Monday, April 10, 2006
HW #3 - P7 - Hypothetical Process Paths and the Latent Heat of Vaporization
Use the hypothetical process path shown here to help you determine the change in enthalpy in Joules for 20.0 g of heptane (C7H16) as it changes from a saturated liquid at 300 K to a temperature of 370 K and a pressure of 58.7 kPa. Calculate the DH for each step in the path. Do not use tables of thermodynamic properties, except to check your answers. Instead, use the Antoine Equation to estimate the heat of vaporization of heptane at 300 K. Use the average heat capacity of heptane gas over the temperature range of interest. Assume heptane gas is an ideal gas at the relevant temperatures and pressures.
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11 comments:
So I think im retarded. I know we went over this in class an all but im still having issues. So I use antoine to get two pressures at temperatures around 300K and then use Claudius Clapeyron to solve for dH and i get like 2039.54 J/mol and then i convert and do good things and i get like 407.1 J. yea....
Anon 9:50 PM
LOL. I think you are smart, or I wouldn't answer the question !
The problem is your value of deltaH. Something is wrong and it is difficult for me to say what it might be. I would guess units or problems with LN and LOG10. But that is just a guess. DeltaHvap should be about to 37000 J/mole.
Post again if you are stuck. Give me more info about how you got these values.
I figured it out... i think i am bad at algebra, thanks!
anon 9:33 PM
Multo bene !
I am a bit lost on this entire problem. I think I got dH(12) by taking H^ at 300kPa from the saturation tables. After that I don't have much idea how to use the Antoine equation to get the dHvap. Where do I get the A,B,C for this?
A little direction with this problem would be great.
Ok, if someone does respond to this problem, I think I figured out everything but the first part in calculating dH(12). This again stems from the fact that I don't know how to apply the antoine equation.
To find A B and C you get on the NIST name search and look up heptane and I just check a bunch of the boxes in the left column and it gives you the parameters for the antoine equation. Then you go to the other NIST page and do a heptane saturation temperature increments search and do it for like 299.5-300.5 K and use those two pressures and temperatures in the antoine equation. Then you take your two sets of temps and pressures into your clapeyron equation and plug them in, and since they have the same C, you can set them equal to each other and solve using an R of 8.134.
For part H34, I thought that U34 would be zero (since for an IG, U does NOT depend upon pressure changes, only temperature changes) but since Delta H = Delta U + delta (pressure * volume), delta h would NOT = 0. However, our ta said in the chat last night that it is. Uh, could you enlighten me to why that is so?
For deltH12, I am getting 36,773.49454 J/mole and for deltaH23 I'm getting 1,094.66 J/mole. Are these right? And if they are, where do I go from here so that my deltaH12 becomes 7,000J and deltaH23 becomes 2,500J like on the ans. key on the homwork?
anon @12:45
For H34, it is zero becuase heptane can be treated as an Ideal gas. If you treat it as a non ideal gas or a liquid, then the other equation applies. Hope this helps.
Audrey:
I don't know if those are correct but to get the answer in joules you have to use the molar mass of heptane and the fact that it says in the problem you are given 20 grams to covert it to just joules and not joules per mole.
You cannot count on help with your HW this late in the game. I suggest you start working on the problems sooner.
I know it is too late, but here are my thoughts on the last 6 comments that came in after I stopped working last night...
You get the Antoine constants (A, B & C) from the Name Search section of the NIST webBook. Antoine gives P* = fxn(T). Use this and Clausius-Clapeyron Eqn to solve for dHvap (which is dH12).
Wow. Thanks Anonymous for spelling out how to use the NIST Name Search page to solve this problem !
Anon 12:45 AM
dH = du + d(PV)
For an IG, d(PV) = d(RT) = R dT
Since only P changes, dT = 0
You know dU = 0.
Therefore, dH = 0
H = fxn(T) ONLY, not P FOR AN IG.
Audrey:
Yes, your values are correct. Nice work. DKRESS explained exactly how to get from your values to my answers. Thank you, DKRESS !
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