Chemical Engineering Thermodynamics

[sean/mac]

One mole of supercooled liquid tin (Sn) is adiabatically contained at 495K. Given:

the melting temperature of Sn, Tm, Sn = 505 [K]
the enthalpy of fusion of Sn, hfus, Sn = 7070 [J/mol]
the heat capacity of liquid Sn, Cp, Sn(l) = 34.7 – 9.2 x10-3 T [J / (mol K)]
the heat capacity of solid Sn, Cp, Sn(s) = 18.5 + 2.6 x 10-2 T [J / (mol K)]

(a) Briefly explain what will happen to the supercooled liquid tin.

(b) Calculate the fraction of liquid tin which spontaneously freezes. Provide a clearly labeled
hypothetical path used in your calculation. State any assumption(s) you make.

I know (a), I'm just unsure how to draw a hypothetical path and unsure where to even start the calculation, any help is appreciated

 

[Mapes]

I'm thinking that some of the supercooled liquid tin will freeze. (If it doesn't, then the problem is pretty uninteresting.) And I'm thinking that the freezing process will give off some heat, yes? Enough heat to raise the temperature of this adiabatic system to the freezing point? Does this help?

 

[piercas]

(a) The supercooled liquid tin will eventually reach its melting temperature of 505K and undergo a phase change from liquid to solid. This is because at 495K, the liquid tin is below its melting temperature and is considered to be in a metastable state, meaning it is temporarily stable but not the most stable state for the substance. As the temperature continues to decrease, the liquid tin will eventually reach its melting temperature and undergo a spontaneous phase change to become a solid.

(b) To calculate the fraction of liquid tin that will spontaneously freeze, we can use the Clausius-Clapeyron equation:

ln(P2/P1) = -ΔHfus/R * (1/T2 - 1/T1)

where P1 and P2 are the initial and final pressures, T1 and T2 are the initial and final temperatures, ΔHfus is the enthalpy of fusion, and R is the gas constant.

Assuming that the pressure remains constant and using the given values, we can rearrange the equation to solve for the fraction of liquid tin that will freeze:

(P2/P1) = e^(-ΔHfus/R * (1/T2 - 1/T1))

Using T1 = 495K and T2 = 505K, we can calculate the fraction of liquid tin that will freeze as:

(P2/P1) = e^(-7070 J/mol / (8.314 J/mol K * (1/505K - 1/495K)))

(P2/P1) = e^(-7070 J/mol / (8.314 J/mol K * 0.00202 K^-1))

(P2/P1) = e^(-7070 J/mol / 16.8 J/mol)

(P2/P1) = e^-421.43

(P2/P1) = 4.7 x 10^-184

This means that only a very small fraction of the liquid tin will spontaneously freeze, as the majority of it will remain in its liquid state.

As for the hypothetical path, we can imagine a constant pressure line on a temperature-entropy diagram (T-S diagram) starting at 495K and ending at 505K. This line represents the path that the supercooled liquid tin will take as it reaches its melting temperature and undergoes a phase change. The assumption made in this calculation is that the pressure remains constant during the process.