Question

At -3ºC vapor pressure of ice is 3.566mm and for supercooled water is 3.669mm.
For a transition of 1 mol of supercooled water into 1 mol of ice at -3ºC and 3.669 mm, compute delta G. (Ice density = 0.917 g/cm^3) (Solution: -63.9 J)

Answer 1

@.Sam. ??

Answer 2

do u know the formula
\[dg=-s \ dT+ v \ dP\]
?

Answer 3

Yes

Answer 4

well T is constant for this transition so dT=0 and formula becomes to
\[dg=v \ dP\]

Answer 5

Yes, I've done that but I don't get -63.9 J (the solution in the book), maybe I'm wrong with my calculus but I dont' know where

Answer 6

sorry i lost my connection
let me check it

Answer 7

Ok, np, thank you!

Answer 8

what was your answer?

Answer 9

A non-sense I thinkk, 2.02 J

Answer 10

\[\Delta P=-0.113 \ \ mmHg\]
am i right?

Answer 11

Yep

Answer 12

and we assume that v is constant
ok?

Answer 13

Ok

Answer 14

\[v=\frac{1}{0.917}\frac{cm^3}{gr}18\frac{gr}{mol}=19.63\frac{cm^3}{mol}\]

Answer 15

\[\Delta P=-0.113 \ \ mmHg=-0.113 \frac{101325}{760}=15.065 \ \ Pa\]

Answer 16

-15.065*

Answer 17

my answer is a very small number
o.O

Answer 18

-0.002 J

Answer 19

what shall we do?

Answer 20

And what about using Clausius-Clapeyron equation? We can find out latent heat's

Answer 21

but this transition starts with a supercooled water for which we dont have any information about its state

Answer 22

\[\Delta G=-SdT+VdP\]
as T=constant

\[\Delta G =VdP\]

\[\Delta G=\int\limits_{3.669}^{3.566} VdP = RTln(3.566/3.669)=8.31(-3+273.15)\ln(3.566/3.669)=-63.9 (J)\]

Answer 23

If you considered water vapor as an ideal gas you can make this substitution

Answer 24

Anyway thanks a lot @mukushla! ^^

Answer 25

thats ok
thank u for letting me know the correct answer