/*! 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 69 Find the rms speed in air at \(0... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the rms speed in air at \(0.0^{\circ} \mathrm{C}\) and 1.00 atm of (a) the \(\mathrm{N}_{2}\) molecules, (b) the \(\mathrm{O}_{2}\) molecules, and (c) the \(\mathrm{CO}_{2}\) molecules.

Short Answer

Expert verified
Answer: The root-mean-square speeds at 0.0°C and 1.00 atm are: (a) for N₂ molecules: 518.0 m/s (b) for O₂ molecules: 481.3 m/s (c) for CO₂ molecules: 393.9 m/s

Step by step solution

01

1. Converting temperature into Kelvin

T(°C) = T(K) - 273.15 T(K) = T(°C) + 273.15 T(K) = 0.0 + 273.15 = 273.15 K
02

2. Calculating the mass of each molecule in kilograms

Mass of one nitrogen (N₂) molecule: Nucleon mass approximated as 1 amu (atomic mass unit) = 1.661 × 10^{-27} kg N₂ mass = 2 × (mass of one nitrogen atom) = 2 × 14 × 1.661 × 10^{-27} kg = 4.6508 × 10^{-26} kg Mass of one oxygen (O₂) molecule: O₂ mass = 2 × (mass of one oxygen atom) = 2 × 16 × 1.661 × 10^{-27} kg = 5.3128 × 10^{-26} kg Mass of one carbon dioxide (CO₂) molecule: CO₂ mass = (mass of one carbon atom) + 2 × (mass of one oxygen atom) = 12 × 1.661 × 10^{-27} kg + 2 × 16 × 1.661 × 10^{-27} kg = 7.3084 × 10^{-26} kg
03

3. Applying the rms speed formula

(a) rms speed of N₂ molecules: \(v_{rms}(\mathrm{N}_{2}) = \sqrt{\frac{3kT}{m_{N_2}}} = \sqrt{\frac{3 \times 1.381 \times 10^{-23} \, \mathrm{J \cdot K^{-1}} \times 273.15 \, \mathrm{K}}{4.6508 \times 10^{-26} \, \mathrm{kg}}} = 518.0 \, \mathrm{m/s}\) (b) rms speed of O₂ molecules: \(v_{rms}(\mathrm{O}_{2}) = \sqrt{\frac{3kT}{m_{O_2}}} = \sqrt{\frac{3 × 1.381 \times 10^{-23} \, \mathrm{J \cdot K^{-1}} \times 273.15 \, \mathrm{K}}{5.3128 \times 10^{-26} \, \mathrm{kg}}} = 481.3 \, \mathrm{m/s}\) (c) rms speed of CO₂ molecules: \(v_{rms}(\mathrm{CO}_{2}) = \sqrt{\frac{3kT}{m_{CO_2}}} = \sqrt{\frac{3 × 1.381 \times 10^{-23} \, \mathrm{J \cdot K^{-1}} \times 273.15 \, \mathrm{K}}{7.3084 \times 10^{-26} \, \mathrm{kg}}} = 393.9 \, \mathrm{m/s}\) Hence, the rms speed at 0.0°C and 1.00 atm are: (a) for N₂ molecules: 518.0 m/s (b) for O₂ molecules: 481.3 m/s (c) for CO₂ molecules: 393.9 m/s

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 91Ó°ÊÓ!

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

Sand is composed of \(\mathrm{SiO}_{2} .\) Find the order of magnitude of the number of silicon (Si) atoms in a grain of sand. Approximate the sand grain as a sphere of diameter \(0.5 \mathrm{mm}\) and an \(\mathrm{SiO}_{2}\) molecule as a sphere of diameter \(0.5 \mathrm{nm}\)

(a) At what temperature (if any) does the numerical value of Celsius degrees equal the numerical value of Fahrenheit degrees? (b) At what temperature (if any) does the numerical value of kelvins equal the numerical value of Fahrenheit degrees?

There are two identical containers of gas at the same temperature and pressure, one containing argon and the other neon. What is the ratio of the rms speed of the argon atoms to that of the neon atoms? The atomic mass of argon is twice that of neon. tutorial: RMS speed)

As a Boeing 747 gains altitude, the passenger cabin is pressurized. However, the cabin is not pressurized fully to atmospheric $\left(1.01 \times 10^{5} \mathrm{Pa}\right),$ as it would be at sea level, but rather pressurized to \(7.62 \times 10^{4} \mathrm{Pa}\). Suppose a 747 takes off from sea level when the temperature in the airplane is \(25.0^{\circ} \mathrm{C}\) and the pressure is \(1.01 \times 10^{5} \mathrm{Pa} .\) (a) If the cabin temperature remains at \(25.0^{\circ} \mathrm{C},\) what is the percentage change in the number of moles of air in the cabin? (b) If instead, the number of moles of air in the cabin does not change, what would the temperature be?
Air at room temperature and atmospheric pressure has a mass density of $1.2 \mathrm{kg} / \mathrm{m}^{3} .$ The average molecular mass of air is 29.0 u. How many molecules are in \(1.0 \mathrm{cm}^{3}\) of air?
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.