# Confused about use of external vs internal pressure in thermodynamics

• etotheipi
In summary, the conversation discusses the use of external and internal pressure in thermodynamic equations such as work done, enthalpy change, and specific heat. It also addresses the concept of isobaric processes and the behavior of ideal gases in an irreversible process. Through the analysis of an example scenario, it is concluded that the change in internal energy of an ideal gas is zero when the temperature returns to its initial value, indicating that the heat transferred is equal to the work done by the external pressure.
etotheipi
I've learned that ##W = -\int{P_{ext} dV}##, and only during a reversible/quasi-static process where ##P_{int} = P_{ext}## can we write the work done on the gas in terms of the internal pressure (and consequently use ##PV=nRT## etc. which apply to the internal gas).

However, a lot of sources, when showing derivations of various basic thermodynamic equations, don't refer to ##P_{ext} dV## and use a general ##P dV##, which I assume refers to the internal pressure.

As an example, the change in enthalpy (at constant pressure) ##dH## is defined as$$dH = dU + d(PV) = dU + P_{int} dV$$Knowing that ##dQ = dU + P_{ext} dV##, this apparently proves that ##dH = dQ## at constant pressure. However, the two pressure are clearly different so there must be more to it?

Another example is specific heat at constant pressure. If ##dQ = C_{P}dT## then ##dU = C_{P}dT - P_{ext}dV##. They then quote that from the ideal gas law, at constant pressure, ##PdV = nRdT##, which when substituted in gives Meyer's relation. Except this is only the case if ##P = P_{ext}##, which doesn't appear to be true in this example considering the ideal gas law deals with the internal pressure.

Both of these have used the ##P## term quite ambiguously and I can't follow why. I was wondering if someone could clear things up - do these equations only "work" (no pun intended...) if the internal and external pressures are identical? Thank you!

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Would you consider it an isobaric process if the gas in the cylinder were initially at equilibrium at pressure p, and you then suddenly changed the external pressure from p to P and held it at that value until the system re-equilibrated?

Chestermiller said:
Would you consider it an isobaric process if the gas in the cylinder were initially at equilibrium at pressure p, and you then suddenly changed the external pressure from p to P and held it at that value until the system re-equilibrated?

I'm guessing that since the internal pressure is going to eventually increase to P upon re-equilibration, the process will not be isobaric (referring to the system)? Though I'm not too sure how I can apply the same sort of reasoning to the above processes.

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etotheipi said:
I'm guessing that since the internal pressure is going to eventually increase to P upon re-equilibration, the process will not be isobaric (referring to the system)? Though I'm not too sure how I can apply the same sort of reasoning to the above processes.
Are you aware that, in an irreversible process (like a rapid expansion or compression), an ideal gas does not obey the ideal gas law, and you can't determine the "internal pressure" using the ideal gas law? Are you also aware that, for a massless frictionless piston, the internal force per unit area exerted by the gas on the piston must exactly match the external applied pressure (by Newton's 2nd law)?

Let's analyze the above process using the first law of thermodynamics. Suppose that the system is in contact with a constant temperature reservoir held at the same temperature as the initial temperature of the gas. What is the change in internal energy of this system? What is the change in enthalpy? How much work is done? How much heat is transferred? How does the change in enthalpy compare with the heat transferred?

Chestermiller said:
Are you aware that, in an irreversible process (like a rapid expansion or compression), an ideal gas does not obey the ideal gas law, and you can't determine the "internal pressure" using the ideal gas law? Are you also aware that, for a massless frictionless piston, the internal force per unit area exerted by the gas on the piston must exactly match the external applied pressure (by Newton's 2nd law)?

Let's analyze the above process using the first law of thermodynamics. Suppose that the system is in contact with a constant temperature reservoir held at the same temperature as the initial temperature of the gas. What is the change in internal energy of this system? What is the change in enthalpy? How much work is done? How much heat is transferred? How does the change in enthalpy compare with the heat transferred?

