Why is PV work in thermodynamics so difficult to understand?

In summary, the conversation discusses the concept of PV work in thermodynamics and its applications. It is mentioned that while the equation for work seems simple, it does not take into account external factors and may not be applicable in irreversible processes. The ideal gas law also only applies to equilibrium states, making it difficult to calculate work in irreversible processes. The concept of isothermal and adiabatic processes is also touched upon, with the understanding that these processes may not be uniform in temperature or pressure. The conversation concludes by acknowledging that the difficulty in understanding thermodynamics may be due to poorly written textbooks.
  • #1
Est120
54
3
Homework Statement
Expression for "Work" in thermodynamics
Relevant Equations
Newton's 2nd law, ideal gas equation of state
i can't manage to grasp the concept of PV work in thermodynamics, for example we all know that W= integral(F*dx) like here
1612194728300.png

but this says that, at the end, W doesn't really depend on the gas temperature or reversible process crap
at the end W is simply a constant, atmospheric pressure is constant, piston weight is constant (and since V2=V1=0, and we neglect friction)
so that's absolutely understandable for me but in thermodynamics books:
1612194924626.png

and this doesn't agree with the first equation because here we need to have a reversible process in such a way that we can always associate a definite value for P to the gas
there is absolutely NO book that explains PV work in an understandable way , in thermodynamics work isn't just "Force x distance"
 
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  • #2
This assumes that you don't know how to calculate the force exerted by the gas in an irreversible process. However, you can use the other forces acting on the piston to determine what the work is that the gas does. Do you think that, in an irreversible expansion or compression, an ideal gas satisfies the ideal gas law?
 
  • #3
then why do we need all the reversible path crap also the value for W according to the "normal" formula (F*dist) at the end is always the same regardless of the path (mg and Pa are always the same in te process)
using that formula you can't differentiate from an isothermal or adiabatic or x process.
there are some things in thermodynamics that i will assume i will never understand
 
  • #4
The work is not always the same. This is a specific case in which you know the external force but may not necessarily directly know the force of the gas. So the formula still enables you to calculate the work.

The thing to remember is that, by Newton's 3rd law, the force per unit area exerted by the gas on the piston is always equal in magnitude and opposite in direction to the force per unit area exerted by the piston on the gas. However, the ideal gas law only applies to thermodynamic equilibrium states of the gas (and to reversible processes, which all consist of a continuous sequence of thermodynamic equilibrium states). So, for an irreversible process, you can't get the force per unit area exerted by the gas on the piston using the ideal gas law (or other equation of state). You need to hope that you can determine the work indirectly using the external forces. Sometimes, you can do that, but many times you can't. In the example they gave, where the external pressure and the friction force are known, you can determine the work ( at least in the end where the piston kinetic energy has been dissipated).

In terms of differentiating processes, you can always identify an adiabatic process because, in such a process, the cylinder is insulated and there is no heat transferred to or from the gas.

If the process is irreversible, the temperature of the gas may not be uniform spatially within the cylinder, so saying that it is isothermal is problematic. Some people regard an irreversible isothermal process as one in which the cylinder is held in contact with a constant temperature reservoir throughout the process, at the same temperature as the initial temperature of the gas. This certainly allows heat transfer to occur to the gas, so that the process is not adiabatic. But, during this irreversible change, the temperature of the gas inside the cylinder is not uniform at the reservoir temperature.

Be aware that, in an irreversible process, the gas pressure depends not only on the gas volume but also on the time rate of change of gas volume. So it is a rate process.

The reason that you have so much trouble understanding thermodynamics is not your fault. It's just that most of the textbooks out there are so poorly written.
 
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Related to Why is PV work in thermodynamics so difficult to understand?

1. Why is PV work important in thermodynamics?

PV work is important in thermodynamics because it helps us understand the relationship between pressure, volume, and energy in a system. This work is a measure of the energy that is transferred when a system changes its volume against an external pressure. It is a crucial concept in thermodynamics as it helps us calculate the change in internal energy of a system.

2. What makes PV work difficult to understand?

PV work can be difficult to understand because it involves multiple variables and their interactions. It requires a thorough understanding of thermodynamic principles and equations, as well as a strong grasp of mathematical concepts such as integration. Additionally, PV work is often visualized through graphs and diagrams, which can be challenging for some individuals to interpret.

3. How does PV work relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. PV work is a form of energy transfer, as it involves the transfer of energy between a system and its surroundings. Therefore, PV work is closely related to the first law of thermodynamics and is often used to calculate changes in the internal energy of a system.

4. Can you provide an example of PV work in action?

One example of PV work in action is the expansion of a gas in a piston-cylinder system. As the gas expands, it pushes against the piston, causing the volume to increase. This work done by the gas is equal to the force applied by the gas multiplied by the distance the piston moves, which is the change in volume. This work can be calculated using the equation W = PΔV, where W is the work done, P is the external pressure, and ΔV is the change in volume.

5. How can one better understand PV work in thermodynamics?

To better understand PV work in thermodynamics, it is important to have a strong foundation in the fundamental principles and equations of thermodynamics. It can also be helpful to visualize PV work through diagrams and graphs and to practice solving problems involving PV work. Additionally, seeking out additional resources such as textbooks, online tutorials, or consulting with a thermodynamics expert can also aid in understanding this concept.

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