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For an ideal gas constant volume heat addition process, change of entropy equation is:

$$\Delta S= c_v \ln\frac{T_2}{T_1}+ R \ln\frac{v_2}{v_1}= \int \frac{dq}{T}+S_{gen}$$

The term $R \ln\frac{v_2}{v_1}$ equals zero, since it’s a constant volume process.

For ideal gas $\int \frac{dq}{T} =c_v \ln\frac{T_2}{T_1}$ .

Then:

$$\Delta S= c_v \ln\frac{T_2}{T_1}= c_v \ln\frac{T_2}{T_1}+S_{gen}$$

Therefore, the $S_{gen}$ term equals zero and the process is reversible.

The question is: why does the $S_{gen}$ 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 $T_1 =400$ K and $T_2 =430$ K.

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    $\begingroup$ Please format math using MathJax $\endgroup$ – Aaron Stevens Jun 13 at 12:59
  • $\begingroup$ It seems that you have the cart before the horse as the saying goes. Taking the assumption of a reversible or an irreversible process tells us whether we set $S_{gen}$ zero or not. The explanation is not to be taken the other way around. $\endgroup$ – Jeffrey J Weimer Jun 13 at 13:58
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Foundations

The approach is to start from the definition of entropy change for a system $dS$ according to the laws of thermodynamics.

$$ dS \equiv \frac{\delta q}{T} $$

The heat flow $\delta q$ 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.

$$ \Delta S = \int \frac{\delta q_{rev}}{T} + \Delta S_{irr}$$

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 $\Delta S_{rev}$ and the entropy change of a real world process is the irreversible entropy generation $\Delta S_{irr}$ or $S_{gen}$.

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:

$$ dU^\star_{rev} = C_V dT = \delta q $$

$$ dS^\star_{rev} = C_v \frac{dT}{T} $$

$$ \Delta S^\star_{rev,V} = C_V \ln(T_f/T_i) $$

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

$$ \Delta S^\star_{rev,T} = R \ln(V_f/V_i) $$

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

$$ \Delta S^\star_{rev} = C_V \ln(T_f/T_i) + R \ln(V_f/V_i) $$

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 $S_{gen}$ to obtain

$$ \Delta S^\star_{univ} = \Delta S^\star_{sys} + \Delta S^\star_{surr} + S_{gen} = S_{gen}$$

The $S_{gen}$ 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 $dS_{gen} = \delta q/T_{surr}$.

The Case at Hand

The founding equation starts with $\Delta S$. This is ambiguous. Is it to be $\Delta S_{univ}$, $\Delta S_{sys}$, or $\Delta S_{surr}$? 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 $\Delta S$ as $\Delta S_{sys}$. We cannot set it to $\Delta S_{univ}$. We can only set $\Delta S$ to $\Delta S_{surr}$ 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 $\Delta S_{sys}$ or $\Delta S_{surr}$. In the former case, $T = T_{sys}$. In the latter case, $T = T_{surr}$.

Let's now drop the ambiguous $\Delta S$ to obtain this.

$$C_V \ln(T_f/T_i) + R\ln(V_f/V_i) = \int \frac{\delta q}{T} + S_{gen}$$

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 $S_{gen} \neq 0$ on the right side means that we are defining an irreversible process in the surroundings. The remaining term $\int \ldots$ must be the reversible entropy change of the surroundings. Therefore, $\delta q = \delta q_{rev,surr}$ and $T = T_{surr}$.

Specific Example

When the system is rigid, the term for $V_f/V_i$ drops from the left side. With temperatures $T_i = 400$ K, $T_f = 430$ K, and $T_{surr} = 500$ K, the expression becomes as below.

$$C_V \ln(430/400) = \int \frac{\delta q_{rev,surr}}{500} + S_{gen}$$

With the assumption that you know the material, you know $C_V$. 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 ($\delta q_{rev,surr}$) as a constant. An immediate option is to say that this value is the reversible heat given to the system $\delta q_{rev,surr} = -\delta q_{rev,sys} = - C_V dT$. From this, you obtain

$$C_V \ln(430/400) = C_V \frac{400 - 430}{500} + S_{gen}$$

This allows you to solve for $S_{gen}$.

Alternatively, you must state the amount of irreversible entropy generated $S_{gen}$. When you also say that $\delta q_{rev,surr}$ 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 $S_{gen} \equiv 0$. This means, you must ignore the temperature of the surroundings in your considerations. This gives the following:

$$C_V \ln(430/400) = \int \frac{\delta q_{rev,surr}}{T_{surr}}$$

With knowledge of the material, you use this to determine the reversible entropy change in the system or in the surroundings.

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  • $\begingroup$ Your final equation is incorrect. In an irreversible process, the entropy change for the ideal gas is still given by the previous equation. Entropy is a function of state (i.e., a physical property of the system), independent of any process path. $\endgroup$ – Chet Miller Jun 13 at 15:07
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The $S_{gen}$ term is not equal to zero. The equation you used to calculate $\int{dq/T}$ was applied improperly. The T in this equation should be the temperature at the boundary interface between the gas and its surroundings (where the heat transfer dq is occurring), $T_B$, not the temperature of the gas T: $$\int{\frac{dq}{T_B}}$$ For example, if the gas is being heated by contact with a reservoir at $T_{res}$, then $T_B=T_{res}$. Unfortunately, this critical requirement is almost always omitted in thermodynamics textbooks. For more details on this, see Fundamentals of Engineering Thermodynamics by Smith and van Ness.

