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Chapter 3: Problem 24

Show that, for a perfect gas, \((\partial U / \partial S)_{v}=T\) and \((\partial U / \partial V)_{\mathrm{S}}=-p\).

Short Answer

Expert verified
For a perfect gas, \(\frac{\partial U}{\partial S}\bigg|_V = T\) as temperature is the mode of energy change at constant volume, and \(\frac{\partial U}{\partial V}\bigg|_S = -p\) as pressure contributes to energy change at constant entropy.

Step by step solution

01

Title - Understand the Concepts

We start by recalling that the internal energy change dU for a perfect gas can be represented in terms of entropy S and volume V as a differential equation: dU = TdS - pdV. The partial derivatives \(\frac{\partial U}{\partial S}\bigg|_V\) and \(\frac{\partial U}{\partial V}\bigg|_S\) represent how internal energy U changes with entropy S and volume V, respectively. The former is taken at constant volume and the latter at constant entropy.
02

Title - Calculate \(\frac{\partial U}{\partial S}\bigg|_V\)

Differentiate the internal energy U with respect to entropy S at constant volume (V). From the fundamental thermodynamic relation for a perfect gas, dU = TdS - pdV, we note that at constant volume, the term - pdV drops out as dV=0, leaving us with dU = TdS. Therefore, the partial derivative of U with respect to S at constant volume is \(\frac{\partial U}{\partial S}\bigg|_V = T\), where T is the temperature.
03

Title - Calculate \(\frac{\partial U}{\partial V}\bigg|_S\)

Similarly, when differentiating the internal energy U with respect to volume V at constant entropy (S), the term TdS drops since dS=0, leaving dU = -pdV. The partial derivative of U with respect to V at constant entropy is therefore \(\frac{\partial U}{\partial V}\bigg|_S = -p\), where p is the pressure.

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Key Concepts

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

Internal Energy and Entropy
Understanding the relationship between internal energy and entropy is fundamental for students tackling thermodynamics, especially within the context of perfect gases. Internal energy, denoted as U, is the total energy contained within a system, which in a perfect gas only depends on temperature. Entropy, symbolized by S, is a measure of a system's disorder - how energy is spread out among particles.

For a perfect gas, an increase in entropy typically means energy is more dispersed within the system. Since the internal energy of a perfect gas is dependent on temperature, which in turn is influenced by how energy is distributed (entropy), there is a direct link between the two. To clarify, when entropy increases at a constant volume, the system must absorb heat - hence, the temperature and internal energy increase correspondingly.
Thermodynamic Partial Derivatives
Thermodynamic partial derivatives are immensely valuable in understanding how changing one variable affects another while holding a different variable constant. In the context of perfect gases, they highlight how changes in entropy or volume can influence internal energy.

A partial derivative of the internal energy with respect to entropy or volume tells us how internal energy will change as entropy or volume change, independently of one another. The formula \( (\frac{\text{\textpartial} U}{\text{\textpartial} S})_{V} = T \) reveals that at a constant volume, any change in entropy directly alters the temperature. Conversely, \( (\frac{\text{\textpartial} U}{\text{\textpartial} V})_{S} = -p \) explains that at a constant entropy, any change in volume inversely changes the pressure within the system.
Fundamental Thermodynamic Relation
The fundamental thermodynamic relation for a perfect gas, expressed as \( dU = TdS - pdV \) serves as an essential equation in understanding thermodynamics. This equation ties together internal energy (U), temperature (T), entropy (S), pressure (p), and volume (V).

It illustrates how changes in entropy and volume can result in changes to the internal energy of a gas. If the volume remains constant, only the entropy part \( TdS \) contributes to the change in internal energy, whereas if the entropy remains constant, only the volume part \( -pdV \) affects the change. This relation helps in painting a clearer picture of how thermodynamic processes occur at the molecular level.
Entropy and Volume Relationship
The relationship between entropy and volume is particularly intriguing in the realm of perfect gases. As volume increases while temperature remains constant, it can be deduced that the system's entropy must also increase. Why? Because the gas molecules have more space to occupy, which increases the number of ways the molecules can be arranged.

This contributes to the 'disorder' of the system, a core aspect of entropy. When we talk about partial derivatives in this context, it's about how tweaking volume鈥攚hile maintaining constant entropy鈥攃an impact other thermodynamic properties. This relationship emphasizes the interconnectivity of thermodynamic variables and is critical for students to grasp for a deeper understanding of how gases behave under different conditions.

