Asked by Jason
The normal melting point of H2O is 273.15 K, and deltaH fusion is 6010 J/mol. Calculate the change in the normal melting point at 175 and 625 bar assuming that the density of the liquid and solid phases remains constant at 997 and 917 kg/m^3 respectively. Explain the sign of your answer.
So I had no problem calculating:
when P= 175 bar, deltaT= -1.25 K
when P= 625 bar, deltaT= -4.48 K
Assuming my answers are correct, can someone explain the negative signs in the context of the question?
So I had no problem calculating:
when P= 175 bar, deltaT= -1.25 K
when P= 625 bar, deltaT= -4.48 K
Assuming my answers are correct, can someone explain the negative signs in the context of the question?
Answers
Answered by
DrBob222
I think the negative sign simply means that at increased pressure the melting point decreases by 1.25 and 4.48 K respectively but I don't know if this is sufficient explanation or not.
Answered by
Joe
I think the negative sign implies that the slope between S and L is negative so as you increase Pressure you will get a small T. But what formula did you use to find delta T.
Answered by
Jason
I went
(dP/dT)fusion=deltaSm(fusion)/deltaVm(fusion)
=deltaHm(fusion)/Tfusionm(1/pliq. - 1/psolid)
=-140 barK^-1
So then deltaT=-deltaP/140
What is L? volume (litres?)
(dP/dT)fusion=deltaSm(fusion)/deltaVm(fusion)
=deltaHm(fusion)/Tfusionm(1/pliq. - 1/psolid)
=-140 barK^-1
So then deltaT=-deltaP/140
What is L? volume (litres?)
Answered by
DrBob222
I think Joe is referring to the phase diagram and the slope of the S(solid)/L(liquid) line which is negative.
Answered by
Joe
Yeah I'm referring to the phase diagram and the Solid/Liquid line
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