to fill their containers, known as diffusion.
  • The speed and energy of translational motion impact how gases exert force and respond to celestial bodies or applied pressures.
  • The average kinetic energy derived from this motion provides a baseline for understanding gas temperature.
When studying gases, considering translational motion helps clarify how particles interact and react to different forces and changes. Acknowledging translational energy is vital to predicting and measuring gas behavior in various scientific applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(III) How well does the ideal gas law describe the pressurized air in a scuba tank? (a) To fill a typical scuba tank, an air compressor intakes about 2300 \(\mathrm{L}\) of air at 1.0 \(\mathrm{atm}\) and compresses this gas into the tank's \(12-\mathrm{L}\) internal volume. If the filling process occurs at \(20^{\circ} \mathrm{C},\) show that a tank holds about 96 \(\mathrm{mol}\) of air. \((b)\) Assume the tank has 96 \(\mathrm{mol}\) of air at \(20^{\circ} \mathrm{C}\) . Use the ideal gas law to predict the air's pressure within the tank. (c) Use the van der Waals equation of state to predict the air's pressure within the tank. For air, the van der Waals constants are \(a=0.1373 \mathrm{N} \cdot \mathrm{m}^{4} / \mathrm{mol}^{2}\) and \(b=3.72 \times 10^{-5} \mathrm{m}^{3} / \mathrm{mol} .\) (d) Taking the van der Waals pressure as the true air pressure, show that the ideal gas law predicts a pressure that is in error by only about 3\(\% .\)

(II) Oxygen diffuses from the surface of insects to the interior through tiny tubes called tracheae. An average trachea is about \(2 \mathrm{~mm}\) long and has cross-sectional area of \(2 \times 10^{-9} \mathrm{~m}^{2}\). Assuming the concentration of oxygen inside is half what it is outside in the atmosphere, \((a)\) show that the concentration of oxygen in the air (assume \(21 \%\) is oxygen) at \(20^{\circ} \mathrm{C}\) is about \(8.7 \mathrm{~mol} / \mathrm{m}^{3},\) then \((b)\) calculate the diffusion rate \(J,\) and \((c)\) estimate the average time for a molecule to diffuse in. Assume the diffusion constant is \(1 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\).

(II) When using a mercury barometer, the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 \(\mathrm{mm}\) -Hg. At sea level, the height \(h\) of mercury in a barometer is about 760 \(\mathrm{mm} .\) (a) If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? (b) What is the percent error? (c) What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?

(III) A 0.5-mol sample of \(\mathrm{O}_{2}\) gas is in a large cylinder with a movable piston on one end so it can be compressed. The initial volume is large enough that there is not a significant difference between the pressure given by the ideal gas law and that given by the van der Waals equation. As the gas is slowly compressed at constant temperature (use \(300 \mathrm{~K}\) ), at what volume does the van der Waals equation give a pressure that is \(5 \%\) different than the ideal gas law pressure? Let \(a=0.14 \mathrm{~N} \cdot \mathrm{m}^{4} / \mathrm{mol}^{2}\) and \(b=3.2 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{mol}\).

A scuba tank has a volume of \(3100 \mathrm{~cm}^{3}\). For very deep dives, the tank is filled with \(50 \%\) (by volume) pure oxygen and \(50 \%\) pure helium. (a) How many molecules are there of each type in the tank if it is filled at \(20^{\circ} \mathrm{C}\) to a gauge pressure of \(12 \mathrm{~atm} ?(b)\) What is the ratio of the average kinetic energies of the two types of molecule? (c) What is the ratio of the rms speeds of the two types of molecule?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.