/*! 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 47 The isothermal compressibility, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The isothermal compressibility, \(\kappa_{T},\) is defined by \(\kappa_{T}=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\) Because \((\partial P / \partial V)_{T}=0\) at the critical point, \(\kappa_{T}\) diverges there. A question that has generated a great deal of experimental and theoretical research is the question of the manner in which \(\kappa_{T}\) diverges as \(T\) approaches \(T_{\mathrm{c}} .\) Does it diverge as \(\ln \left(T-T_{\mathrm{c}}\right)\) or perhaps \(\operatorname{as}\left(T-T_{\mathrm{c}}\right)^{-\gamma}\) where \(\gamma\) is some critical exponent? An early theory of the behavior of thermodynamic functions such as \(\kappa_{T}\) very near the critical point was proposed by van der Waals, who predicted that \(\kappa_{T}\) diverges as \(\left(T-T_{\mathrm{c}}\right)^{-1} .\) To see how van der Waals arrived at this prediction, we consider the (double) Taylor expansion of the pressure \(P(\bar{V}, T)\) about \(T_{\mathrm{c}}\) and \(V_{\mathrm{c}}\): \(P(\bar{V}, T)=P\left(\bar{V}_{\mathrm{c}}, T_{\mathrm{c}}\right)+\left(T-T_{\mathrm{c}}\right)\left(\frac{\partial P}{\partial T}\right)_{\mathrm{c}}+\frac{1}{2}\left(T-T_{\mathrm{c}}\right)^{2}\left(\frac{\partial^{2} P}{\partial T^{2}}\right)_{\mathrm{c}}\) \(+\left(T-T_{\mathrm{c}}\right)\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)\left(\frac{\partial^{2} P}{\partial V \partial T}\right)_{\mathrm{c}}+\frac{1}{6}\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{3}\left(\frac{\partial^{3} P}{\partial \bar{V}^{3}}\right)_{\mathrm{c}}+\cdots\) Why are there no terms in \(\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)\) or \(\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{2} ?\) Write this Taylor series as \(P=P_{\mathrm{c}}+a\left(T-T_{\mathrm{c}}\right)+b\left(T-T_{\mathrm{c}}\right)^{2}+c\left(T-T_{\mathrm{c}}\right)\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)+d\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{3}+\cdots\) Now show that \(\left(\frac{\partial P}{\partial \bar{V}}\right)_{T}=c\left(T-T_{\mathrm{c}}\right)+3 d\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{2}+\cdots \quad\left(\begin{array}{c}T \rightarrow T_{\mathrm{c}} \\ \bar{V} \rightarrow \bar{V}_{\mathrm{c}}\end{array}\right)\) and that \(\kappa_{T}=\frac{-1 / \bar{V}}{c\left(T-T_{\mathrm{c}}\right)+3 d\left(\bar{V}-\bar{V}_{\mathrm{c}}\right)^{2}+\cdots}\) Now let \(\bar{V}=\bar{V}_{\mathrm{c}}\) to obtain \(\kappa_{T} \propto \frac{1}{T-T_{\mathrm{c}}} \quad T \rightarrow\left(T_{\mathrm{c}}\right)\) Accurate experimental measurements of \(\kappa_{T}\) as \(T \rightarrow T_{\mathrm{c}}\) suggest that \(\kappa_{T}\) diverges a little more strongly than \(\left(T-T_{\mathrm{c}}\right)^{-1}\). In particular, it is found that \(\kappa_{T} \rightarrow\left(T-T_{\mathrm{c}}\right)^{-\gamma}\) where \(\gamma=1.24\) Thus, the theory of van der Waals, although qualitatively correct, is not quantitatively correct.

Short Answer

Expert verified
\(\kappa_{T}\) diverges as \((T-T_{\mathrm{c}})^{-1}\) according to van der Waals, but experiments show \(\gamma = 1.24\).

Step by step solution

01

Understand the Taylor Expansion

The pressure function, \(P(\bar{V}, T)\), is expanded about the critical values \(T_{\mathrm{c}}\) and \(V_{\mathrm{c}}\) using a double Taylor series. The given equation includes coefficients representing different partial derivatives evaluated at the critical point. It does not include terms with \((\bar{V}-\bar{V}_{\mathrm{c}})\) or \((\bar{V}-\bar{V}_{\mathrm{c}})^{2}\). This indicates that first and second-order derivatives of pressure with respect to \(\bar{V}\) vanish at the critical point.
02

