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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:\[T = \frac{2 \times 8.0 \times 10^{-24} \text{ J}}{3 \times 1.38 \times 10^{-23} \text{ J/K}} \approx 38.5 \text{ K}\]

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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.

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Most popular questions from this chapter

Point charges \(q_1 = -\)4.5 nC and \(q_2 = +\)4.5 nC are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of 36.9\(^\circ\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9}\) N \(\cdot\) m ?

The earth has a net electric charge that causes a field at points near its surface equal to 150 N\(/\)C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth's electric field a feasible means of flight? Why or why not?

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.4 \(\times \space 10^{-8}\) g. The drops will leave the nozzle and travel toward the paper at 50 m\(/\)s, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2.0 cm long, where there is a uniform vertical electric field with magnitude 8.0 \(\times \space 10^4 \space N/C\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 \(m/s\), what \(q\) value is needed to achieve the same 0.30-mm deflection?

Three point charges are arranged on a line. Charge \(q_3 = +\)5.00 nC and is at the origin. Charge \(q_2 = -\)3.00 nC and is at \(x = +\)4.00 cm. Charge \(q_1\) is at \(x = +\)2.00 cm. What is \(q_1\) (magnitude and sign) if the net force on \(q_3\) is zero?

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