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Chapter 3: Problem 25

What is the generalization of $$ d E=T d S-p d V+\mu d N $$ for systems that can be lifted and tossed? (That is, for systems in which height \(h\) and velocity \(\mathbf{v}\) are mechanical parameters.)

Short Answer

Expert verified
Include terms for height and velocity: \( dE = T dS - p dV + \mu dN + mg dh + m\mathbf{v} \cdot d\mathbf{v} \).

Step by step solution

01

Understanding the Given Equation

The given equation \(dE = T dS - p dV + \mu dN\) represents the change in energy for a thermodynamic system in terms of temperature \(T\), entropy \(S\), pressure \(p\), volume \(V\), chemical potential \(\mu\), and the number of particles \(N\).
02

Identifying Additional System Parameters

For systems that can be lifted and tossed, two new mechanical parameters are introduced: height \(h\), which is associated with gravitational potential energy, and velocity \(\mathbf{v}\), which is associated with kinetic energy.
03

Including the Gravitational Potential Energy Term

The gravitational potential energy change is represented by the term \(mgh\), where \(m\) is the mass and \(g\) is the acceleration due to gravity. When considering changes, this becomes \(d(mgh) = mg dh\).
04

Including the Kinetic Energy Term

The change in kinetic energy is represented by the term \(\frac{1}{2} m v^2\). The differential form is \(d\left( \frac{1}{2} m v^2 \right) = m\mathbf{v} \cdot d\mathbf{v}\), which can be obtained using velocity \(\mathbf{v}\) and its change \(d\mathbf{v}\).
05

Generalizing the First Law of Thermodynamics

Including both gravitational potential energy and kinetic energy terms, the generalized energy change for the system becomes:\[dE = TdS - pdV + \mu dN + mg dh + m\mathbf{v} \cdot d\mathbf{v}\]

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Key Concepts

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

mechanical parameters in thermodynamics
In thermodynamics, mechanical parameters help describe additional forms of energy that a system might possess or interact with. Traditional thermodynamic variables include temperature (T), entropy (S), pressure (p), and volume (V), which are used to express changes in a system's internal energy. However, certain systems also involve mechanical aspects like height and velocity, which contribute to broader interpretations of energy such as gravitational potential energy and kinetic energy.

When analyzing such systems, it's essential to incorporate these mechanical parameters to accurately describe the total energy changes. For instance, when objects are lifted, thrown, or move within a gravitational field, both the potential and kinetic forms of mechanical energy become significant.
gravitational potential energy
Gravitational potential energy is the energy held by an object because of its position relative to a gravitational source. It is fundamentally about the energy stored due to an object's height.

The formula to calculate gravitational potential energy is:
  • \[ PE_{gravity} = mgh \]
  • Where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and \( h \) is the height above a reference point.
As a system changes in height, its gravitational potential energy changes, which corresponds to the term \( mg\, dh \) in differential form. This addition allows us to account for shifts in energy when a system is elevated within a gravitational field.
kinetic energy in thermodynamics
Kinetic energy is the energy a system possesses due to its motion. It's an integral part of mechanical parameters when considering energy changes in a thermodynamic context.

The kinetic energy of an object can be calculated using the formula:
  • \[ KE = \frac{1}{2} mv^2 \]
  • Where \( m \) is the mass and \( v \) is the velocity of the object.
In a thermodynamic system where velocity can change, this energy change is represented by the term \( m\mathbf{v} \cdot d\mathbf{v} \) in the equation. This differential expression elaborates how the object's energy transforms as its velocity varies, encompassing the dynamic nature of kinetic energy in thermodynamics.

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Most popular questions from this chapter

The entropy of a newly discovered gas is determined to be $$ S(E, V, N)=\left(\frac{k_{B}}{T_{0}}\right)^{1 / 2}\left(N E+\frac{k_{B} T_{0} V^{2}}{v_{0}^{2}}\right)^{1 / 2} $$ where the constants \(T_{0}\) and \(v_{0}\) are positive, intensive quantities with the dimensions of temperature and volume, respectively. a. Verify that this equation is acceptable in its dimensions and in its extensivity. b. Find \(T(S, V, N)\) and \(p(S, V, N)\) for this gas. c. Find \(C_{V}(T, V, N)\) for this gas.

