Density of a solid and Clapeyron equation

[LCSphysicist]

Homework Statement

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Relevant Equations

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A long vertical column is closed at the bottom and open at the top; it is
partially filled with a particular liquid and cooled to -5°C. At this temperature
the fluid solidifies below a particular level, remaining liquid above this level. If the
temperature 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 of
the liquid phase is 1 g/cm3. Find the density of the solid phase. Neglect thermal
expansion of all materials.

So i think i am missing something pretty dumb, but anyway:

$$|\Delta P_{ressure}| = \rho_{s} g \Delta H$$

Claperyon equation:

$$\frac{\Delta P}{\Delta T} = \frac{\Delta L_{m}}{\Delta V_{m} T}$$

Equally both:

$$|\rho_{s}| = \frac{|\Delta T \Delta L_{m}|}{|\Delta H \Delta V_{m} g T|}$$

My problem is, how do i find $\Delta V_{m}$?? Shouldn't it be given?

 

[TSny]

Homework Statement:: .
Relevant Equations:: .

So i think i am missing something pretty dumb, but anyway:

$$|\Delta P_{ressure}| = \rho_{s} g \Delta H$$

The phase change takes place at the interface between the liquid and solid phases. What determines the pressure at this location?

Also, I would recommend not taking absolute values in this problem. You will need to be careful with signs.

Claperyon equation:

$$\frac{\Delta P}{\Delta T} = \frac{\Delta L_{m}}{\Delta V_{m} T}$$
...
My problem is, how do i find $\Delta V_{m}$?? Shouldn't it be given?

You have the ratio $\frac{\Delta L_{m}}{\Delta V_{m}}$ where, presumably, the subscript $m$ refers to molar values. However, the ratio is the same if you think of the $m$ as indicating per unit mass instead of per mole. (The data in the problem is given in terms of unit mass.) Can you express the change in specific volume, $\Delta V_{m}$, in terms of $\rho_s$ and $\rho_l$?

 

[Chestermiller]

The trick to solving this problem is to tentatively assume that the density of the solid phase is very nearly equal to the density of the liquid phase.

 

[LCSphysicist]

Yes, i just got $$\rho_s = \frac{\Delta L |\Delta T|} {|\Delta H| g T} (\frac{1}{\rho_L}-\frac{1}{\rho_s})^{-1}$$, unfortunatelly substituing the numerical values give something very wrong ;P

 

[Chestermiller]

Yes, i just got $$\rho_s = \frac{\Delta L |\Delta T|} {|\Delta H| g T} (\frac{1}{\rho_L}-\frac{1}{\rho_s})^{-1}$$, unfortunatelly substituing the numerical values give something very wrong ;P

Let's see exactly what you did. Incidentally, the equation should read: $$\rho_L = \frac{\Delta L |\Delta T|} {|\Delta H| g T} (\frac{1}{\rho_L}-\frac{1}{\rho_s})^{-1}$$

 

[TSny]

Yes, i just got $$\rho_s = \frac{\Delta L |\Delta T|} {|\Delta H| g T} (\frac{1}{\rho_L}-\frac{1}{\rho_s})^{-1}$$, unfortunatelly substituing the numerical values give something very wrong ;P

This looks correct to me. I'm assuming that $\Delta H$ represents the increase in height of the solid and $\Delta L$ is the (positive) latent heat. When you solve this symbolically for $\rho_s$, what expression do you get?

 

[Chestermiller]

This looks correct to me. I'm assuming that $\Delta H$ represents the increase in height of the solid and $\Delta L$ is the (positive) latent heat. When you solve this symbolically for $\rho_s$, what expression do you get?

Since the liquid is on top, why isn't the pressure change at the phase change interface $\rho_L g \Delta H$?

 

[TSny]

Since the liquid is on top, why isn't the pressure change at the phase change interface $\rho_L g \Delta H$?

That is a correct expression for the change in pressure if $\Delta H = \Delta h_L$, where $\Delta h_L$ is the change in vertical length of the liquid phase, $h_{L,f}-h_{L,0}$. But, mass conservation implies $\rho_L \Delta h_L =-\rho_S \Delta h_S$ where $\Delta h_S$ is the change in vertical length of the solid phase, $h_{S,f}-h_{S,0}$. So you can express the pressure change in terms of $\rho_S$ and $\Delta h_S$.