/*! 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} Q 5.34 Below 0.3 K the slope of the °H... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Below 0.3 K the slope of the °He solid-liquid phase boundary is negative (see Figure 5.13).

(a) Which phase, solid or liquid, is more dense? Which phase has more entropy (per mole)? Explain your reasoning carefully.

(b) Use the third law of thermodynamics to argue that the slope of the phase boundary must go to zero at T = 0. (Note that the *He solid-liquid phase boundary is essentially horizontal below 1 K.)

(c) Suppose that you compress liquid *He adiabatically until it becomes a solid. If the temperature just before the phase change is 0.1 K, will the temperature after the phase change be higher or lower? Explain your reasoning carefully.

Short Answer

Expert verified

The slope must be zero.

The temperature should be lower after the phase transition.

Step by step solution

01

Given information

At constant temperature, the partial derivative of Gibbs free energy with respect to pressure equals volume.

∂G∂PT=V

Where,

G is Gibbs free energy

V is volume

P is pressure

The denser phase is always stable at greater pressures. The shallow slope of Gibbs free energy vs pressure curve may be seen in the bottom volume. As a result, under high pressure, the value of G for the other phase will be reduced.

The solid phase of 3He is stable at greater pressures, as seen in figure 5.34. As a result, the solid phase of 3He" is denser than the liquid phase.

02

Part (a) Step 2: Explanation

Find the slope of the boundary line using the Clausius-Clapeyron equation.

dPdT=ΔSΔV

Where,

∆Sis the change in entropy

∆Vis the change in volume

The slope of the pressure and temperature curve below 0.3 K is negative, as seen in figure 5.34. As a result, the ∆Sand ∆Vvalues have the same sign. As a result, if the phase has a smaller volume, it will have a higher entropy.

As a result, the solid phase of 3He must have more entropy than the liquid phase around the phase boundary below 0.3K.

03

Part (b) Step 2: Explanation 

When the temperature approaches zero, the entropy of either phase must go to zero, according to the third law of thermodynamics.

As a result, the entropy difference between the two phases equals zero.

Find the slope of the boundary line using the Clausius-Clapeyron equation.

dPdT=ΔSΔV=0ΔV=0

Hence, the slope must be zero

04

Part (c) Step 3: Explanation

The adiabatic process involves no heat entering or leaving the system. As a result, the entropy does not alter as a result of this activity. When 3He is crushed adiabatically, it becomes solid 3He. As a result, the entropy of liquid 3He does not change during adiabatic compression.

Even at T = 0.1 K, entropy will grow as a result of portion (a). As a result, the only way to keep entropy constant is to lower the temperature.

As a result, the temperature should be lower after the phase transition.

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!

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

In the previous section I derived the formula (∂F/∂V)T=-P. Explain why this formula makes intuitive sense, by discussing graphs of F vs. V with different slopes.

Compare expression 5.68 for the Gibbs free energy of a dilute solution to expression 5.61 for the Gibbs free energy of an ideal mixture. Under what circumstances should these two expressions agree? Show that they do agree under these circumstances, and identify the function f(T, P) in this case.

Graphite is more compressible than diamond.

(a) Taking compressibilities into account, would you expect the transition from graphite to diamond to occur at higher or lower pressure than that predicted in the text?

(b) The isothermal compressibility of graphite is about 3 x 10-6 bar-1, while that of diamond is more than ten times less and hence negligible in comparison. (Isothermal compressibility is the fractional reduction in volume per unit increase in pressure, as defined in Problem 1.46.) Use this information to make a revised estimate of the pressure at which diamond becomes more stable than graphite (at room temperature).

Suppose that an unsaturated air mass is rising and cooling at the dry adiabatic lapse rate found in problem 1.40. If the temperature at ground level is 25 C and the relative humidity there is 50%, at what altitude will this air mass become saturated so that condensation begins and a cloud forms (see Figure 5.18)? (Refer to the vapor pressure graph drawn in Problem 5.42)

Let the system be one mole of argon gas at room temperature and atmospheric pressure. Compute the total energy (kinetic only, neglecting atomic rest energies), entropy, enthalpy, Helmholtz free energy, and Gibbs free energy. Express all answers in SI units.

See all solutions

Recommended explanations on Physics 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.