Work done in reversible and irreversible processes

In summary, the conversation discusses confusion over working with reversible and irreversible processes in thermodynamics. The speaker wants to calculate the entropy change for an ideal gas undergoing an isothermal process with two different conditions. They use the equation dU = TdS - PdV and find different results for the entropy change. The expert clarifies that the equation holds for both reversible and irreversible processes, but cannot be applied directly to the latter. They also provide a resource for understanding how to determine entropy change for irreversible processes.
  • #1
Decimal
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Hello,

I am encountering some confusion understanding the difference in working with reversible and irreversible processes in thermodynamics. Let's say I have a process where an ideal gas at a certain starting temperature ##T_i## expands from volume ##V_i## to ##V_f##. The temperature of the surroundings stays constant at ##T_i##. There are two ways in which this process can happen, namely as an isothermal reversible process, and an isothermal process with a constant external pressure ##P_{ext}##. I would like to calculate the entropy change of the gas in both of these processes.

I will use the following equation, which according to my book holds for both reversible and irreversible processes: $$ dU = TdS - PdV $$ Both processes are isothermal so ##dU = 0##. This gives the following relation: $$dS = \frac {P} {T} dV$$ Working this out for the first process gives me: $$\Delta S = n*R *\int{\frac {dV} {V}}$$ Here I used the ideal gas equation. Now for the second process I again do the same calculation, with a constant pressure, I get: $$\Delta S = \frac {P_{ext}} {T} \int dV$$ So $$\Delta S = P_{ext}*(V_f - V_i)$$

However this answer is much different from the first answer, whilst the entropy should be a state variable for the gas. My book says that since the state is the same at the end of the process, the entropy should also be the same at the end for both processes. In my calculation this isn't a case. I feel like I am either missing a difference between the work done in reversible and irreversible processes, or I am not using the constant pressure correctly. Any help would be much appreciated!
 
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  • #2
Decimal said:
Hello,

I am encountering some confusion understanding the difference in working with reversible and irreversible processes in thermodynamics. Let's say I have a process where an ideal gas at a certain starting temperature ##T_i## expands from volume ##V_i## to ##V_f##. The temperature of the surroundings stays constant at ##T_i##. There are two ways in which this process can happen, namely as an isothermal reversible process, and an isothermal process with a constant external pressure ##P_{ext}##. I would like to calculate the entropy change of the gas in both of these processes.

I will use the following equation, which according to my book holds for both reversible and irreversible processes: $$ dU = TdS - PdV $$
It really gives me a pain when books say that this relationship holds for both reversible and irreversible process. This equation actually expresses the relationship between dU, dS, and dV between two closely neighboring (differentially separated) thermodynamic equilibrium states of a material, so it is independent of any specific process. Since a reversible process involves differential changes along a continuous sequence of thermodynamic equilibrium states, it can also be used for reversible process between thermodynamic equilibrium states that are separated by large changes in U, S, and V at the end states. However, since an irreversible process does not involve differential changes along a continuous sequence of thermodynamic equilibrium states, the equation cannot be applied directly to relate the changes between end states the are separated by large changes. However, if you can identify a reversible path between the same two end states as the irreversible process, the equation can be applied to that.

Both processes are isothermal so ##dU = 0##. This gives the following relation: $$dS = \frac {P} {T} dV$$ Working this out for the first process gives me: $$\Delta S = n*R *\int{\frac {dV} {V}}$$ Here I used the ideal gas equation. Now for the second process I again do the same calculation, with a constant pressure, I get: $$\Delta S = \frac {P_{ext}} {T} \int dV$$ So $$\Delta S = P_{ext}*(V_f - V_i)$$
Like I said, you can't apply the equation directly to the irreversible process. However, since the two end states for the reversible and irreversible paths are the same for this process, your result for the entropy change for the reversible path also applies to the irreversible path; they are equal.

If you want to get an understanding of how to determine the entropy change for an irreversible process process, there is a cookbook recipe for doing this. I've discussed this in a Physics Forums Insights article I wrote a while ago (which also discusses the rationale for the methodology): https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Enjoy!
 
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  • #3
Thanks a lot for the elaborate answer! I will definitely read your article. Looking back I also notice that in my calculation I assumed the work done by the gas, dependent on the external force, to be equal to ##P dV##, which of course only holds for reversible processes.
 
  • #4
Decimal said:
Thanks a lot for the elaborate answer! I will definitely read your article. Looking back I also notice that in my calculation I assumed the work done by the gas, dependent on the external force, to be equal to ##P dV##, which of course only holds for reversible processes.
But, remember also that, irrespective of whether a process is reversible or irreversible, the volumetric work done by a system on its surroundings is always ##W=\int{P_{ext}dV}##, where ##P_{ext}## is the force per unit area at the interface between the system and its surroundings.
 
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  • #5
Yeah I understand. Where I went wrong is equating ##P_{ext}dV## to ##P_{gas}dV##which is only true in a reversible process. Thanks for the help!
 

FAQ: Work done in reversible and irreversible processes

1. What is the definition of work done in reversible and irreversible processes?

The work done in a process is defined as the product of the force applied and the distance moved in the direction of the force. In reversible processes, the force applied is always equal and opposite to the force of the surroundings, resulting in a reversible change in the system. In irreversible processes, the force applied is not equal and opposite to the force of the surroundings, leading to an irreversible change in the system.

2. How does work done in reversible and irreversible processes affect the system?

In reversible processes, the work done is equal to the change in internal energy of the system, and therefore the system can return to its original state without any loss of energy. In irreversible processes, the work done is less than the change in internal energy, and the system cannot return to its original state, resulting in a loss of energy.

3. Is work done in reversible and irreversible processes always positive?

No, work done can be both positive and negative in both reversible and irreversible processes. In reversible processes, if the force and displacement are in the same direction, the work done is positive, and if they are in opposite directions, the work done is negative. In irreversible processes, the work done is always negative as there is a loss of energy.

4. How is the efficiency of a reversible process compared to an irreversible process?

The efficiency of a reversible process is always greater than that of an irreversible process. This is because a reversible process has no loss of energy, and therefore the work done is equal to the change in internal energy, resulting in a higher efficiency. In contrast, an irreversible process has a loss of energy, and therefore the work done is less than the change in internal energy, leading to a lower efficiency.

5. Can work done in reversible and irreversible processes be converted into other forms of energy?

Yes, work done in both reversible and irreversible processes can be converted into other forms of energy, such as heat or electrical energy. In reversible processes, the work done can be fully converted into other forms of energy without any loss. In irreversible processes, some of the work done is lost as heat, resulting in a lower conversion efficiency.

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