Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 1255

The pressure is exerted by the gas on the walls of the container because (A) It sticks with the walls (B) It is accelerated towards the walls (C) It loses kinetic energy (D) On collision with the walls there is a change in momentum

Short Answer

Expert verified
The pressure exerted by the gas on the walls of the container is due to the change in momentum during collisions of gas particles with the walls (Option D). These collisions result in a net force applied to the wall, thus creating pressure.

Step by step solution

01

Pressure and Gas Particles

Pressure is defined as force applied per unit area, mathematically represented as: \(P = \frac{F}{A}\) In the case of gases, pressure is exerted on the walls of a container due to the continuous collision of gas particles with the wall. The force created due to these collisions results in pressure being exerted on the walls. Step 2: Evaluate option (A)
02

Gas Particles Sticking with the Walls

Gas particles are in constant motion and continuously collide with the walls of a container. However, they do not stick to the walls. They bounce off the walls and move in another direction after each collision. Thus, option (A) is incorrect. Step 3: Evaluate option (B)
03

Gas Particles Accelerated Towards the Walls

Gas particles move randomly in all directions inside the container. They are not specifically accelerated towards the walls. Although some gas particles might move towards the walls at any time, others are moving away from them. So, option (B) is incorrect. Step 4: Evaluate option (C)
04

Gas Particles Losing Kinetic Energy

The kinetic energy of gas particles is not lost during the collisions with the walls, as these collisions are considered elastic. In elastic collisions, the total kinetic energy of the system remains constant. So, option (C) is incorrect. Step 5: Evaluate option (D)
05

Change in Momentum During Collision

When gas particles collide with the walls of the container, their momentum changes. As per Newton's third law of motion, an equal and opposite force is applied to the wall during the collision, which results in a net force on the wall, contributing to the pressure exerted by the gas. This change in momentum is the main reason for the pressure exerted by gas particles on the container walls. Thus, option (D) is correct.

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

\(2 \mathrm{~g}\) of \(\mathrm{O}_{2}\) gas is taken at \(27^{\circ} \mathrm{C}\) and pressure \(76 \mathrm{~mm} \mathrm{Hg}\). Find out volume of gas (ln liter) (A) \(3.08\) (B) \(44.2\) (C) \(2.05\) (D) \(2.44\)

The average kinetic energy of a gas molecule at \(27^{\circ} \mathrm{C}\) is \(6.21 \times 10^{-21} \mathrm{~J}\). Its average kinetic energy at $227^{\circ} \mathrm{C}$ will be (A) \(5.22 \times 10^{-21} \mathrm{~J}\) (B) \(11.35 \times 10^{-21} \mathrm{~J}\) (C) \(52.2 \times 10^{-21} \mathrm{~J}\) (D) \(12.42 \times 10^{-21} \mathrm{~J}\)

The pressure and temperature of two different gases \(P\) and T having the volumes \(\mathrm{V}\) for each. They are mixed keeping the same volume and temperature, the pressure of the mixture will be, (A) \(\mathrm{P}\) (B) \((\mathrm{P} / 2)\) (C) \(4 \mathrm{P}\) (D) \(2 \mathrm{P}\)

If the molecular weight of two gases are \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\), then at a given temperature the ratio of root mean square velocity \(\mathrm{v}_{1}\) and \(v_{2}\) will be (A) \(\sqrt{\left(\mathrm{M}_{1} / \mathrm{M}_{2}\right)}\) (B) \(\sqrt{\left(M_{2} / M_{1}\right)}\) (C) $\left.\sqrt{[}\left(\mathrm{M}_{1}-\mathrm{M}_{2}\right) /\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right)\right]$ (D) $\left.\sqrt{[}\left(\mathrm{M}_{1}+\mathrm{M}_{2}\right) /\left(\mathrm{M}_{1}-\mathrm{M}_{2}\right)\right]$

The average kinetic energy per molecule of a gas at \(-23^{\circ} \mathrm{C}\) and \(75 \mathrm{~cm}\) pressure is \(5 \times 10^{-14}\) erg for \(\mathrm{H}_{2}\). The mean kinetic energy per molecule of the \(\mathrm{O}_{2}\) at \(227^{\circ} \mathrm{C}\) and \(150 \mathrm{~cm}\) pressure will be (A) \(80 \times 10^{-14} \mathrm{erg}\) (B) \(10 \times 10^{-14}\) erg (C) \(20 \times 10^{-14}\) erg (D) \(40 \times 10^{-14}\) erg
See all solutions

Recommended explanations on English Textbooks

Single Paragraph Essay

Read Explanation

Language Analysis

Read Explanation

Free Response Essay

Read Explanation

English Language Study

Read Explanation

Cues and Conventions

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