In this problem you will investigate the behavior of a van der Waals fluid near the critical point. It is easiest to work in terms of reduced variables throughout.

(a) Expand the van der Waals equation in a Taylor series in , keeping terms through order . Argue that, for T sufficiently close to Tc, the term quadratic in $(V-V_C)$becomes negligible compared to the others and may be dropped.

(b) The resulting expression for P(V) is antisymmetric about the point V = Ve. Use this fact to find an approximate formula for the vapor pressure as a function of temperature. (You may find it helpful to plot the isotherm.) Evaluate the slope of the phase boundary,$dP/dT$

( c) Still working in the same limit, find an expression for the difference in volume between the gas and liquid phases at the vapor pressure. You should find $\left(V_g-V_l\right)\propto {\left(T_c-T\right)}^{\beta }$.8, where (3 is known as a critical exponent. Experiments show that (3 has a universal value of about 1/3, but the van der Waals model predicts a larger value.

(d) Use the previous result to calculate the predicted latent heat of the transformation as a function of temperature, and sketch this function.

The shape of the T = Tc isotherm defines another critical exponent, called $\left(P-P_c\right)\propto {\left(V-V_c\right)}^{\delta }$Calculate 5 in the van der Waals model. (Experimental values of 5 are typically around 4 or 5.)

A third critical exponent describes the temperature dependence of the isothermal compressibility, K=-t This quantity diverges at the critical point, in proportion to a power of (T-Tc) that in principle could differ depending on whether one approaches the critical point from above or below. Therefore the critical exponents 'Y and -y' are defined by the relations

$\kappa \propto \left\{\begin{array}{l}{\left(T-T_c\right)}^{-\gamma }\\ {\left(T_c-T\right)}^{-{\gamma }^{\text{'}}}\end{array}\right.$

Calculate K on both sides of the critical point in the van der Waals model, and show that 'Y = -y' in this model.

Step by step solution

01

Part(a) Step 1: Given information

We have been given$p=8t(3v-1{)}^{-1}-3v^{-2}$

02

Part(a) Step 2: Simplify

The terms used in:

$\frac{{\partial }^{3}p}{\partial v^3}=-1296t(3v-1{)}^{-4}+72v^{-5}$

Therefore, to plot isotherm and perform the Maxwell construction

$\text{Series}\left[8t/(3v-1)-3/v^{~}2,\{v,1,3\}\right]$

03

Part(b) Step 1: Given information

We have been given, As $V-1$even gets closer to. the two curves becoming indistinguishable over the range.

04

Part(b) Step 2; Simplify

Constant-pressure line that results in equal area enclosed by two loops

05

Part(c) Step 1: Given information

We have been given$v_g-v_l=(1+2\sqrt{1-t})-(1-2\sqrt{1-t})=4\sqrt{1-t}$

The volume of the liquid and gas at transition pressure are just the values of $v$at transition pressure found in the above parts.

06

Part(c) Step 2: Simplify

The term we will get:

$4t-3=4t-3-6(t-1)(v-1)-\frac{3}{2}(9t-8)(v-1{)}^3$

07

Part(d) Step 1;Given information

We have been given$L=T\left(V_g-V_l\right)\frac{dP}{dT}=P_{\mathrm{c}}{V}_c\left(v_g-v_l\right)\frac{dp}{dt}=\frac{3}{8}Nk{T}_c\left(v_g-v_l\right)\frac{dp}{dt}$

This equation describes a parabola opening to the left descresing to left as $t\to 1$

08

Part(d) Step 2: Simplify

The term we get:

$\frac{L}{Nk{T}_c}=\frac{3}{8}·4\sqrt{1-t}·4=6\sqrt{1-t}$

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