Why is solid density used in this thermodynamics problem?

In summary: In equilibrium, the solid below the initial interface has to exert an upward force on the liquid so that the net force on the liquid vanishes. If the liquid undergoes a phase change, that doesn't change its total mass, so the force exerted from below doesn't change.
  • #1
EE18
112
13
Homework Statement
See below.
Relevant Equations
Clapeyron equation
Callen asks us the following question in his famous textbook:
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
I have answered as follows:
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),
$$\Delta P = \left(\frac{dP}{dT}\right)_{cc} \Delta T = \frac{l(T_0,P_0)}{T_0 \Delta v}\Delta T = \frac{l(T_0,P_0)}{T_0 \Delta (1/\rho)}\Delta T \implies \Delta (1/\rho) = \frac{\Delta T}{\Delta P}\frac{l(T_0,P_0)}{T_0} $$
$$\implies \rho_s = \frac{1}{1/\rho_l - \frac{\Delta T}{\Delta P}\frac{l(T_0,P_0)}{T_0}} = \frac{\rho_l}{1 + \frac{\Delta T}{gh}\frac{l(T_0,P_0)}{T_0}}.$$
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?
 
Physics news on Phys.org
  • #2
EE18 said:
Homework Statement: See below.
Relevant Equations: Clapeyron equation

Callen asks us the following question in his famous textbook:

I have answered as follows:

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?
I think you misread the hint. It says the pressure at the original position of the solid-liquid interface remains constant, which makes sense since the total mass above that point doesn't change when the temperature decreases.
 
  • #3
vela said:
I think you misread the hint. It says the pressure at the original position of the solid-liquid interface remains constant, which makes sense since the total mass above that point doesn't change when the temperature decreases.
I am a bit confused about 1) what the hint is saying, and about 2) why the hint is true.

1) Is the idea here that if the original interface is at some level ##x##, then ##P(x)## doesn't change throughout? And from that, once the temperature is dropped and given that the new interface is at ##x+h## and given that there is solid between ##x+h## and ##x##, we have ##P(x+h) = P(x) - \rho_s g h##?

2) Why is it true that ##P(x)## doesn't change? I am not really familiar with pressures in solids and have only used ##\rho g h## in the context of liquids.
 
  • #4
In equilibrium, the solid below the initial interface has to exert an upward force on the liquid so that the net force on the liquid vanishes. If the liquid undergoes a phase change, that doesn't change its total mass, so the force exerted from below doesn't change.

You may find it helpful to review the derivation of the expression ##\rho g h## and ask yourself if the reasoning relies on whether you're analyzing a fluid or a solid.
 
  • #5
vela said:
In equilibrium, the solid below the initial interface has to exert an upward force on the liquid so that the net force on the liquid vanishes. If the liquid undergoes a phase change, that doesn't change its total mass, so the force exerted from below doesn't change.

You may find it helpful to review the derivation of the expression ##\rho g h## and ask yourself if the reasoning relies on whether you're analyzing a fluid or a solid.
I have unfortunately never really seen the derivation of ##\rho g h## and just know it as a rule of thumb. Perhaps I need to study some more mechanics as your first paragraph there evades me. In particular, your comment about the total mass not changing. If you're able to flesh out in some more detail I would greatly appreciate it!
 

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Classical Physics
Replies
6
Views
473
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
5
Replies
170
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
675
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top