Question

A substance is found to have a melting point of 80°C. The vapour pressure
of the liquid is 40 Torr at 108.2°C and 60 Torr at 119°C. The vapour pressure of the solid is 1
Torr at 52.6°C. Using only this information, calculate

Answer 1

a)the boiling point of the substance and its enthalpy and entropy of vaporization at the boiling point
b)the vapour pressure at the melting point;
c) the enthalpy of sublimation and fusion of the solid, assuming that the melting
and triple point temperatures are the same

Answer 2

A little more work on this one, couldn't quite get c)

Answer 3

Hmm, this definitely looks trickier. Give me a few minutes!

Answer 4

Alright i'll be here attempting to wrap my head around c)

Answer 5

Alright i'm going to focus on c with you, the first two seem okay, you just need to rearrange the equation we used last time as both of these are talking about vaporization.

Answer 6

I'm looking this up and I came up with a helpful tidbit:

To estimate sublimation use the vapor pressures of the solid phase, (we only have 1? I suppose the melting point of 80 degrees is a hint to find a second value.

Also, brb something came up.

Answer 7

to estimate vaporization use the vapor pressures of the liquid phase. (have two of those, that's good.)

and fusion = sublimation - vaporization

Answer 8

oh alright that's a lot of helpful tidbits

Answer 9

I think the main problem is finding sublimation, we can get vaporization and hence fusion after that.

Answer 10

Because we just input the given values in the P1/2 and T1/2 variables.

I suppose we could consider using 80 degrees or a bit less, just because it's melting doesn't mean it's not a solid. It must be important otherwise they would not have given us that info.

Answer 11

Wait, what am I saying. Ignore my previous statement let me see...

Answer 12

Aha alright, I'm reading what you're writing I guess I'll forget it all

Answer 13

if we use the equation using the solid value.

\[Ln (760/760) =- \frac{ H }{ R } (\frac{ 1 }{ 325.75 } - \frac{ 1 }{ 353 })\] (because it has a vapor pressure of 1 at 52.6°c)

So we solve the equation for H and we get vaporization!

Answer 14

I think I was on the right track? but maybe not explaining myself in the right way.

Answer 15

1 torr, lmao i'm so tired.

it would be 1/760 , not 1/1

Answer 16

it's alright I understood what you meant. I'm starting to understand it better now (y) , thanks alot

Answer 17

Good! happy to help. c:

I should go back to my studies, if you want any help just give me a ping and I'll reply when I can. ;u;

Answer 18

Alright good luck my man, I'll probably pass out soon it's pretty late here

Answer 19

Alright, cheers, and good luck with the rest of your studies!