Entropy, Enthelpy and Supercooled liquids problem

[knowlewj01]

Homework Statement

Calculate the difference in entropy between solid cyclohexane at -20°C and supercooled liquid cyclohexane at -20°C.

Verify that $\frac{\Delta H_{ fusion.supercooled}}{T_{fusion.supercooled}}$ is not equal to the change in entropy.

Homework Equations

The Attempt at a Solution

This is the final part of a question, i have already determined the enthalpy of fusion of supercooled cyclohexane at -20°C $\Delta H$= 192 kJ/kg

Supercooling a liquid is not a reversable process, ie. once the supercooled liquid is frozen it cannot melt at that same temperature.

so the formula:

$\int dS = \int\frac{dQ}{T}$

cannot be used because this is for reversable processes only.

for the change in entropy would i have to find a reversable route? maybe the sum of entropy changes from: heating the substance from -20 to melting point, melting the substance, supercooling the substance back to -20.

 

[Andrew Mason]

https://www.physicsforums.com/showthread.php?t=284432" may help you.

AM

 

[knowlewj01]

Yeah, that's good, however, in the explanation by Andrew, second post from the bottom.
he uses the argument that the integrals cancel, but the C's arent equal because one is heat capacity of water and the other is of ice, am I right?

Also, if I'm considering just the change in entropy between the two states do i need to consider the change in entropy of the surroundings? I guess i don't because I'm not given any information about them.

Finally, in my notes i have the general equation for change in entropy for a reversale process:

$\Delta S = C_V ln\left[\frac{T_2}{T_1}\right] + nRln\left[\frac{V_2}{V_1}\right]$

where $C_V$ is heat capacity at constant volume. can i assume that:

$C_V \approx C_P$ for liquids and solids?

thanks

 

[Andrew Mason]

Yeah, that's good, however, in the explanation by Andrew, second post from the bottom.
he uses the argument that the integrals cancel, but the C's arent equal because one is heat capacity of water and the other is of ice, am I right?

You are right. It will not make a material difference though because it increases the entropy change of the 100 g of water, so when you add the change in entropy of the 100g of water to the change in entropy of the surroundings total entropy change is even greater (and still greater than 0).

Also, if I'm considering just the change in entropy between the two states do i need to consider the change in entropy of the surroundings? I guess i don't because I'm not given any information about them.

Correct.

Finally, in my notes i have the general equation for change in entropy for a reversible process:

$\Delta S = C_V ln\left[\frac{T_2}{T_1}\right] + nRln\left[\frac{V_2}{V_1}\right]$

where $C_V$ is heat capacity at constant volume. can i assume that:

$C_V \approx C_P$ for liquids and solids?

Your formula is correct for ideal gases only. The difference between Cp and Cv for non-gases will depend on the substance. For water the difference is about 1%.

AM

 

[rwisz]

Yes, in order to calculate the change in entropy for this process, you would need to find a reversible route. This could involve heating the substance from -20°C to its melting point, melting the substance, and then cooling it back down to -20°C. This would allow you to use the equation \Delta S = \int_{T_1}^{T_2} \frac{C_p}{T} dT to calculate the change in entropy, where C_p is the specific heat capacity of the substance.

Alternatively, you could also use statistical thermodynamics to calculate the change in entropy. This involves considering the number of microstates (possible arrangements of particles) for the solid and supercooled liquid at -20°C and using the equation S = k_B \ln{W} where k_B is the Boltzmann constant and W is the number of microstates. However, this method may be more complex and may require more information about the molecular structure of cyclohexane.