acetone latent heat at reduced pressure 1

[saleda]

Dear All

I wanted to know the latent heat of vaporization of acetone at 680mmHg for my condenser design, but could not find it. as latent heat decreases with increase in pressure and vice versa.

would somebody refer to some source or provide related equation.

Thanks

Saleda

 

[ione]

It should be approx 39436 J/mol.
You can use the equation below:

ln p = - ΔHvap/RT + C

where
p is the vapour pressure
ΔHvap is the heat of vaporization
R is the gas constant (8.314 J/mole K)
T is the equilibrium temperature at vapour pressure p
C is a constant of integration

You can use known value of ΔHvap (at given vapour pressure and saturation temperature) to calculate C. Then, considering that at 680 mmHg the saturation pressure is 348.67 K, you can find ΔHvap at your conditions.

 

[Latexman]

This is a good application for the Watson correlation:

Watson correlation on page 3

Good luck,
Latexman

 

[ione]

I was mistaken. For the following conditions
p = 680 mmHg
T = 326.2 K
it should be
ΔHvap = 29488 J/mol

 

[saleda]

Dear ione,

Your result shows lower Hvap value than at ATM pressure. whereas it should increse.

My calculation results 125.82 cal/gm at 680 mmHg than 124.3 cal/gm at ATM.

Plz validate.

Thanks,
Saleda

 

[25362]

From the Antoine equation in Perry's VIII ed.:

log Psat = A - B/(T+C)

P in torr, T in oC
A=7.11714
B =1210.595
C = 229.664

Range of applicability: from -13 to +55oC.

 

[ione]

saleda,

at p = 760 mmHg and T = 329.4 K, the enthalpy of vaporization ΔHvap is approx 29281 J/mol, while at p =680 mmHg and T = 326.2 K ΔHvap is 29488 J/mol.

You might want to check at webbook Nist

http://webbook.nist.gov/cgi/cbook.cgi?ID=C67641&Mask=4&Type=HVAP-FORM2&Plot=on#HVAP-FORM2