Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 45

In a period of \(6.00 \mathrm{~s}, 9.00 \cdot 10^{23}\) nitrogen molecules strike a section of a wall with an area of \(2.00 \mathrm{~cm}^{2}\). If the molecules move with a speed of \(400.0 \mathrm{~m} / \mathrm{s}\) and strike the wall head on in elastic collisions, what is the pressure exerted on the wall? (The mass of one \(\mathrm{N}_{2}\) molecule is \(4.68 \cdot 10^{-26} \mathrm{~kg}\).)

Short Answer

Expert verified
Answer: The pressure exerted on the wall is 2.81 x 10^4 Pa.

Step by step solution

01

Calculate the momentum change for one molecule after collision

After the elastic collision, the direction of the nitrogen molecule is reversed. The momentum change for one molecule can be calculated by subtracting the initial momentum from the final momentum. This can be expressed as: Momentum change \(= m(v_{final} - v_{initial})\) Since the molecule changes its direction, \(v_{final} = -v_{initial}\), and the momentum change becomes: Momentum change \(= 2mv_{initial}\) Substituting the given values, we have: Momentum change \(= 2(4.68\cdot 10^{-26}\mathrm{~kg})(400.0\mathrm{~m} / \mathrm{s}) = 3.744\cdot 10^{-23}\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)
02

Calculate the momentum change for all molecules

We are given that there are \(9.00\cdot 10^{23}\) nitrogen molecules striking the wall. To find the total momentum change for all molecules, multiply the momentum change for one molecule by the number of molecules: Total momentum change \(= \text{number of molecules} \times \text{momentum change per molecule}\) Total momentum change \(= (9.00\cdot 10^{23})(3.744\cdot 10^{-23}\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})= 3.37\cdot 10^1\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\)
03

Use the impulse-momentum theorem to find the force

The impulse-momentum theorem states that the force times the time interval is equal to the momentum change: \(F\Delta t = \Delta p\) Where \(F\) is the force, \(\Delta t\) is the time interval, and \(\Delta p\) is the momentum change. We are given a time interval of \(6\mathrm{~s}\). Solving for the force, we have: \(F=(\Delta p)/(\Delta t) = (3.37\cdot 10^1\mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})/(6\mathrm{~s}) = 5.62\mathrm{~N}\)
04

Calculate the pressure

Now, we can find the pressure exerted on the wall by dividing the force by the given area (\(2.00\mathrm{~cm}^2\)). First, convert the area from \(\mathrm{cm}^2\) to \(\mathrm{m}^2\): \(2.00\mathrm{~cm}^{2} = 2.00\cdot 10^{-4}\mathrm{~m}^{2}\) Now, calculate the pressure: Pressure \(= \frac{F}{A} = \frac{5.62\mathrm{~N}}{2.00\cdot 10^{-4}\mathrm{~m}^{2}} = 28100\mathrm{~Pa}\) So, the pressure exerted on the wall is \(2.81\cdot 10^4\mathrm{~Pa}\).

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

At a temperature of \(295 . \mathrm{K},\) the vapor pressure of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right)\) is \(60.7 \mathrm{kPa}\). Suppose \(1.000 \mathrm{~g}\) of gaseous pentane is contained in a cylinder with diathermal (thermally conducting) walls and a piston to vary the volume. The initial volume is \(1.000 \mathrm{~L},\) and the piston is moved in slowly, keeping the temperature at \(295 \mathrm{~K}\). At what volume will the first drop of liquid pentane appear?

Liquid bromine \(\left(\mathrm{Br}_{2}\right)\) is spilled in a laboratory accident and evaporates. Assuming the vapor behaves as an ideal gas, with a temperature \(20.0^{\circ} \mathrm{C}\) and a pressure of \(101.0 \mathrm{kPa}\), find its density.

When you blow hard on your hand, it feels cool, but when you breathe softly, it feels warm. Why?

Interstellar space far from any stars is usually filled with atomic hydrogen (H) at a density of 1 atom/cm \(^{3}\) and a very low temperature of \(2.73 \mathrm{~K}\). a) Determine the pressure in interstellar space. b) What is the root-mean-square speed of the atoms? c) What would be the edge length of a cube that would contain atoms with a total of \(1.00 \mathrm{~J}\) of energy?

A closed auditorium of volume \(2.50 \cdot 10^{4} \mathrm{~m}^{3}\) is filled with 2000 people at the beginning of a show, and the air in the space is at a temperature of \(293 \mathrm{~K}\) and a pressure of \(1.013 \cdot 10^{5} \mathrm{~Pa}\). If there were no ventilation, by how much would the temperature of the air rise during the \(2.00-\mathrm{h}\) show if each person metabolizes at a rate of \(70.0 \mathrm{~W} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Atoms and Radioactivity

Read Explanation

Waves Physics

Read Explanation

Solid State Physics

Read Explanation

Classical Mechanics

Read Explanation

Measurements

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