Simplify the Pressure Equation

The equation is simplified to \(P = P_{\mathrm{c}} + a(T-T_{\mathrm{c}}) + b(T-T_{\mathrm{c}})^2 + c(T-T_{\mathrm{c}})(\bar{V}-\bar{V}_{\mathrm{c}}) + d(\bar{V}-\bar{V}_{\mathrm{c}})^3 + \cdots\). Here, \(a, b, c,\) and \(d\) represent the coefficients of the Taylor series expansion, derived from partial derivatives of the pressure at the critical point.
03

Calculate the Partial Derivative \(\left(\frac{\partial P}{\partial \bar{V}}\right)_{T}\)

Differentiate the simplified pressure equation concerning \(\bar{V}\) at constant temperature \(T\), which results in \(\left(\frac{\partial P}{\partial \bar{V}}\right)_{T} = c(T-T_{\mathrm{c}}) + 3d(\bar{V}-\bar{V}_{\mathrm{c}})^2 + \cdots\). As both \(T\to T_{\mathrm{c}}\) and \(\bar{V}\to \bar{V}_{\mathrm{c}}\), the dominant term involves \(c(T-T_{\mathrm{c}})\).
04

Derive the Expression for \(\kappa_{T}\)

Using the definition of isothermal compressibility, \(\kappa_{T} = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\), substitute the calculated partial derivative. This gives \(\kappa_{T} = \frac{-1 / \bar{V}}{c(T-T_{\mathrm{c}}) + 3d(\bar{V}-\bar{V}_{\mathrm{c}})^2 + \cdots}\). By simplifying for \(\bar{V} = \bar{V}_{\mathrm{c}}\), you obtain \(\kappa_{T} \propto \frac{1}{T-T_{\mathrm{c}}}\).
05

Compare with Experimental Observations

Experimental findings indicate \(\kappa_{T} \rightarrow (T-T_{\mathrm{c}})^{-\gamma}\), where \(\gamma = 1.24\). Although van der Waals predicted a divergence similar to \((T-T_{\mathrm{c}})^{-1}\), experiments suggest a more pronounced divergence, with \(\gamma > 1\). The theory of van der Waals aligns qualitatively but not quantitatively with the experimental results.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Critical Point
In the study of phase transitions, the critical point represents a set of conditions—specifically a temperature, pressure, and volume—at which distinct liquid and gas phases cease to exist. It is a unique point where the properties of a substance undergo dramatic changes. As you approach the critical point, the properties of substances become indistinguishable, transitioning seamlessly from one phase to another.

The critical point is crucial in understanding the behavior of isothermal compressibility, \(\kappa_{T}\), which measures how much the volume of a substance decreases under pressure, at constant temperature. At the critical point, the compressibility becomes infinite, which means the substance is extremely responsive to pressure changes. This infinite compressibility is due to \(\left(\frac{\partial P}{\partial V}\right)_{T} = 0\), indicating a horizontal tangent on the P-V curve at this point.

Knowing how to analyze behavior around the critical point is vital for comprehending complex phase transformations, including those used in designing industrial processes and equipment.
Van der Waals Theory
Dutch physicist Johannes Diderik van der Waals developed an equation of state to describe the properties of real gases, which are more accurate than the ideal gas law. His theory considers molecular attraction and finite molecular size, offering insights into phase transitions between liquids and gases.

The Van der Waals equation provides a framework not only for predicting the existence of a critical point but also for understanding how physical properties like isothermal compressibility behave near this critical point. According to van der Waals, as the temperature \(\(T\)\) approaches the critical temperature \(\(T_{c}\)\), the isothermal compressibility diverges following a \(\left(T - T_{c}\right)^{-1}\) pattern.

Despite its qualitative accuracy, van der Waals' predictions do not always match experimental observations. For instance, the divergence exponent he predicted differs slightly from what is observed in laboratory conditions, highlighting the complexity of accurately modeling critical phenomena. This divergence from reality sparks continual research and refinement of theories to better capture the nuances of critical behavior.
Taylor Series Expansion
The Taylor series is a powerful mathematical tool that allows functions to be expressed as infinite sums of terms calculated from derivatives at a single point. In the context of isothermal compressibility and critical points, a Taylor series expansion helps us approximate the pressure function \(\(P(\bar{V}, T)\)\) around its critical values \(\(T_c\)\) and \(\(V_c\)\).

The double Taylor expansion of pressure involves expressing the changes in pressure in terms of changes from these critical temperatures and volumes. This approach highlights that certain terms, like those with \(\(\bar{V} - \bar{V}_c\)\) or \(\(\bar{V} - \bar{V}_c\)^2\), disappear at the critical point, simplifying the calculations.

