Thermodynamics - Change in the free energy

In summary, the conversation discusses a Van der Waals gas in a recipient with a mobile wall at equilibrium. The system is in contact with a heat reservoir and reversible work is done to increase the volume of one part. The work done by the system is equal to the decrease of the Helmholtz free energy, which can be calculated by integrating PdV for both parts and adding them. The correct solution is given by W=aN2(\frac{2}{V}-\frac{1}{V+ΔV}-\frac{1}{V-ΔV})+NKBT log(\frac{(V-bN)^2}{(V+ΔV-bN)(V-ΔV-bN
  • #1
Jalo
120
0
1. Homework Statement [/b]

Consider a Van der Waals gas.
Consider a recipient of volume 2V, with a mobile wall (with no friction) that divides the recipient in two, each part having exactally N particles. The system is at equilibrium and the mobile wall is exactally in the middle of the recipient.

Consider that the system is in contact with a heat reservoir of temperature T.
Now imagine reversible work is realized on the system so that the volume of oe of the parts increases ΔV.


Find the work done by the system. (The work given in a reversible process to a system is equal to the decrease of the Helmholtz free energy)


Homework Equations



F=U-TS

U=[itex]\frac{3}{2}[/itex]NKbT - a[itex]\frac{N^2}{V}[/itex]

F(T,V,N)=-[itex]\frac{aN^2}{V}[/itex]-NKbT [ log(V-bN) + [itex]\frac{3}{2}[/itex]log([itex]\frac{3}{2}[/itex]KbT)-log(N)+log(c)-[itex]\frac{3}{2}[/itex] ]

The Attempt at a Solution



I found the function F(T,V,N) in the first exercise. Since the work done by the system is equal to the decrease of the Helmholtz free energy ( dW = -dF ) I just calculated

W=ΔF=F(T,V+ΔV,N) - F(T,V,N)

The answer I got was incorrect tho... The correct answer is:

W=aN2([itex]\frac{2}{V}[/itex]-[itex]\frac{1}{V+ΔV}[/itex]-[itex]\frac{1}{V-ΔV}[/itex])+NKBT log([itex]\frac{(V-bN)^2}{(V+ΔV-bN)(V-ΔV-bN}[/itex])

I'm a little confused as to why my resolution isn't correct.. If anyone could give me a hand I'd appreciate.

Thanks.
 
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  • #2
Calculate the work done by the gas by integrating PdV for both parts and adding them.

ehild
 

1. What is the concept of free energy in thermodynamics?

Free energy is a measure of the energy available to do work in a system. In thermodynamics, it is denoted by the symbol G and is a combination of the system's internal energy (U), entropy (S), and pressure (P). It represents the maximum amount of useful work that can be extracted from a system under constant temperature and pressure.

2. How is the change in free energy related to spontaneity of a process?

The change in free energy (ΔG) is a measure of the spontaneity of a process. If ΔG is negative, the process is spontaneous and can occur without any external input of energy. If ΔG is positive, the process is non-spontaneous and requires an input of energy to occur. If ΔG is zero, the process is at equilibrium.

3. What factors affect the change in free energy?

The change in free energy is affected by temperature, pressure, and the initial and final states of the system. It is also affected by the concentration of reactants and products, as well as the presence of catalysts or inhibitors.

4. How is the change in free energy related to enthalpy and entropy?

The change in free energy (ΔG) is related to enthalpy (ΔH) and entropy (ΔS) through the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin. This equation is known as the Gibbs free energy equation and is used to calculate the change in free energy for a given process.

5. Can the change in free energy be negative?

Yes, the change in free energy can be negative. A negative ΔG value indicates that the process will occur spontaneously and release energy. This is often seen in exothermic reactions (ΔH < 0) and processes with an increase in entropy (ΔS > 0).

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