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Topic: Gibbs Free Energy Calculation when temperature and pressure increased  (Read 3856 times)

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Offline cyt_91

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1. The problem statement, all variables and given/known data

Calculate the change in entropy and Gibb’s free energy of the system when 2
moles of water goes from 300 K and 1 atm to 310 K and 40 atm.

2. Relevant equations

delta S(system) = n*Cp*ln(T2/T1)
delta G = delta H - T(delta S)

3. The attempt at a solution

I calculated the entropy change to be 4.94 J/K.
First I calculated entropy change involved in the heating of liquid water at constant 1 atm from 300 to 310 K using the formula

delta S(system) = n*Cp*ln(T2/T1) = 2*75.3*ln(310/300) = 4.94 J/K

Since this is liquid water, can we assume that the work done on the system during compression from 1 atm to 40 atm is negligible as the change in volume is insignificant?
Therefore, the energy transferred to the system during compression is not significant.
Can we say that the entropy change of the system would be 4.94 J/K ?

How do we calculate the Gibbs' Free energy change? Is the equation

delta G = delta H - T(delta S) valid in this case?
(as the the Temperature and Pressure is not constant).

Thanks for any help given...I've been cracking my head over this question...

Offline Kemi

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Re: Gibbs Free Energy Calculation when temperature and pressure increased
« Reply #1 on: April 25, 2012, 07:14:21 AM »
Are you familiar with the equation dG = Vdp - SdT? At constant temperature you can write dG = Vdp. If the change in volume is insignificant, as you suggest, this equation integrates to ΔG = VΔp.

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