- #1

EE18

- 112

- 13

- Homework Statement
- See below.

- Relevant Equations
- Clapeyron equation

Callen asks us the following question in his famous textbook:

I have answered as follows:A long vertical column is closed at the bottom and open at the top; it ispartially filled with a particular liquid and cooled to - 5°C. At this temperaturethe fluid solidifies below a particular level, remaining liquid above this level. If thetemperature is further lowered to -5.2°C the solid-liquid interface moves upward by 40 cm. The latent heat (per unit mass) is 2 cal/g, and the density ofthe liquid phase is 1 g/cm 3• Find the density of the solid phase. Neglect thermal expansion of all materials.

Hint: Note that the pressure at the original position of the interface remainsconstant.

Answer: 2.6 g/cm 3

However, I get the wrong answer and, in fact, the correct answer obtains from using ##\Delta P = -\rho_s g h##; that is, using the solid density for the change in pressure. Now why on earth should this be? Shouldn't the liquid column's height difference dictate the pressure?Note that, per the hint, the pressure on top of the liquid is constant (atmospheric pressure) throughout, but the pressure in the liquid at the bottom is higher (by ##\rho_l gh##, where ##\rho_l## is the density of the fluid). Thus, when we decrease the temperature to ##T_0 = - 5 C## and observe as per the question that "the fluid solidifies below a particular level [in the liquid]", what is happening is that the liquid is solidifying at all points whereat the pressure is greater than the pressure on the coexistence curve ##P(T_0)##. When we decrease the temperature somewhat further, we find that ##P(T_1)## is lower (since the solid grows, and the liquid-solid interface has less liquid above it -- a lower pressure). We can use the Clapeyron equation to make the determination about ##\rho_s##.

The change in pressure at the equilibrium interface (i.e. ##P(T_1) - P(T_0)##) is ##\Delta P = -\rho_l g h## where ##h = 40 cm## is the (magnitude of) change in height of the interface. Thus we have, using that ##\Delta v = \Delta (1/\rho)## if ##v## is the volume per unit mass (since ##l## is so expressed),

$$\Delta P = \left(\frac{dP}{dT}\right)_{cc} \Delta T = \frac{l(T_0,P_0)}{T_0 \Delta v}\Delta T = \frac{l(T_0,P_0)}{T_0 \Delta (1/\rho)}\Delta T \implies \Delta (1/\rho) = \frac{\Delta T}{\Delta P}\frac{l(T_0,P_0)}{T_0} $$

$$\implies \rho_s = \frac{1}{1/\rho_l - \frac{\Delta T}{\Delta P}\frac{l(T_0,P_0)}{T_0}} = \frac{\rho_l}{1 + \frac{\Delta T}{gh}\frac{l(T_0,P_0)}{T_0}}.$$