Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 1272

The average kinetic energy of hydrogen molecules at \(300 \mathrm{~K}\) is \(E\). At the same temperature the average kinetic energy of oxygen molecules will be (A) [E/(16)] (B) \(E\) (C) \(4 \mathrm{E}\) D) \([E /(4)]\)

Short Answer

Expert verified
The average kinetic energy of oxygen molecules at \(300~K\) is the same as that of hydrogen molecules, which is represented by \(E\). So, the correct option is (B) \(E\).

Step by step solution

01

Identify the given information

We are given: 1. The temperature of both hydrogen and oxygen gases: \(T = 300~K\) 2. The average kinetic energy of hydrogen molecules: \(E\) Our goal is to find the average kinetic energy of oxygen molecules at the same temperature.
02

Recall the equation for the average kinetic energy

For an ideal gas, the average kinetic energy (\(E_k\)) of a molecule is given by: \[E_k = \dfrac{3}{2}kT\] where \(k\) is Boltzmann's constant, and \(T\) is the temperature in Kelvin.
03

Use the equation to determine the average kinetic energy of the hydrogen molecules

We are given that the average kinetic energy of hydrogen molecules at \(300~K\) is \(E\). So, using the equation, we can write: \[E = \dfrac{3}{2}kT\]
04

Determine the average kinetic energy of the oxygen molecules

Since the temperature is the same for both the hydrogen and oxygen gas molecules (\(T = 300~K\)), the average kinetic energies of oxygen molecules, \(E_o\), must also be given by: \[E_o = \dfrac{3}{2}kT\]
05

Compare the average kinetic energies of hydrogen and oxygen molecules

As per step 3 and 4, we know that \(E = \dfrac{3}{2}kT\) and \(E_o=\dfrac{3}{2}kT\). Comparing both expressions, we get: \[E_o = E\] So, the average kinetic energy of oxygen molecules at the same temperature is the same as that of hydrogen molecules.
06

Answer

The correct option is: (B) \(E\) The average kinetic energy of oxygen molecules at \(300~K\) is the same as that of hydrogen molecules, which is represented by \(E\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Calculate the temperature at which \(\mathrm{rms}\) velocity of \(\mathrm{SO}_{2}\) molecules is the same as that of \(\mathrm{O}_{2}\) molecules at \(27^{\circ} \mathrm{C}\). Molecular weights of Oxygen and \(\mathrm{SO}_{2}\) are \(32 \mathrm{~g}\) and \(64 \mathrm{~g}\) respectively (A) \(327^{\circ} \mathrm{C}\) (B) \(327 \mathrm{~K}\) (C) \(127^{\circ} \mathrm{C}\) (D) \(227^{\circ} \mathrm{C}\)

For a gas \(\left(R / C_{V}\right)=0.67\). This gas is made up of molecules which are (A) Diatomic (B) monoatomic (C) polyatomic (D) mixture of diatomic and polyatomic molecules

The molar specific heat at constant pressure for a monoatomic gas is (A) \((3 / 2) \mathrm{R}\) (B) \((5 / 2) \mathrm{R}\) (C) \(4 \mathrm{R}\) (D) \((7 / 2) \mathrm{R}\)

At constant temperature on increasing the pressure of a gas \(5 \%\) its volume will decrease by (A) \(5 \%\) (B) \(5.26 \%\) (C) \(4.76 \%\) (D) \(4.26 \%\)

When the temperature of a gas is raised from \(27^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\), the percentage increase in the rms velocity of the molecules will be (A) \(15 \%\) (B) \(17.5 \%\) (C) \(10 \%\) (D) \(20 \%\)
See all solutions

Recommended explanations on English Textbooks

Syntax

Read Explanation

Text Comparison

Read Explanation

Cues and Conventions

Read Explanation

Argumentative Essay

Read Explanation

Prosody

Read Explanation

Language and Social Groups

Read Explanation
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.

Sign-up for free