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  • $\begingroup$ Chet, as usual, a better answer than mine. Bob $\endgroup$ – Bob D Jun 13 at 13:03
  • $\begingroup$ This is incorrect and fully confusing the fundamentals. When we determine the entropy change of a system, the temperature is the temperature of the system. Boundaries have no physical meaning other than to be between the system and the surroundings. They are NOT to be assigned a temperature, and they are not to be used to move the system temperature to some "reservoir", which is then by definition the temperature in the surroundings. In a reversible process, the reservoir (surroundings) and the system are at the same temperature. $\endgroup$ – Jeffrey J Weimer Jun 13 at 13:54
  • $\begingroup$ @JeffreyJWeimer. I'm sorry you feel that way. As you said, when we consider the entropy change of a system that has experienced an irreversible process, we devise an alternative reversible process path in which the temperature of the system and boundary are the same. But, when we are evaluating the entropy generated in the irreversible process, we employ the integral of $dq/T_B$ to determine the entropy transferred from the surroundings to the system, and we subtract this from the entropy change of the system (for the reversible path) to get the entropy generated in the system. $\endgroup$ – Chet Miller Jun 13 at 14:18
  • $\begingroup$ @JeffreyJWeimer If you want to learn more about this, please see the Moran et al reference I provided and also see Thermodynamics by Enrico Fermi, which likewise emphasizes using the boundary temperature in calculating the entropy transfer from the surroundings to the system. $\endgroup$ – Chet Miller Jun 13 at 15:03
  • $\begingroup$ The insult not to me, it is to the need to be precise in the language used to explain the process. The significant fallacy in your explanation is to use the word boundary as you use it. In thermodynamics, a boundary is a location that is neither the system nor the surroundings. Heat does not flow into or out of a boundary. Heat flows into or out of the system. Heat flows into or out of the surroundings. Heat flows THROUGH the boundary. $\endgroup$ – Jeffrey J Weimer Jun 13 at 15:06
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The question is: why does the S(gen) term equal zero and the process is reversible when this is a heat addition through a finite temperature difference?

S(gen)=0 only if the process is carried out quasi-statically. That means the temperature difference between the system and surroundings during the heat transfer is, in the limit, zero. That is what makes the process reversible.

Of course all REAL processes are irreversible. You can't have heat transfer without a temperature difference. The reversible transfer of heat is an idealization that requires the temperature difference to be infinitesimally small. That necessitates the process is carried out extremely slowly (quasi-statically). This means entropy generation approaches zero in the limit.

In my question I assume that the process in not quasi-static and there is a finite temperature difference between the system and the surroundings, but the problem is that the formulas I mentioned show that the process is reversible and s(gen) is zero, even when the process is not quasi-static.

The equation you provided assumes a reversible process. In your formula, $T_1$ is the initial temperature of the gas at the start of the process and $T_2$ is the final temperature of the gas at the end of the process. These are not equal. It is assumed that the difference between the gas temperature and the temperature of the surroundings is kept infinitely small throughout the process. For an ideal gas the relationship between temperature, pressure and volume of the gas between any two equilibrium states 1 and 2 is given by:

$$\frac{P_{2}V_{2}}{T_{2}}=\frac{P_{1}V_{1}}{T_{1}}$$

For a constant volume process, $V_{2}=V_{1}$ and therefore

$$\frac{P_2}{T_2}=\frac{P_1}{T_1}$$

So for a heat addition, both the temperature and pressure increase such that the ratio is constant. So in your equation $T_{2}>T_{1}$ and there is an increase in entropy of the system. But this increase is not entropy generation. Entropy generation only occurs if the process is irreversible (non quasi-static).

The entropy of the system can increase in a reversible process. This is not referred to as entropy generated. But for a reversible process the entropy of the surroundings decreases an equal amount such that the total entropy change, system + surroundings, equals zero. This means entropy generation is zero. That is what is meant by a reversible process.

Hope this helps.

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  • $\begingroup$ In my question I assume that the process in not quasi-static and there is a finite temperature difference between the system and the surroundings, but the problem is that the formulas I mentioned show that the process is reversible and s(gen) is zero, even when the process is not quasi-static. $\endgroup$ – صهيب أبو ريدة Jun 13 at 12:16
  • $\begingroup$ @صهيبأبوريدة See the revision to my answer to address your follow up comment. $\endgroup$ – Bob D Jun 13 at 12:47
  • $\begingroup$ @صهيبأبوريدة The minute that we state that a process is reversible, we state that system and surroundings are in mechanical, thermal, and chemical equilibrium at all points in time during the process. $\endgroup$ – Jeffrey J Weimer Jun 13 at 15:37

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