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Get Full Access to An Introduction To Thermal Physics - 1 Edition - Chapter 5 - Problem 55p
Get Full Access to An Introduction To Thermal Physics - 1 Edition - Chapter 5 - Problem 55p

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# In this problem you will investigate the behavior of a van

ISBN: 9780201380279 40

## Solution for problem 55P Chapter 5

An Introduction to Thermal Physics | 1st Edition

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Problem 55P

Problem 55P

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 (V − Vc), keeping terms through order (V − Vc)3. Argue that, for T sufficiently close to Tc, the term quadratic in (V − Vc) becomes negligible compared to the others and may be dropped.

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

(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 (Vg − Vl) α (Tc − T)β where β is known as critical exponent, Experiments show that β 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.

(e) The shape of the T = TC isotherm defines another critical exponent, called δ: (P − PC) α (V − Vc)δ Calculate δ in the van der Waals model. (Experimental values of δ are typically around 4 or 5.)

(f) A third critical exponent describes the temperature dependence of the isothermal compressibility,

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 γ and γ′ are defined by the relations

Calculate κ on both sides of the critical point in the van der Waals model, and show that γ = γ′ in this model.

Step-by-Step Solution:
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MATH 2450 WEEK 6 Chain Rule in Higher Dimensions g(t) = df(x(t) , y(t))= ∂f * dx + ∂f * dy dt ∂x dt ∂y dt R(t) = < x(t), y(t), f( x(t), y(t) ) df(g(t)) = df * dg dt dg dt 2 EX. Use the chain rule to find the df/dt of the function f(x,y) = x + 3xy where x(t) = cos(t) and y(t) = sin(t) Fx= 2x + 3y Fx= 3x dx/dt = -sin(t) dy/dt = cos(t) Now plug in the values you found above into the chain rule equation g(t) = ∂f * dx + ∂f * dy ∂x dt ∂y dt g(t) = (2x + 3y) (-sin(t)) + (3x) (cos(t)) Now make everything is terms of ‘t’. Remember to replace the ‘x’ and ‘y’ with their ORIGINAL values(x(t) = cos(t) and y(t

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