Heat capacity at constant pressure Equation (3.131) shows that the heat capacity at constant volume, which is defined in terms of an entropy derivative, is also equal to an energy derivative. You might suspect a similar relation between \(C_{p}(T, p, N)\) and $$ \left.\frac{\partial E(T, p, N)}{\partial T}\right)_{p, N} $$ Show that such a suspicion is not correct, and instead find an expression for \(C_{p}\) in terms of a derivative of enthalpy.

Creationists sometimes claim that the second law of thermodynamics prohibits biological evolution. a. The surface of the Sun (mean temperature \(5778 \mathrm{~K}\) ) heats the surface of the Earth (mean temperature \(288 \mathrm{~K}\) ) through visible and near-infrared radiation. The solar energy absorbed by the Earth each second is \(1.732 \times 10^{17} \mathrm{~J}\). What is the entropy change per second (due to this process) of the Sun? The Earth? Does the entropy of "Sun plus Earth" increase or decrease? b. Yet the mean temperature of the Earth changes slowly, if at all. This is because almost all of the solar energy absorbed by the Earth is then emitted through far-infrared radiation which in turn heats "outer space" - the cosmic microwave background (CMB; temperature \(2.728 \mathrm{~K}\) ). What is the entropy change per second (due to this process) of the Earth? The CMB? Does the entropy of "Earth plus CMB" increase or decrease? c. Now refine the model by supposing that, due to evolution, the entropy of the Earth is not exactly constant, but is decreasing. (In this case the entropy of the CMB would have to be increasing faster than rate predicted in part (b).) Suppose that, due to evolution, each individual organism is 1000 times "more improbable" than the corresponding individual was 100 years ago. In other words, if \(\Omega_{i}\) is the number of microstates consistent with the specification of an organism 100 years ago, and if \(\Omega \mathrm{f}\) is the number of microstates consistent with the specification of today's "improved and less probable" organism, then \(\Omega_{f}=10-3 \Omega_{i}\). What is the corresponding change in entropy per organism? d. The population of Earth is about 1018 eukaryotic individuals and 1032 prokaryotic individuals. If the estimate of part (c) holds for each one of them, what is the change in entropy due to evolution each second? e. How accurately would you have to measure the entropy flux of part (b) in order to notice the diversion of entropy flux calculated in part (d)? Has any scientific quantity ever been measured to this accuracy? f. It is generally agreed that the greatest rate of evolution fell during the Cambrian period, from 542 million years ago to 488 million years ago. During this so-called "Cambrian explosion" multicellular organisms first formed and then radiated into remarkable variety. Suppose that during the Cambrian period entropy was diverted into the evolution of living things at the rate calculated in part (d). And suppose that at the end of the Cambrian there were \(10^{18}\) multicellular individuals. How much "improved and less probable" would each organism be, relative to its single-celled ancestor at the beginning of the Cambrian period? The moral of the story? There's plenty of entropy to go around.

Stumbling in the thermodynamic dance a. From the thermodynamic assembly with variables \(T, p\), and \(N\), with master equation $$ d G=-S d T+V d p+\mu d N $$ one is tempted to produce an assembly with variables \(T, p\), and \(\mu\), using master function $$ \Phi=G-\mu N $$ Why does this fail? b. Show that what is really desired is an intensive-only description in terms of the variables \(\mathrm{T}\) and \(\mathrm{p}\), for which the master equation is $$ d \mu=-s d T+v d p $$ where \(s=S / N\) and \(v=V / N\). (This result is called the "Gibbs-Duhem equation".)

Which variables to exchange? In carrying out the thermodynamic dance, we have exchanged the variable \(S\) for the function \(T(S, V, N)\), and we have exchanged the variable \(\mathrm{V}\) for the function \(p(T, V, N)\), and more. Why would it never be appropriate to exchange the volume for the temperature? (Clue: Consider the volume of water at temperatures just above and below freezing.)
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