About Reversible vs Irreversible Gas Compression and Expansion Work

In summary, In the insight, there is a lack of information about the Figure 8 Work Done By Viscous Stresses on Piston. The sentence should actually read:Fig. 9 shows a plot of the ratio of the irreversible work to the reversible work ##W_{Irr} / W_R## for the case of an ideal gas as a function of the expansion/compression volume ratio ##V_f /...##
  • #1
cianfa72
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Hi,

reading the interesting Reversible vs Irreversible Gas Compression and Expansion Work insight by @Chestermiller I would like to ask for clarification on some points.

In the second bullet at the beginning
Since, by Newton’s third law, the external force per unit area exerted by the surroundings on the gas is equal to the gas thermodynamic pressure at the interface P.
my understanding is as follows: consider an ideal gas contained in a cylinder featuring a massless piston. An external force per unit area ##P_{ext}## is applied to the piston through a bunch of (external) bodies.

Applying the Newton's third laws to the massless piston we get two things:
  • the force per unit area the piston applies to the external bodies through the piston's outer edge is the same (in module) as the external force per unit area ##P_{ext}##
  • the force per unit area the thermodynamic pressure of the gas at the piston's inner edge ##P_I## (i.e. the thermodynamic pressure ##P_I## at the piston internal interface) applies to the piston is the same (in module) as the force the piston applies per unit area on the gas inside the cylinder
The above hold since otherwise the acceleration of the massless piston would be infinite.

Does it make sense ? Thank you.
 
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  • #2
cianfa72 said:
Hi,

reading the interesting Reversible vs Irreversible Gas Compression and Expansion Work insight by @Chestermiller I would like to ask for clarification on some points.

In the second bullet at the beginning

my understanding is as follows: consider an ideal gas contained in a cylinder featuring a massless piston. An external force per unit area ##P_{ext}## is applied to the piston through a bunch of (external) bodies.

Applying the Newton's third laws to the massless piston we get two things:
  • the force per unit area the piston applies to the external bodies through the piston's outer edge is the same (in module) as the external force per unit area ##P_{ext}##
  • the force per unit area the thermodynamic pressure of the gas at the piston's inner edge ##P_I## (i.e. the thermodynamic pressure ##P_I## at the piston internal interface) applies to the piston is the same (in module) as the force the piston applies per unit area on the gas inside the cylinder
The above hold since otherwise the acceleration of the massless piston would be infinite.

Does it make sense ? Thank you.
Yes it does make sense. But, in the way that I define things, I would consider ##P_{ext}## to be the force per unit area that the inside face of the piston applies to the gas. Otherwise, based on the way you have defined ##P_{ext}##, I would have to include the piston as part of the system. Of course, in this situation it doesn't matter, but, if the piston had mass and thermal inertia, by my way of doing things, the piston would have to be included as part of the system.

Also, I should mention that, if the process is irreversible, the force per unit area applied by the gas to the inside face of the piston would conceptually have to include the viscous normal stress.
 
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  • #3
Chestermiller said:
Yes it does make sense. But, in the way that I define things, I would consider ##P_{ext}## to be the force per unit area that the inside face of the piston applies to the gas. Otherwise, based on the way you have defined ##P_{ext}##, I would have to include the piston as part of the system. Of course, in this situation it doesn't matter, but, if the piston had mass and thermal inertia, by my way of doing things, the piston would have to be included as part of the system.
Yes, definitely.

Chestermiller said:
Also, I should mention that, if the process is irreversible, the force per unit area applied by the gas to the inside face of the piston would conceptually have to include the viscous normal stress.
Of course, as you explained in the 'Spring-Damper Model Analysis' and 'Ideal Gas Analysis' sections.
 
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  • #4
cianfa72 said:
Yes, definitely.Of course, as you explained in the 'Spring-Damper Model Analysis' and 'Ideal Gas Analysis' sections.
You are wonderful. I hope that you continue to stay involved with Physics Forums. It's so nice to encounter someone like you who is on the same wavelength as I.
 
  • #5
By the way: in the insight I do not see 'Figure 8 Work Done By Viscous Stresses on Piston' and the sentence
Fig. 9 shows a plot of the ratio of the irreversible work to the reversible work ##W_{Irr} / W_R## for the case of an ideal gas as a function of the expansion/compression volume ratio ##W_f / W_i##
should actually read:

Fig. 9 shows a plot of the ratio of the irreversible work to the reversible work ##W_{Irr} / W_R## for the case of an ideal gas as a function of the expansion/compression volume ratio ##V_f / V_i##
 
Last edited:

1. What is reversible gas compression and expansion work?

Reversible gas compression and expansion work refers to the process of compressing or expanding a gas in a reversible manner, meaning that the system can be returned to its original state without any loss of energy or increase in entropy.

2. What is irreversible gas compression and expansion work?

Irreversible gas compression and expansion work refers to the process of compressing or expanding a gas in an irreversible manner, meaning that the system cannot be returned to its original state without some loss of energy and increase in entropy.

3. How do reversible and irreversible gas compression and expansion work differ?

The main difference between reversible and irreversible gas compression and expansion work is that reversible processes are carried out slowly and without any friction or heat transfer, while irreversible processes are carried out quickly and with some friction and heat transfer, resulting in a loss of energy and increase in entropy.

4. What are the practical applications of reversible and irreversible gas compression and expansion work?

Reversible gas compression and expansion work is often used in thermodynamic cycles, such as the Carnot cycle, to maximize the efficiency of energy conversion. Irreversible gas compression and expansion work is commonly used in everyday devices, such as refrigerators and air conditioners, where efficiency is less important than speed and cost.

5. How can reversible and irreversible gas compression and expansion work be calculated?

The reversible gas compression and expansion work can be calculated using the equation W = -PextΔV, where Pext is the external pressure and ΔV is the change in volume. The irreversible gas compression and expansion work can be calculated using the equation W = -∫PdV, where P is the pressure and dV is the infinitesimal change in volume.

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