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Most popular questions from this chapter

The cycle involved in the operation of an internal combustion engine is called the Otto cycle. Air can be considered to be the working substance and can be assumed to be a perfect gas. The cycle consists of the following steps: (1) reversible adiabatic compression from \(\mathrm{A}\) to \(\mathrm{B},(2)\) reversible constantvolume pressure increase from \(\mathrm{B}\) to \(\mathrm{C}\) due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from \(\mathrm{C}\) to \(\mathrm{D}\), and \((4)\) reversible and constant-volume pressure decrease back to state A. Determine the change in entropy (of the system and of the surroundings) tor each step of the cycle and determine an expression for the efficiency of the cycle, assuming that the heat is supplied in Step \(2 .\) Evaluate the efficiency for a compression ratio of \(10: 1\). Assume that, in state \(\mathrm{A}, \mathrm{V}=4.00 \mathrm{dm}^{3}, p=1.00 \mathrm{~atm}\), and \(\mathrm{T}=300 \mathrm{~K}\), that \(\mathrm{V}_{\mathrm{A}}==10 \mathrm{~V}_{\mathrm{B}}, \mathrm{p}_{\mathrm{C}} / \mathrm{p}_{\mathrm{B}}=5\), and that \(C_{p m}=\frac{2}{2} R\).

The adiabatic compressibility, \(\kappa_{S}\), is defined like \(\kappa_{T}\) (eqn 2.44) but at constant entropy. Show that for a perfect gas \(p \gamma \kappa_{\mathrm{S}}=1\) (where \(\gamma\) is the ratio of heat capacities).

Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into two sections \(\mathrm{A}\) and \(\mathrm{B}\). All changes in \(\mathrm{B}\) is isothermal; that is, a thermostat surrounds \(\mathrm{B}\) to keep its temperature constant. There is \(2.00 \mathrm{~mol}\) of the gas in each section. Initially, \(T_{\mathrm{A}}==T_{\mathrm{B}}=300 \mathrm{~K}, V_{\mathrm{A}}=\) \(V_{\mathrm{B}}\) \(=2.00 \mathrm{dm}^{3}\). Energy is supplied as heat to Section A and the piston moves to the right reversibly until the final volume of Section B is \(1.00 \mathrm{dm}^{3}\). Calculate (a) \(\Delta S_{\mathrm{A}}\) and \(\Delta S_{\mathrm{B}}\), (b) \(\Delta A_{\mathrm{A}}\) and \(\Delta \mathrm{A}_{\mathrm{B}}\), (c) \(\Delta G_{\mathrm{A}}\) and \(\Delta G_{\mathrm{B}}\), (d) AS of the total system and its surroundings. If numerical values cannot be obtained, indicate whether the values should be positive, negative, or zero or are indeterminate from the information given. (Assume \(C_{\mathrm{v}, \mathrm{m}}=20 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\).)

To calculate the work required to lower the temperature of an object, we need to consider how the coefficient of performance changes with the temperature of the object. (a) Find an expression for the work of cooling an object from \(\mathrm{T}\), to \(\mathrm{T}_{\mathrm{r}}\) when the refrigerator is in a room at a temperature \(\mathrm{T}_{\mathrm{h}}\). Hint. Write \(d w=d y l c\left(T\right.\) relate \(d q\) to \(\mathrm{dT}\) through the heat capacity \(\mathrm{C}_{p}\), and integrate the resulting expression. Assume that the heat capacity is independent of temperature in the range of interest. (b) Use the result in part (a) to calculate the work needed to freeze \(250 \mathrm{~g}\) of water in a refrigerator at \(293 \mathrm{~K}\). How long will it take when the refrigerator operates at \(100 \mathrm{~W}\) ?

A gaseous sample consisting of \(1.00 \mathrm{~mol}\) molecules is described by the equation of state \(p \mathrm{~V}_{\mathrm{m}}=\mathrm{RT}(1+B p)\). Initially at \(373 \mathrm{~K}\), it undergoes JouleThomson expansion from \(100 \mathrm{~atm}\) to \(1.00 \mathrm{~atm}\). Given that \(C_{p, \mathrm{~m}}=R, \mu=0.21\) \(\mathrm{K} \mathrm{atm}^{-1}, B=-0.525(\mathrm{~K} / T) \mathrm{atm}^{-1}\), and that these are constant over the temperature range involved, calculate \(\Delta T\) and \(\Delta S\) for the gas.
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