# Constant pressure or isobaric expansion of gas

1. Apr 1, 2014

### Soumalya

It is said that in an isobaric expansion of a gas pressure remains constant throughout the expansion process.

Suppose we have a quantity of gas at initial pressure P1 and volume V2 in a piston cylinder arrangement.We heat it slowly such that it expands to obtain a state with pressure P2 and volume V2. Now if the process was isochoric P1=P2 and V2>V1.So the work done is said to be the displacement work done by the gas on the piston as ∫PdV=P(V2-V1) [as pressure remains constant].

But on the other hand for a system to expand there must be an imbalance in the external(surroundings) and internal(system)pressures.But in an isochoric process expansion takes place even if the pressure is said to be same at all instances of the process!

That is, if P1=P2=P(system pressure) how does the gas push the piston back?

Since the system was under equilibrium with the applied external pressure at state point 1 when external force was balanced by the internal pressure(force) by the gas on the piston multiplied by piston area.At state point 2 there is no change in pressure and hence internal pressure of the system is the same while the external pressure is still the same!Then how does the piston move back?

2. Apr 2, 2014

### Staff: Mentor

Hi Soumalya, welcome to PF!

You are right that for the piston to move, you need a $\delta P$. But conceptually, you can make it infinitesimally small, such that you can consider the pressure inside the piston as constant and equal to the outside pressure. It is similar to considering the reversible processes, where you take that the changes take place on a timescale that is slow with respect to the internal re-equilibration.

By the way, be careful not to confuse isobaric = constant pressure with isochoric = constant volume.

3. Apr 2, 2014

### 256bits

Suppose the pressure of the gas was a result of the weight of a mass resting on the (frictionless ) piston.
And all forces are in balance - the weight of the mass pushing down on the piston counterbalancing the pressure P1 of the gas pushing up on the piston area.

If we heat the gas, for the piston to move upwards, the mass has to accelerate to a certain velocity in relation to how quickly the heat is being added. Once a steady state velocity of the mass ( and piston ) is achieved, you will have to agree that the forces again balance - the weight of the mass pushes down with as much force as the pressure P2 of the gas pushes up on the piston ( since the velocity is constant ). P1 must equal P2 throughout the constant velocity segment of the isobaric process.

4. Apr 5, 2014

### nayanm

Also, some of the new-fangled experimental physical chemistry textbooks define an "infinitesimal" change in pressure to be such that the imprecision of your instrumentation is large enough to where it can't pick up such small changes/oscillations associated with the actually finite (though small) change in pressure.

This loosens the definition of such things as reversible and quasi-static processes to be applicable experimentally.

5. Apr 5, 2014

### yuiop

It might help to imagine that the piston is initially locked in place before the heating takes place. After heating, the pressure in the cylinder is noticeably greater than the external pressure. The piston is released and the volume expands rapidly accelerating the piston. If the piston has mass, it will gain momentum and may even overshoot to the extent that the pressure inside the cylinder drops below the external pressure, but then the piston reverse direction and oscillates until it comes to its final equilibrium position at the new volume with pressure equalised. Now this is not how we normally do thermodynamics, as it usually requires very slow "quasi-static" changes, but visualised like this you can see that the average pressure inside the cylinder is the same as the pressure outside the cylinder and with a frictionless piston the pressure differential is infinitesimal. In practice it is impossible to have a perfectly frictionless piston and there will be a small dp.

6. Apr 24, 2014

### Soumalya

What you say is the system would re-balance itself to a new equilibrium position after departing from the initial state properties where the weight of the mass is again balanced by the pressure force of the gas on the piston.

Since the process involves a transient phase in between two equilibrium states my precise question is how do we say the process is under constant pressure at all instances of observation?

Note: During the transient phase pint>pext until pint=pext when we say a new equilibrium state has been achieved.

Hence,a reversible process involves intermediate episodes of irreversibility when pint>pext.

