For an ideal gas constant volume heat addition process, change of entropy equation is:
ΔS=cvlnT2T1+Rlnv2v1=∫dqT+Sgen
The term Rlnv2v1 equals zero, since it’s a constant volume process.
For ideal gas ∫dqT=cvlnT2T1 .
Then:
ΔS=cvlnT2T1=cvlnT2T1+Sgen
Therefore, the Sgen term equals zero and the process is reversible.
The question is: why does the Sgen term equal zero and the process is reversible when this is a heat addition through a finite temperature difference?
To give a numerical example, imagine that an ideal gas is put in a rigid tank of uniform temperature where its initial temperature is 400 K, and a hot reservoir at 500 K. then heat is transferred from the hot reservoir to the rigid tank until the temperature of the rigid tank is 430 K.
referring to the expressions above T1=400 K and T2=430 K.
Answer
The approach is to start from the definition of entropy change for a system dS according to the laws of thermodynamics.
dS≡δqT
The heat flow δq is path dependent and T is the temperature of the system. We take a reversible path to find the reversible entropy change. We include an irreversible term as needed. Therefore, we write the expression below for the entropy change of a system under any process.
ΔS=∫δqrevT+ΔSirr
Reversible processes are those where the system and surroundings are in exact mechanical (equal pressures), thermal (equal temperatures), and chemical (equal chemical potentials) equilibrium at all points in time during the process. Such processes do not exist in the real world. They are hypothetical processes that allow us to make fundamental insights.
The difference between the entropy change of a system undergoing a reversible process ΔSrev and the entropy change of a real world process is the irreversible entropy generation ΔSirr or Sgen.
Ideal gases are also non-existent in the real world. They are however closely approximated by real gases to the point that we allow for the assumption even in practice.
For an ideal gas that undergoes a reversible change in temperature at constant volume, we obtain the following:
dU⋆rev=CVdT=δq
dS⋆rev=CvdTT
ΔS⋆rev,V=CVln(Tf/Ti)
The last step requires that we assume that heat capacity is constant (the heat capacity of an ideal gases can depend only on temperature).
For an ideal gas that undergoes a reversible change in volume at constant temperature, we can also prove
ΔS⋆rev,T=Rln(Vf/Vi)
Combining the two expressions, we obtain the entropy change of an ideal gas with constant heat capacity undergoing any reversible change in temperature and volume as
ΔS⋆rev=CVln(Tf/Ti)+Rln(Vf/Vi)
The total entropy change of the universe is the sum of the system and the surroundings. In a reversible process, the system and surroundings have the same entropy change. The total of the universe is therefore zero.
For an irreversible process, the entropy change of an ideal gas at constant heat capacity will still be the same as above. The irreversible entropy change is included and assigned to the surroundings. We use the term Sgen to obtain
ΔS⋆univ=ΔS⋆sys+ΔS⋆surr+Sgen=Sgen
The Sgen term accounts for the fact that system and surroundings are not at perfect mechanical, thermal, or chemical equilibrium at all points in time during the process.
The entropy change of the system is found using a reversible path. The irreversibility is assigned to the surroundings. Using the definition of entropy, we can make the comparable statement dSgen=δq/Tsurr.
The founding equation starts with ΔS. This is ambiguous. Is it to be ΔSuniv, ΔSsys, or ΔSsurr? This ambiguity should be clarified first.
The first expression after the equal sign is the entropy change for an ideal gas with constant heat capacity that undergoes a change in temperature and volume. Using only this term, we would intuitively set ΔS as ΔSsys. We cannot set it to ΔSuniv. We can only set ΔS to ΔSsurr when we make the statement that the surroundings is an ideal gas.
The second expression is the entropy change of any generic irreversible process for any type of material. It could be viewed as ΔSsys or ΔSsurr. In the former case, T=Tsys. In the latter case, T=Tsurr.
Let's now drop the ambiguous ΔS to obtain this.
CVln(Tf/Ti)+Rln(Vf/Vi)=∫δqT+Sgen
The left side is the entropy change for an ideal gas under any process. This is the system. The right side is therefore the surroundings.
The fact that we include Sgen≠0 on the right side means that we are defining an irreversible process in the surroundings. The remaining term ∫… must be the reversible entropy change of the surroundings. Therefore, δq=δqrev,surr and T=Tsurr.
When the system is rigid, the term for Vf/Vi drops from the left side. With temperatures Ti=400 K, Tf=430 K, and Tsurr=500 K, the expression becomes as below.
CVln(430/400)=∫δqrev,surr500+Sgen
With the assumption that you know the material, you know CV. You have one equation and one two unknowns. You have one of two approaches to an answer. You must state the amount of heat flow out of the surroundings (δqrev,surr) as a constant. An immediate option is to say that this value is the reversible heat given to the system δqrev,surr=−δqrev,sys=−CVdT. From this, you obtain
CVln(430/400)=CV400−430500+Sgen
This allows you to solve for Sgen.
Alternatively, you must state the amount of irreversible entropy generated Sgen. When you also say that δqrev,surr is constant, you can solve for it.
Finally, with the parameters as given, the process is not reversible because the temperatures of the system and the surroundings are not the same at all stages. You could presume to call the process reversible so that Sgen≡0. This means, you must ignore the temperature of the surroundings in your considerations. This gives the following:
CVln(430/400)=∫δqrev,surrTsurr
With knowledge of the material, you use this to determine the reversible entropy change in the system or in the surroundings.