Let's say that ##P > p##. When the external pressure is increased suddenly, the system will undergo an irreversible compression. The work done on the system will be ##W = -\int{P_{ext} dV} = -P_{ext}\Delta V##. I don't see why any heat could have been transferred, so I'm guessing ##\Delta Q = 0##. So we have ##\Delta U = -P_{ext}\Delta V##.

Now the change in enthalpy ##\Delta H = \Delta U + \Delta(P_{int}V)##, and since neither ##P_{int}## nor ##V## are constants, I'm not sure exactly what ##\Delta(P_{int}V)## is though I can say that it is definitely non zero, which leads to ##\Delta H \neq \Delta Q##? That's interesting.

etotheipi said:
Let's say that ##P > p##. When the external pressure is increased suddenly, the system will undergo an irreversible compression. The work done on the system will be ##W = -\int{P_{ext} dV} = -P_{ext}\Delta V##. I don't see why any heat could have been transferred, so I'm guessing ##\Delta Q = 0##. So we have ##\Delta U = -P_{ext}\Delta V##.
The gas would have to equilibrate to the same final temperature as its initial temperature if the system is in contact with a constant temperature reservoir at the initial temperature. What does this tell you about ##\Delta U## of your ideal gas?

Chestermiller said:
The gas would have to equilibrate to the same final temperature as its initial temperature if the system is in contact with a constant temperature reservoir at the initial temperature. What does this tell you about ##\Delta U## of your ideal gas?

I'm not too confident about this answer, but if for an ideal gas ##U = \frac{3}{2} nRT##, then if ##T## returns to its original value we must have ##\Delta U = 0##, implying then that ##\Delta Q = P_{ext} \Delta V##?

etotheipi said:
I'm not too confident about this answer, but if for an ideal gas ##U = \frac{3}{2} nRT##, then if ##T## returns to its original value we must have ##\Delta U = 0##, implying then that ##\Delta Q = P_{ext} \Delta V##?
That is correct. Now what is the change in enthalpy? How does the change in enthalpy compare with Q?

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etotheipi and anorlunda
Chestermiller said:
That is correct. Now what is the change in enthalpy? How does the change in enthalpy compare with Q?

Yes sorry about this, I don't have a textbook and am trying to learn the concepts over the internet which isn't a very good way of doing so since there's so much contradictory information out there. I was wondering if you knew of any good introductory texts that I could try getting hold of?

The change in enthalpy is ##\Delta (P_{int}V) = PV_{2} - pV_{1}##. The initial and final states are equilibrium so the ideal gas law applies, with Boyle's relation giving ##PV_{2} - pV_{1} = 0##. So we have ##\Delta H = 0##, which equals ##\Delta Q##?

etotheipi said:
Yes sorry about this, I don't have a textbook and am trying to learn the concepts over the internet which isn't a very good way of doing so since there's so much contradictory information out there. I was wondering if you knew of any good introductory texts that I could try getting hold of?

Introduction to Engineering Thermodynamics by Moran et al. I highly recommend this book.
The change in enthalpy is ##\Delta (P_{int}V) = PV_{2} - pV_{1}##. The initial and final states are equilibrium so the ideal gas law applies, with Boyle's relation giving ##PV_{2} - pV_{1} = 0##. So we have ##\Delta H = 0##, which equals ##\Delta Q##?
Your result for the enthalpy change is correct. Nicely done. But you already correctly showed that Q is not equal to zero: Q = W

etotheipi
Chestermiller said:
Introduction to Engineering Thermodynamics by Moran et al. I highly recommend this book.
Your result for the enthalpy change is correct. Nicely done. But you already correctly showed that Q is not equal to zero: Q = W

I had an idea relating to the problem right at the beginning; it seems as though we require constant internal pressure for ##dQ = dH## (since it certainly didn’t apply to the scenario we just analysed where internal pressure was variable), and if internal pressure is constant this forces ##P_{int} = P_{ext}## (since we can assert the internal pressure is uniform if it is constant, and so ##P_{int} = P_{on-boundary} = P_{ext}## etc.)?

