I think a rough answer would be to subtract the heat of vaporization
(10.514 kcal/mole) and enthalpy to heat it to the temperature of interest
from the heat of formation of the >gas<. But I don't think this works at
temps over 373, because then the pressure is over 1 atm.
Here are some data applying to 400K (pressure is 2.369 atm(a)):
Temp 298K 400K
---- -----
Sensible heat (gas) H-H(298) 0 0.825 kcal/mole
Sensible heat (liquid) H-H(298) 0 1.960 "
Heat of vap'n (liq-gas) H(L) 10.514 9.408 "
Heat of formation (gas) delta H(f) -57.798 -58.042 "
Boy, this stuff is hard for non-thermodynamicists. Thanks for any help.
Ed Luinstra
luin...@freenet.calgary.ab.ca
You are not using a Pentium-based PC, are you? ;-)
Tony
---------------------
Antoni S. Gozdz
to...@nyquist.bellcore.com
Hahahahaha..best joke I've heard this week! Good one toni!
__________________________
Azma...@uwwvax.uww.edu
PeeCees are crap, and you know it!
PeeCees
Enthalpy is a state function; therefore, the result will be path
independent. You have the heat of formation of the gas at (400,2.369).
Use the heat of vap to go to liquid at (400,2.369); then use the sensible
heat to go to liquid at (T,2.369). If the pressure is different then 2.369
atm., you can use the Poynting correction term. This will be in most texts
on thermo, but not in many P-Chem texts. It is usually negligible. Also,
note that the sensible heat is a function of temp and pressure. Assuming
that they are constant will probably introduce more error than neglecting
the enthalpy of the compressed liquid.
John