/*! 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} Problem 56 The ammonia molecule \(\left(\ma... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 56

The ammonia molecule \(\left(\mathrm{NH}_{3}\right)\) has a dipole moment of \(5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) Ammonia molecules in the gas phase are placed in a a uniform electric field \(\vec{\boldsymbol{E}}\) with magnitude \(1.6 \times\) \(10^{6} \mathrm{N} / \mathrm{C} .\) (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\vec{\boldsymbol{E}}\) from parallel to perpendicular? (b) At what absolute temperature \(T\) is the average translational kinetic energy \(\frac{3}{2} k T\) 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) \(8.0 \times 10^{-24}\) J; (b) \(3.9\) K.

Step by step solution

01

Calculate the Change in Potential Energy

The potential energy of a dipole in an electric field is given by: \( U = -oldsymbol{p} imes oldsymbol{E} \cos\theta \), where \( \theta \) is the angle between the dipole moment \( \boldsymbol{p} \) and the electric field \( \boldsymbol{E} \). For parallel orientation, \( \theta = 0 \) and \( \cos\theta = 1 \), thus \( U_{parallel} = -pE \). For perpendicular orientation, \( \theta = 90^\circ \) and \( \cos\theta = 0 \), thus \( U_{perpendicular} = 0 \). The change in potential energy is therefore: \( \Delta U = U_{perpendicular} - U_{parallel} = 0 + pE = pE. \) Substitute \( p = 5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} \) and \( E = 1.6 \times 10^{6} \mathrm{N}/\mathrm{C} \) to find \( \Delta U: \) \[ \Delta U = 5.0 \times 10^{-30} \times 1.6 \times 10^6 = 8.0 \times 10^{-24} \mathrm{J} \].
02

Set Up the Equation for Temperature

The average translational kinetic energy of a molecule is given by \( \frac{3}{2} kT \), where \( k \) is the Boltzmann constant, \( 1.38 \times 10^{-23} \mathrm{J}/\mathrm{K} \). Set \( \frac{3}{2} kT = \Delta U \) from Step 1. Solve for \( T \): \[ \frac{3}{2} (1.38 \times 10^{-23}) T = 8.0 \times 10^{-24} \].
03

Solve for Absolute Temperature

Rearrange the equation from Step 2 to solve for \( T \): \( T = \frac{8.0 \times 10^{-24}}{\left(\frac{3}{2} \times 1.38 \times 10^{-23}\right)} \). Simplifying, we find: \[ T = \frac{8.0 \times 10^{-24}}{2.07 \times 10^{-23}} = 0.39 \times 10^1 \approx 3.9 \mathrm{K} \]. Thus, the absolute temperature is approximately 3.9 Kelvin.

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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 vector quantity that measures the separation of positive and negative charges within a system. It represents how much electric polarity a molecule has.
Think of it like a tiny magnet within the molecule.
  • In molecules, dipole moments arise from differences in electronegativity between atoms, causing one end to be slightly positive and the other slightly negative.
  • The dipole moment is calculated using the formula: \( \mathbf{p} = q \cdot \mathbf{d} \), where \( q \) is the charge and \( \mathbf{d} \) is the separation distance.
  • For ammonia (\(\text{NH}_3\)), the dipole moment is given as \(5.0 \times 10^{-30} \text{C} \cdot \text{m}\).
The direction of the dipole moment is from the negative charge to the positive charge, much like the direction of an electric field created by these charges.
Potential Energy
Potential energy in the context of an electric dipole placed in an electric field refers to the energy stored due to the dipole's orientation regarding the field.
The potential energy depends on the angle \( \theta \) between the dipole moment \( \mathbf{p} \) and the electric field \( \mathbf{E} \).
  • The formula for potential energy \( U \) is: \( U = - \mathbf{p} \cdot \mathbf{E} \cos\theta \).
  • When the dipole is aligned parallel to the field, \( \theta = 0 \) making \( U = -pE \).
  • When perpendicular, \( \theta = 90^\circ \), yielding \( U = 0 \).
The change in potential energy when moving from a parallel (minimum energy) to a perpendicular orientation is critical in understanding the dipole's behavior in varying fields and is calculated to be \(8.0 \times 10^{-24} \text{J}\).
Electric Field
An electric field is a vector field around a charged object where a force would be exerted on other charged objects. It is experienced by charged entities, like dipoles, whenever they are placed within it.
  • An electric field \( \mathbf{E} \) quantifies the electric force per unit charge exerted on a tiny positive test charge placed at that point.
  • In this scenario, the field strength is \(1.6 \times 10^{6} \text{N/C}\).
  • It influences the orientation and energy of dipoles like \(\text{NH}_3\) molecules within it.
Understanding the electric field's impact on dipoles helps predict how molecules behave in external electric fields.
Translational Kinetic Energy
Translational kinetic energy in molecular terms refers to the energy possessed due to the movement of the molecules from one location to another.
It is a measure of the thermal energy of a system and increases with higher temperatures.
  • The average translational kinetic energy \( KE \) is given by: \( KE = \frac{3}{2} kT \), where \( k \) is the Boltzmann constant \(1.38 \times 10^{-23} \text{J/K}\).
  • In this exercise, we solve for the temperature \( T \) at which the kinetic energy equals the potential energy change (\(8.0 \times 10^{-24} \text{J}\)).
  • Solving yields \( T \approx 3.9 \text{K} \), suggesting that above this temperature, thermal energy disrupts dipole alignment.
The balance between kinetic energy and potential energy is crucial for understanding when external factors, like temperature, can overcome orientation forces in a molecule.

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Most popular questions from this chapter

BI0 Signal Propagation in Neurons. Neurons are components of the nervous system of the body that transmit signals as electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the neuron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e\) , enter the axon. How many coulombs of charge enter a \(1.5-\mathrm{cm}\) length of the axon during this process?

CP A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\) . The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{m} / \mathrm{s},\) and \(\alpha=30.0^{\circ} .\)

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) -coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm}\) find the \(x\) - and \(y\) -components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

\(\mathrm{A}-5.00\) -nC point charge is on the \(x\) -axis at \(x=1.20 \mathrm{m.}\) A second point charge \(Q\) is on the \(x\) -axis at \(-0.600 \mathrm{m} .\) What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -x-direction, (b) 45.0 \(\mathrm{N} / \mathrm{C}\) in the - \(x\) -direction?

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) . (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torque the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?
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