Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The ammonia molecule (NH\(_3\)) has a dipole moment of \(5.0 \times 10^{30}\) C \(\cdot\) m. Ammonia molecules in the gas phase are placed in a uniform electric field \(\overrightarrow{E}\) with magnitude \(1.6 \times 10^6\) N\(/\)C. (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\overrightarrow{E}\) from parallel to perpendicular? (b) At what absolute temperature \(T\) is the average translational kinetic energy \(\frac{3}{2} kT\) of a molecule equal to the change in potential energy calculated in part (a)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)

Short Answer

Expert verified
(a) 8.0 x 10^-24 J, (b) 38.5 K.

Step by step solution

01

Understanding the Change in Orientation

The molecule initially has its dipole moment, \(\vec{p}\), aligned with the electric field \(\vec{E}\), which is considered parallel. The dipole then moves to a perpendicular orientation. The electric potential energy of a dipole \(\vec{p}\) in an electric field \(\vec{E}\) is given by \(U = -\vec{p} \cdot \vec{E}\). Initially, when parallel, \(\theta = 0^\circ\) and \(U_i = -pE\). Then, when perpendicular, \(\theta = 90^\circ\) and \(U_f = 0\). The change in potential energy, \(\Delta U\), is \(U_f - U_i = 0 - (-pE)\).
02

Calculating Change in Potential Energy

We calculate \(\Delta U\) using the formula from Step 1. With \(p = 5.0 \times 10^{-30} \text{ C}\cdot\text{m}\) and \(E = 1.6 \times 10^6 \text{ N/C}\):\[\Delta U = pE = (5.0 \times 10^{-30} \text{ C}\cdot\text{m})(1.6 \times 10^6 \text{ N/C}) = 8.0 \times 10^{-24} \text{ J}\]
03

Relating Kinetic Energy and Potential Energy

To find the temperature \(T\) where the average translational kinetic energy equals the change in potential energy \(\Delta U\), equate \(\frac{3}{2} kT\) to \(\Delta U\). Solving for \(T\), we have:\[T = \frac{2 \Delta U}{3k}\]Where \(k = 1.38 \times 10^{-23} \text{ J/K}\) is Boltzmann's constant.
04

Calculating Absolute Temperature

Substitute \(\Delta U = 8.0 \times 10^{-24} \text{ J}\) and \(k = 1.38 \times 10^{-23} \text{ J/K}\) into the equation from Step 3:\[T = \frac{2 \times 8.0 \times 10^{-24} \text{ J}}{3 \times 1.38 \times 10^{-23} \text{ J/K}} \approx 38.5 \text{ K}\]

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dipole Moment
A dipole moment is a measure of the separation of positive and negative charges within a molecule, leading to a polar molecule. It is a vector quantity with both magnitude and direction. The units of a dipole moment are Coulomb-meters \(\text{C} \cdot \text{m}\). The direction points from the negative center to the positive center.

In molecules like ammonia (NH\(_3\)), there is a net dipole moment due to the difference in electronegativity between nitrogen and hydrogen atoms, causing the electrons to be pulled more towards the nitrogen atom. This results in an unequal distribution of electric charge.

A higher dipole moment indicates a stronger polar characteristic of a molecule. The magnitude of the dipole moment plays an important role in determining the interaction of molecules with external electric fields. This interaction can significantly affect the molecule's energy state, as observed when ammonia molecules change their dipole orientation in an electric field.
Ammonia Molecule
The ammonia molecule (NH\(_3\)) is a common example of a polar molecule with a significant dipole moment. It has a trigonal pyramidal shape with nitrogen at the apex and three hydrogen atoms forming the base of the pyramid.

Ammonia's polarity arises from the lone pair of electrons on the nitrogen atom, which repels the bonds between nitrogen and hydrogen, resulting in an asymmetrical charge distribution. This gives rise to its dipole moment.

When placed in an electric field, the ammonia molecule orients itself so that its dipole moment aligns with the electric field. This alignment can cause changes in energy states due to interactions between the molecule's dipole moment and the applied electric field. The calculated electric potential energy change when the dipole orientation goes from parallel to perpendicular reflects this.
Electric Field
An electric field is a region around a charged particle or object where electric forces can be experienced by other charged particles. It is represented by the symbol \(\vec{E}\) and measured in newtons per coulomb (\(\text{N/C}\)).

The field direction is defined as the direction in which a positive charge would move if placed in the field. In the case of ammonia in the problem, it is subject to a uniform electric field.

Electric fields interact with polar molecules like ammonia, causing them to align based on their dipole moment. This interaction is important for understanding potential energy changes when the molecule's dipole orientation changes relative to the electric field. The parallel alignment of the dipole with the electric field results in a lower potential energy compared to the perpendicular alignment.
Translational Kinetic Energy
Translational kinetic energy refers to the energy associated with the motion of a molecule as it moves from one place to another. It is given by the expression \(\frac{3}{2} kT\), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature in kelvins.
  • \(k = 1.38 \times 10^{-23} \text{ J/K}\)
  • This formula describes how kinetic energy depends directly on the temperature.
When molecules like ammonia are placed in an electric field, their potential energy can change based on orientation. The temperature at which the average translational kinetic energy equals the change in potential energy is crucial. This temperature can be a threshold where thermal motions prevent dipoles from staying aligned with the field, causing them to become randomized at higher temperatures.

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

Two very large parallel sheets are 5.00 cm apart. Sheet \(A\) carries a uniform surface charge density of \(-8.80 \space \mu\)C\(/m^2\), and sheet \(B\), which is to the right of \(A\), carries a uniform charge density of \(-11.6 \space \mu\)C\(/m^2\). Assume that the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet \(A\); (b) 4.00 cm to the left of sheet \(A\); (c) 4.00 cm to the right of sheet \(B\).

If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00 \(\mu\)s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

A nerve signal is transmitted through a neuron when an excess of \(Na^+\) ions suddenly enters the axon, a long cylindrical part of the neuron. Axons are approximately 10.0 \(\mu\)m in diameter, and measurements show that about 5.6 \(\times \space 10^{11} \space Na^+\)ions per meter (each of charge \(+e\)) enter during this process. Although the axon is a long cylinder, the charge does not all enter everywhere at the same time. A plausible model would be a series of point charges moving along the axon. Consider a 0.10-mm length of the axon and model it as a point charge. (a) If the charge that enters each meter of the axon gets distributed uniformly along it, how many coulombs of charge enter a 0.10-mm length of the axon? (b) What electric field (magnitude and direction) does the sudden influx of charge produce at the surface of the body if the axon is 5.00 cm below the skin? (c) Certain sharks can respond to electric fields as weak as 1.0 \(\mu N/C\). How far from this segment of axon could a shark be and still detect its electric field?

Two particles having charges \(q_1 =\) 0.500 nC and \(q_2 =\) 8.00 nC are separated by a distance of 1.20 m. At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?

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.

Sign-up for free