Thermochemistry:Gibbs' Free Energy and Entropy

[cyt91]

Homework Statement

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.

Homework 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)

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.

 

[nate808]

Thank you for your post. It is great to see you taking an interest in these thermodynamic calculations! I am happy to help you understand the concepts better.

1(a) For ideal gases, the internal energy change during an isothermal process is indeed zero. However, for non-ideal gases, the internal energy change is not necessarily zero. This is because non-ideal gases exhibit intermolecular forces that contribute to the internal energy of the system. Therefore, the internal energy change during an isothermal process for non-ideal gases can be calculated by considering the van der Waals equation and integrating it with respect to volume. The same approach can be used for calculating the enthalpy change and heat transfer during the process.

1(b) For the calculation of entropy change in this case, you have correctly used the formula for the heating of liquid water. However, for the compression process, the work done on the system is not negligible. This is because the work done is not just due to the change in volume, but also due to the change in pressure. Therefore, the entropy change during compression should be calculated separately using the formula ΔS = nRln(P2/P1).

For the calculation of Gibbs free energy change, the equation ΔG = ΔH - TΔS is indeed valid. This is because the process is still occurring at constant temperature and pressure, even though these values are changing. Therefore, the equation can be used to calculate the Gibbs free energy change.

I hope this helps you understand the concepts better. If you have any further questions, please do not hesitate to ask. Keep up the good work!