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.

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

${\left(\frac{\partial G}{\partial P}\right)}_T=V$

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.

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

$\frac{dP}{dT}=\frac{\Delta S}{\Delta V}$

Where,

$∆S$is the change in entropy

$∆V$is 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 $∆S$and $∆V$values 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.

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.

$$\begin{gathered} \frac{dP}{dT}=\frac{\Delta S}{\Delta V} \\ =\frac{0}{\Delta V} \\ =0 \end{gathered}$$

Hence, the slope must be zero

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.