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Chapter 10: Problem 16

(a) Derive the equation $$ \left(\frac{\partial C_{V}}{\partial V}\right)_{T}=T\left(\frac{\partial^{2} P}{\partial T^{2}}\right)_{V} $$ (b) Prove that \(C_{V}\) of an ideal gas is a function of \(T\) only. (c) In the case of a gas obeying the equation of state $$ \frac{P v}{R T}=1+\frac{\beta}{v} $$ where \(B\) is a function of \(T\) only, show that $$ c_{V}=-\frac{R T}{v} \frac{d^{2}}{d T^{2}}(B T)+\left(c_{V}\right)_{0} $$ where \(\left(c_{V}\right)_{0}\) is the value at very large volumes.

Short Answer

Expert verified
The derived expressions are \( \left(\frac{\partial C_{V}}{\box of partial V}\right)_{T}=T\left(\box of partial^2 P/\box of partial T^2 \right)_{V} \), and for an ideal gas \(C_V\) is dependent only on \(T\). For the given state equation, \( c_V = -\frac{R T}{v} \frac{d^2}{d T^2}(B T)+c_{V_{0}}\).

Step by step solution

01

Derive the First Thermodynamic Identity

Start with the thermodynamic identity: \[dU = TdS - PdV\]. For constant volume, \(C_V = \left(\frac{\partial U}{\partial T}\right)_V\). Take the partial derivative with respect to volume at constant temperature to get: \[ \left(\frac{\partial C_V}{\partial V}\right)_T = \left(\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T}\right)_V\right)_T\].
02

Apply Maxwell Relations

Use the Maxwell relation derived from the Helmholtz free energy \(F = U - TS\): \[ dF = -PdV - SdT \]. Applying the Maxwell relation: \[ \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V \]. Take the second derivative: \[ \left( \frac{\partial^2 P}{\partial T^2} \right)_V = \left( \frac{\partial}{\partial T} \left( \frac{\partial S}{\partial V} \right)_T \right)_V \]. We know \( \left( \frac{\partial S}{\partial T} \right)_V = \frac{C_V}{T} \), hence: \[ \left( \frac{\partial}{\partial V} \left( \frac{C_V}{T} \right) \right)_T = \frac{1}{T} \left( \frac{\partial C_V}{\partial V} \right)_T \].
03

Combine and Simplify

Combine these results to get: \[ \frac{1}{T} \left( \frac{\partial C_V}{\partial V} \right)_T = \left (\frac{\partial^2 P}{\partial T^2} \right)_V \]. Hence, \( \left( \frac{\partial C_V}{\partial V} \right)_T = T \left( \frac{\partial^2 P}{\partial T^2} \right)_V \).
04

Prove that \(C_V\) of an Ideal Gas is a function of \(T\) Only

For an ideal gas, internal energy \(U\) is a function of temperature only. Hence, \(C_V = \left(\frac{\partial U}{\partial T}\right)_V\). Since ideal gas assumes no volume dependence on internal energy, it follows that \(C_V\) is a function of the temperature \(T\) only.
05

Derive the Given Equation of State

Given \[ \frac{P v}{R T}=1+\frac{B(T)}{v} \], where \(v\) is the molar volume and \(B(T)\) is a function of temperature. Rewrite as \[ P = \frac{RT}{v} \left( 1 + \frac{B(T)}{v} \right) \].
06

Differentiate and Simplify

First differentiate \(P\) with respect to temperature twice, then use the derived result in step 1. The partial derivatives w.r.t temperature give: \[ \left (\frac{\partial^2 P}{\partial T^2} \right)_V = \frac{R}{v} \frac{d^2}{dT^2} (BT) \]. Using this, we can plug back into the equation from part (a): \[ c_V = - \frac{RT}{v} \frac{d^2}{dT^2} (BT) + c_V(0) \].

