# Thermochemistry:Gibbs' Free Energy and Entropy

1. Apr 22, 2012

### cyt91

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

a) 3 moles of carbon dioxide gas expands reversibly in a piston-container from 20 L
to 40 L in an isothermal process at 400 K. Calculate work, heat, change in internal
energy and change in enthalpy, considering van der Waals behaviour. [a = 3.59
atm L^2 mol^-2 and b = 0.0427 L mol^-1].

b) 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

(p-a(n/v)^2)(V-nb)=nRT
pV=nRT
delta S(system) = n*Cp*ln(T2/T1)
delta G = delta H - T(delta S)

3. The attempt at a solution

1(a)

I calculated the work done by the gas to be 67.8 J.
I obtained an expression for P from VDW equation and integrate the expression for P w.r.t V from 20 L to 40 L.
R is 0.0821 in this case.

However if the gas is behaving non-ideally, how can we calculate the heat,enthalpy change and internal energy change?
Since for non ideal gas,for isothermal expansion delta U is not necessarily zero.

What difference does non-ideal behaviour bring to the calculations for heat, enthalpy change and internal energy change?

For question 1(b), 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.