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Chapter 10: Problem 13

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 $$ T(v-b)^{R / \epsilon r}=\text { const. } $$

Short Answer

Expert verified
For an adiabatic process of a van der Waals gas: \ \boxed{T(v - b)^{R / \epsilon} = const}.

Step by step solution

01

Start with the van der Waals Equation

The van der Waals equation is given by \ \( \bigg(P + \frac{a}{v^2} \bigg)(v - b) = RT \).
02

Understand Adiabatic Process

For an adiabatic process, no heat is exchanged with the surroundings. This means that the change in internal energy is equal to the work done on or by the system.
03

Apply the First Law of Thermodynamics

The first law of thermodynamics is \ \( dU = dQ - PdV \). For an adiabatic process, \ \( dQ = 0 \). Thus, \ \( dU = -PdV \). For a van der Waals gas, \ \( dU = c_V dT \).
04

Express Work Done in Terms of van der Waals Equation

From step 1, we have \ \( P = \frac{RT}{v - b} - \frac{a}{v^2} \). Insert this into \ \( -PdV \).
05

Integrate Both Sides

Integrate \ \( dU = c_V dT = -PdV \) using the expression for \ P from step 4. This gives \ \( c_V dT = - \bigg( \frac{RT}{v - b} - \frac{a}{v^2} \bigg)dV \).
06

Separate the Variables and Integrate

Rearrange and integrate: \ \( \int \frac{c_V dT}{T} = \int -R \frac{dV}{v - b} + \int \frac{a dV}{v^2} \). The integrals yield \ \( c_V \ln{T} = -R \ln{(v - b)} + \frac{a}{v} + const \).
07

Combine and Simplify

Rearrange the terms and exponents to simplify: \ \( T(v - b)^{R / c_V} = constant \). Define \ \( \epsilon = c_V / R \) to match the form given: \ \( \boxed{T(v - b)^{R / \epsilon} = const} \).

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Key Concepts

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

van der Waals Equation
The van der Waals equation of state is an improvement over the ideal gas law. It accounts for the volume occupied by gas molecules and intermolecular forces, which the ideal gas law ignores. The equation is given by:

\( \bigg(P + \frac{a}{v^2} \bigg)(v - b) = RT \).

Here:
  • \(P\) is the pressure of the gas.
  • \(v\) is the molar volume (volume per mole) of the gas.
  • \(T\) is the temperature.
  • \(R\) is the universal gas constant.
  • \(a\) accounts for the attractive forces between molecules.
  • \(b\) accounts for the finite volume occupied by the molecules.
Understanding this equation is crucial for realizing how real gases behave under different conditions.
Adiabatic Process
An adiabatic process is one where no heat is exchanged between the system and its surroundings. In such a process, the system's internal energy changes solely due to work done on or by the system. Mathematically, this is represented as:
  • \(dQ = 0\) (no heat transfer)
  • \(dU = -PdV\) (change in internal energy equals negative work done)
These conditions help derive other important thermodynamic relations and are essential for understanding how gases behave when they expand or compress without heat exchange.
First Law of Thermodynamics
The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or transformed. For any thermodynamic process, the change in internal energy \(dU\) of a system is the sum of the heat added to the system \(dQ\) and the work done on the system \(PdV\):

\(dU = dQ - PdV\).

In an adiabatic process where \(dQ = 0\), this simplifies to:

\(dU = -PdV\).

This simplification is crucial when dealing with reactions and processes involving gases, especially when deriving expressions like the one for the adiabatic process for a van der Waals gas.
Internal Energy
Internal energy \(U\) is the total energy contained within a system, arising from both kinetic and potential energies of molecules. For a van der Waals gas during an adiabatic process, the internal energy change \(dU\) is related to the change in temperature and volume. By the equation

\(dU = c_V dT\),

where \(c_V\) is the specific heat at constant volume. This relation simplifies complicated calculations, letting us express the internal energy changes easily during the adiabatic processes as we've seen in the integration steps in the problem.
Work Done
Work done \(W\) by or on a system during a process can be calculated from the integral of pressure over volume changes: \(W = \int PdV\). For the van der Waals gas, pressure \(P\) is expressed as:

\(P = \frac{RT}{v - b} - \frac{a}{v^2}\).

During an adiabatic process, this forms part of the integral needed to solve for temperature and volume relations. By substituting this pressure back into our original integral and simplifying, we derive the final relationship that ties the system's temperature with its volume, adjusted for real-gas behaviors as expressed by the van der Waals equation.

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Most popular questions from this chapter

(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.

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?

(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 $$ \begin{aligned} &Y=-\frac{H}{T}+S=-\frac{G}{T} \\ &d Y=\frac{H}{T^{2}} d T-\frac{V}{T} d P \end{aligned} $$
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