/*! 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} Q53P At what pressure, in atmospheres... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 41: Q53P (page 1275)

At what pressure, in atmospheres, would the number of molecules per unit volume in an ideal gas be equal to the number density of the conduction electrons in copper, with both gas and copper at temperature T =300K?

Short Answer

Expert verified

The pressure value that would give the number of molecules per unit volume in an ideal gas equal to the number density of conduction electrons in copper is 3.49脳103atm.

Step by step solution

01

The given data

a) The number of conduction electrons in copper,n/V=8.43脳1028m-3

b) The temperature value, T =300 K

02

Understanding the concept of ideal gas equation

The equation that relates the characteristic properties of an ideal gas is known as Ideal Gas Equation. The ideal gas equation can be given as-

PV=nkT

Here P is the pressure,V is the volume, n id the number of moles and k is the Boltzmann constant.

03

Calculation of the pressure

Using the given condition and data in equation, we can get the value of the pressure in atmospheres as follows:

P=nkTV=8.43脳1028m-31.38脳10-23J/K300K=3.49脳108Pa=3.49脳103atm

Hence, the value of the pressure is 3.49脳103atm.

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!

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

In a silicon lattice, where should you look if you want to find (a) a conduction electron, (b) a valence electron, and (c) an electron associated with the 2psubshell of the isolated silicon atom?

The occupancy probability function (Eq. 41-6) can be applied to semiconductors as well as to metals. In semiconductors the Fermi energy is close to the midpoint of the gap between the valence band and the conduction band. For germanium, the gap width is 0.67eV. What is the probability that (a) a state at the bottom of the conduction band is occupied and (b) a state at the top of the valence band is not occupied? Assume that T = 290K. (Note:In a pure semiconductor, the Fermi energy lies symmetrically between the population of conduction electrons and the population of holes and thus is at the center of the gap. There need not be an available state at the location of the Fermi energy.)

An isolated atom of germanium has 32 electrons, arranged in subshells according to this scheme:1s22s22p63s23p63d104s24p2 This element has the same crystal structure as silicon and, like silicon, is a semiconductor. Which of these electrons form the valence band of crystalline germanium?

(a) What maximum light wavelength will excite an electron in the valence band of diamond to the conduction band? The energy gap is 5.50 eV. (b) In what part of the electromagnetic spectrum does this wavelength lie?

A sample of a certain metal has a volume of 4.0脳10-5m3. The metal has a density of 9.0鈥�g/cm3and a molar mass of 60 g/mol. The atoms are bivalent. How many conduction electrons (or valence electrons) are in the sample?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Kinematics Physics

Read Explanation

Quantum Physics

Read Explanation

Fluids

Read Explanation

Electric Charge Field and Potential

Read Explanation

Fundamentals of Physics

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.