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etotheipi said:
I had an idea relating to the problem right at the beginning; it seems as though we require constant internal pressure for ##dQ = dH## (since it certainly didn’t apply to the scenario we just analysed where internal pressure was variable), and if internal pressure is constant this forces ##P_{int} = P_{ext}## (since we can assert the internal pressure is uniform if it is constant, and so ##P_{int} = P_{on-boundary} = P_{ext}## etc.)?
When you use the term internal pressure, it makes me cringe. By this term, do you mean the force per unit area exerted by the gas on the inside face of the piston? Do you think that this can be calculated from the gas temperature and volume using the ideal gas law?

Chestermiller said:
When you use the term internal pressure, it makes me cringe. By this term, do you mean the force per unit area exerted by the gas on the inside face of the piston? Do you think that this can be calculated from the gas temperature and volume using the ideal gas law?

Not exactly, I'm aware that during non-reversible processes the pressure exerted by the gas on the inside face of the piston is not defined using the ideal gas law due to viscous effects etc. In cases such as these, it is required to use the external pressure.

However, I had assumed that if the process is reversible, or in such a case where the pressure of the gas inside the piston is constant, then we can refer to such an internal pressure - i.e. in scenarios where the pressure at the piston face equals the pressure throughout the gas within the piston.

etotheipi said:
Not exactly, I'm aware that during non-reversible processes the pressure exerted by the gas on the inside face of the piston is not defined using the ideal gas law due to viscous effects etc. In cases such as these, it is required to use the external pressure.

However, I had assumed that if the process is reversible, or in such a case where the pressure of the gas inside the piston is constant, then we can refer to such an internal pressure - i.e. in scenarios where the pressure at the piston face equals the pressure throughout the gas within the piston.
In the case that we are examining, the process is not reversible, and viscous stresses do contribute to the force per unit area exerted by the gas on the inside piston face. When we calculate enthalpy, the definition requires that we are use only at a value at thermodynamic equilibrium. That would entail only using the value p in the initial state (not the value P during the irreversible change) and the value P in the final state.

## 1. What is the difference between external and internal pressure in thermodynamics?

External pressure refers to the pressure exerted by the surroundings on a system, while internal pressure refers to the pressure within the system itself. In thermodynamics, external pressure is often used to describe the pressure exerted on a gas or liquid by a piston or other external force, while internal pressure is used to describe the pressure within the gas or liquid due to its own molecular motion.

## 2. How are external and internal pressure related in thermodynamics?

In thermodynamics, external and internal pressure are related through the ideal gas law, which states that the product of pressure and volume is directly proportional to the number of moles of gas and the temperature. This means that as external pressure increases, so does internal pressure, and vice versa.

## 3. When should external pressure be used in thermodynamics calculations?

External pressure should be used in thermodynamics calculations when the system is in contact with its surroundings and is affected by external forces, such as in a piston-cylinder system. In these cases, the external pressure is an important factor in determining the behavior of the system.

## 4. When should internal pressure be used in thermodynamics calculations?

Internal pressure should be used in thermodynamics calculations when the system is isolated and not affected by external forces, such as in an insulated container. In these cases, the internal pressure is the only pressure acting on the system and is used to determine its behavior.

## 5. How does the use of external and internal pressure affect thermodynamic processes?

The use of external and internal pressure can affect thermodynamic processes in different ways. In processes where the external pressure is kept constant, changes in internal pressure can affect the volume and temperature of the system. On the other hand, in processes where the internal pressure is kept constant, changes in external pressure can affect the volume and temperature of the system. Understanding the relationship between external and internal pressure is important in analyzing and predicting the behavior of thermodynamic systems.

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