The simplified Taylor series approximation enables the determination of the dominant behaviors of the system at conditions near the critical point. By capturing only essential terms, it offers clearer insights into how pressure, volume, and temperature interplay. Thus, it is crucial for modulating the physical models that describe critical phenomena in real-world applications.
Critical Exponent
Critical exponents are numerical descriptors that characterize the behavior of physical quantities near critical points. Specifically, they describe how properties like isothermal compressibility, heat capacity, and magnetic susceptibility diverge as the system approaches its critical point.

In the case of isothermal compressibility \(\kappa_{T}\), as the temperature approaches the critical temperature \(\(T_c\)\), it diverges with a critical exponent \({\gamma}\). Van der Waals' theory predicted this exponent as 1, suggesting a \(\left(T - T_{c}\right)^{-1}\) divergence. However, experimental observations suggest a more intense divergence, with a value of \(\gamma = 1.24\).

This slight discrepancy indicates that although Van der Waals' theory qualitatively describes the divergence, it doesn't precisely quantify it correctly. The discovery and calculation of accurate critical exponents are essential for improving theoretical models, allowing them to align more closely with experimental data, ultimately advancing our understanding of critical phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The vapor pressures of solid and liquid hydrogen iodide can be expressed empirically as \(\ln \left(P^{\mathrm{s}} / \mathrm{torr}\right)=-\frac{2906.2 \mathrm{K}}{T}+19.020\) and \(\ln \left(P^{\prime} / \text { torr }\right)=-\frac{2595.7 \mathrm{K}}{T}+17.572\) Calculate the ratio of the slopes of the solid-gas curve and the liquid-gas curve at the triple point.

The pressures at the solid-liquid coexistence boundary of propane are given by the empirical equation $$ P=-718+2.38565 T^{1.283} $$ where \(P\) is in bars and \(T\) is in kelvins. Given that \(T_{\text {fus }}=85.46 \mathrm{~K}\) and \(\Delta_{\text {fus }} \bar{H}=3.53 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\), calculate \(\Delta_{\text {fus }} \bar{V}\) at \(85.46 \mathrm{~K}\).

The molar enthalpy of vaporization of water is \(40.65 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) at its normal boiling point. Use the Clausius-Clapeyron equation to calculate the vapor pressure of water at \(110^{\circ} \mathrm{C}\). The experimental value is 1075 torr.

In this problem, we will show that the vapor pressure of a droplet is not the same as the vapor pressure of a relatively large body of liquid. Consider a spherical droplet of liquid of radius \(r\) in equilibrium with a vapor at a pressure \(P,\) and a flat surface of the same liquid in equilibrium with a vapor at a pressure \(P_{0} .\) Show that the change in Gibbs energy for the isothermal transfer of \(d n\) moles of the liquid from the flat surface to the droplet is \(d G=d n R T \ln \frac{P}{P_{0}}\) This change in Gibbs energy is due to the change in surface energy of the droplet (the change in surface energy of the large, flat surface is negligible). Show that \(d n R T \ln \frac{P}{P_{0}}=\gamma d A\) where \(\gamma\) is the surface tension of the liquid and \(d A\) is the change in the surface area of a droplet. Assuming the droplet is spherical, show that \(\begin{aligned} d n &=\frac{4 \pi r^{2} d r}{\bar{V}^{1}} \\ d A &=8 \pi r d r \end{aligned}\) and finally that \(\ln \frac{P}{P_{0}}=\frac{2 \gamma \bar{V}^{1}}{r R T}\) Because the right side is positive, we see that the vapor pressure of a droplet is greater than that of a planar surface. What if \(r \rightarrow \infty ?\)

When we refer to the equilibrium vapor pressure of a liquid, we tacitly assume that some of the liquid has evaporated into a vacuum and that equilibrium is then achieved. Suppose, however, that we are able by some means to exert an additional pressure on the surface of the liquid. One way to do this is to introduce an insoluble, inert gas into the space above the liquid. In this problem, we will investigate how the equilibrium vapor pressure of a liquid depends upon the total pressure exerted on it. Consider a liquid and a vapor in equilibrium with each other, so that \(\mu^{\prime}=\mu^{8}\). Show that \(\bar{V}^{1} d P^{1}=\bar{V}^{g} d P^{8}\) because the two phases are at the same temperature. Assuming that the vapor may be treated as an ideal gas and that \(\bar{V}^{1}\) does not vary appreciably with pressure, show that \(\ln \frac{P^{\S}\left(\text { at } P^{1}=P\right)}{P^{8}\left(\text { at } P^{1}=0\right)}=\frac{\bar{V}^{1} P^{1}}{R T}\) Use this equation to calculate the vapor pressure of water at a total pressure of 10.0 atm at \(25^{\circ} \mathrm{C} .\) Take \(P^{\mathrm{g}}\left(\text { at } P^{1}=0\right)=0.0313 \mathrm{atm}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.