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posted by  Saito on 8/18/2008 6:24:14 AM  |  status: Live  

change in Gibbs energy

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Question Details:
Estimate the change in the Gibbs energy of 1.00 L of ethanol when the pressure acting on it is increased from 101.3 kPa to 202.6 kPa.

ANS:
Assuming ethanol is a liquid, so there is little dependence of V on P
ΔG=VΔP
ΔG=1 L * 101.3kPa
ΔG=101.3

I think this one is okay, (but my answers haven't been too great so far....).
Is there anything I have done wrong?

Assuming the above working is right Is it safe to assume that there is little dependence of V on P for ethanol at these pressures?

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Sage
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posted by SamiLiLOnE on 8/18/2008 2:11:44 PM  |  status: Live
Asker's Rating: Lifesaver   
Saito's comment:
"thanks for checking that for me."
Response Details:
You need to change kPa to atm.

1 atm = 101.325 kPa

ΔG = VΔP
ΔG = (1.00 L)(2.00 atm - 1.00 atm)
ΔG = 1.00 L atm = 101.325 J

You get the same answer, your error is just in technicalities which may trip you up later on.

Mentor
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posted by astronaut on 8/18/2008 6:16:23 PM  |  status: Live
Asker's Rating: Helpful   
Saito's comment:
"thanks also for verifying the answer."
Response Details:
ΔG=101.3 yes this is the right answer
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