How Accurate is the Freezing Point Calculation for Ethanol at 1000 atm?

[koomanchoo]

ethanol 46g/mol freezes at 3.65degC density changes from 0.789g/cm-3 to 0.801g/cm-3 latent hear of fusion 8.68kJ/mol. freezing point at 1000 atm.

i calculated the ansfer from an example and just want to verify if its correct.

dp/dt=(dfusH)/(t*dfusV) rearanged to dt=(dp*t*dfusV)/dfusH
where dfusV: change in molar volume on melting

okay, with the rearanged equation i changed 1000 atm to pascals = 101325 kPa dfusV was calculated to be 4.875399e-6 m3/mol
i came to a change in temperature of 15.54deg hence final freezing point is 11.8956degC.

could anyone plese help me and tell me if this is correct
also.. in teh equation t=273.15K was used.. could anyone please tell me why this was the case?
thanks
patrick.

 

[vitaly]

I don't know the freezing point, but you can probably find a table that gives you that data so you can check your answers.

Also, you use t = 273.15 Kelvins because that is, in basic terms, the accepted unit of temperature in chemistry/physics, etc. You will use Kelvins in the gas laws, thermochemistry and thermophysics, and pretty much everything else.

 

[thepatientmental]

Your calculation looks correct to me. In the equation, t=273.15K is used because it is the standard temperature at which most thermodynamic calculations are performed. This temperature is equivalent to 0 degrees Celsius or 32 degrees Fahrenheit. It is important to use a standard temperature in thermodynamic calculations to ensure accuracy and consistency.