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Key Concepts

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

thermodynamic identity
The first thermodynamic identity is a fundamental equation that connects various thermodynamic variables. It is written as: \[ dU = TdS - PdV \]. This equation tells us that the change in internal energy (dU) is related to the work done by the system (PdV) and the heat added to the system (TdS). When the volume is constant, the change in internal energy only depends on the temperature. Hence, for constant volume we can write: \[ C_V = \frac{\text{\textpartial} U}{\text{\textpartial} T} \]. This means that the heat capacity at constant volume (\text(C_V\text)) is the derivative of internal energy (\text(U\text)) with respect to temperature (\text(T\text)) at constant volume.
Maxwell relations
Applying the Maxwell relations in thermodynamics helps interrelate different partial derivatives. For the Helmholtz free energy (\text(F\text)), we write: \[ F = U - TS \] and its differential form: \[ dF = -SdT - PdV \]. The Maxwell relation from this is: \[ \frac{\text{\textpartial} S}{\text{\textpartial} V}\textt = \frac{\text{\textpartial} P}{\text{\textpartial} T}\textt \]. This relationship helps translate between entropy (S) and pressure (P) with respect to volume (V) and temperature (T).
ideal gas
An ideal gas is a theoretical concept where the gas particles have no interactions except elastic collisions and occupy no volume. The ideal gas law is described by the equation of state: \[ PV = nRT \]. This implies that for an ideal gas, the internal energy (\text(U\text)) depends only on temperature (\text(T\text)), not on volume. Consequently, the heat capacity at constant volume (\text(C_V\text)) for an ideal gas is purely a function of temperature. This is because: \[ C_V = \frac{\text{\textpartial} U}{\text{\textpartial} T} \text_v \] where U is the internal energy and T is the temperature.
heat capacity
Heat capacity is a measure of the amount of heat energy required to change the temperature of a substance. There are two types of heat capacities:
  • Heat capacity at constant volume (\text(C_V\text)): This measures the heat energy needed to raise the temperature of a substance while keeping its volume constant. It is defined as: \[ C_V = \frac{\text{\textpartial} U}{\text{\textpartial} T} \text_v \].
  • Heat capacity at constant pressure (\text(C_P\text)): This measures the heat energy required to raise the temperature while maintaining constant pressure. It is defined as: \[ C_P = \frac{\text{\textpartial} H}{\text{\textpartial} T} \text_p \] where H is enthalpy.
Both of these capacities are crucial for understanding how substances store and transfer heat.
equation of state
The equation of state is a mathematical model that describes the state of matter under a given set of physical conditions. It relates pressure (P), volume (V), and temperature (T). For an ideal gas, the equation of state is: \[ PV = nRT \]. However, real gases deviate from this ideal behavior, especially under high pressure or low temperature. For in-depth studies, more complex equations like the Van der Waals equation are used. The modified equation in the problem is given by: \[ \frac{Pv}{RT} = 1 + \frac{B(T)}{v} \], where B(T) is a function of temperature only. This equation helps us understand how real gases behave differently compared to the ideal model.

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Most popular questions from this chapter

(a) Derive the equation $$ \left(\frac{\partial C_{P}}{\partial P}\right)_{T}=-T\left(\frac{\partial^{2} V}{\partial T^{2}}\right)_{P} $$ (b) prove that \(C_{P}\) of an ideal gas is a function of \(T\) only. (c) In the case of a gas obeying the equation of state $$ P v=R T+B P $$ where \(B\) is a function of \(T\) only, show that $$ c_{P}=-T \frac{d^{2} B}{d T^{2}} P+\left(c_{P}\right)_{0} $$ where \(\left(c_{P}\right)_{0}\) is the value at very low pressures.

Another set of characteristic functions for a single-substance system can be defined by performing the Legendre transformations on the entropy \(S(U, V)\) rather than on the internal energy \(U(V, S)\). The thermodynamic potentials turn out to be particularly useful in statistical mechanics and the theory of irreversible thermodynamics, in contrast to equilibrium thermodynamics presented in this book. (a) Show that Legendre transformation of \(S(U, V)\) that produces the characteristic function \(J(1 / T, V)\), known as the Massieu function, is given by the transform $$ J=-\frac{U}{T}+S=-\frac{A}{T}, $$ and $$ d J=\frac{U}{T^{2}} d T+\frac{P}{T} d V $$ (b) Show that Legendre transformation of \(J(1 / T, V)\) that produces the thermodynamic potential \(Y(1 / T, P / T)\), known as the Planck function, is defined by the transform and $$ \begin{aligned} &Y=-\frac{H}{T}+S=-\frac{G}{T} \\ &d Y=\frac{H}{T^{2}} d T-\frac{V}{T} d P \end{aligned} $$

From the differential equation for the thermodynamic potential \(A(T, V)\), derive expressions for pressure \(P\), entropy \(S\), internal energy \(U\), heat capacity at constant volume \(C_{y}\), heat capacity at constant pressure \(C_{P}\), volume expansivity \(\beta\), and isothermal compressibility \(\kappa\).

The pressure on \(500 \mathrm{~g}\) of copper is increased reversibly and isothermally from 0 to \(5000 \mathrm{~atm}\) at \(298 \mathrm{~K}\). (Take the density \(\rho=8.96 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\), volume expansivity \(\beta=49.5 \times 10^{-6} \mathrm{~K}^{-1}\), isothermal compressibility \(\kappa=6.18 \times 10^{-12} \mathrm{~Pa}^{-1}\), and specific heat \(c_{P}=385 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) to be constant.) (a) How much heat is transferred during the compression? (b) How much work is done during the compression? (c) Determine the change of internal energy. (d) What would have been the rise of temperature if the copper had been subjected to a reversible adiabatic compression?

Show that for a gas obeying the van der Waals equation \(\left(P+a / v^{2}\right)(v-b)=R T\), with \(c_{V}\) a function of \(T\) only, an equation for an adiabatic process is $$ T(v-b)^{R / \epsilon r}=\text { const. } $$
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