7. Apr 24, 2014

### Soumalya

A small dP however infinitesimal it maybe is an imbalance in force which is mandatory for the piston the move.While we say and assume that for a reversible expansion process pint=pext at all instances of observation it simply doesn't make sense as for pint=pext the piston would never move according to laws of simple mechanics LOL.

What is being done is we are considering only the equilibrium states of the process and eradicating the transient part of the process in between equilibrium states when pint>pext.

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8. Apr 24, 2014

### Staff: Mentor

How do we say it? Easy, we open our mouths and form the words "the process is under constant pressure at all instances of observation".... At least that's what I do .

But I expect that you're asking a different question, namely whether we believe it when we say it, and the answer to that question is "No, but for all practical purposes we can calculate as if we do"; and the other posters to this thread have pretty decent answers as to why this works for all practical purposes.

You'll find similar issues with just about any transition between states of macroscopic systems in equilibrium: Dig deep enough, remove enough of the idealizing assumptions, assume measurements of sufficient unrealistic precision, and you'll often find that the equilibrium is an approximation and the system is in fact moving between disequilibrium on one side and disequilibrium on the other side. Even the simple case of using a 9.8 Newton force to lift a 1kg object at a constant non-zero velocity is subject to these concerns: How did we accelerate the object to that speed in the first place, if the net force is zero?

9. Apr 24, 2014

### Staff: Mentor

To add to what Nugatory wrote, take into account also that there is no such thing as a precisely constant pressure. The pressure is always fluctuating around an average value.

10. Apr 24, 2014

### Soumalya

Hi again,
I have another related doubt that's eating up my mind!Suppose we consider the piston cylinder arrangement again having some mass of gas in it.We observe the expansion of gas for two cases:

1. A reversible isobaric process

2. A reversible polytropic process

Reversible Isobaric Process:

Suppose the gas was at initially at pressure P1 and occupying volume V1.At this stage Pext=Pint=P1 so that the piston is stationary.Upon heating let's say the piston settles to a new position with an increase in volume to V2 and pressure P2 where it balances the external pressure again such that P2=Pext.

Since it's an isobaric process we must have P1=P2=P

Since both are equilibrium states internal gas pressure should also balance the external pressure and it's satisfied as P1=P2=Pext

Note: Pext is constant

Reversible Polytropic Process:

Again if we consider the expansion to follow polytropic path where the initial gas pressure was P1 and volume V1 the piston was stationary as P1=Pext.Now upon heating let's say the gas pressure becomes P2 and volume V2 when it reaches another equilibrium state.But for equilibrium we must have P2=Pext.Since external pressure remains constant we arrive to P1=Pext=P2

But for a polytropic process we must have P1≠P2

So how can equilibrium states be achieved despite what I explained in a polytropic process?

11. Apr 25, 2014

### Staff: Mentor

What is the polytropic process you are considering? It seems to me that you are just again considering an isobaric process, since the pressure is the same.

12. Apr 25, 2014

### Soumalya

I want to observe how an adiabetic process proceeds with equilibrium states against an external pressure.

13. Apr 26, 2014

### Soumalya

Here's what I am confused about:

For a reversible process it should proceed in stages of equilibrium so that properties could be fixed corresponding to each equilibrium state such that a path could be obtained on a a graphical plot.

In an isobaric process all equilibrium states are such that Pint=Pext.That is internal pressure was assumed to balance the exact external pressure at all equilibrium conditions such that the piston was stationary under a net vertical force as zero.

But if we consider a reversible adiabetic expansion of the gas then the change in pressure should follow the relation PVγ=constant such that P1V1λ=P2V2λ.So the volume of the gas increases with an even more aggregative relative increase of pressure.

But if the piston has to proceed in steps of equilibrium for an adiabteic process we must have Pint=pext again for all equilibrium conditions.But in an adiabetic process let's say when the pressure was P1 and the system was under equilibrium i.e, piston was stationary P1=Pext.

Again when another equilibrium state is reached let's say at P2 the condition for the piston to remain stationary should give us P2=Pext.

But P1≠P2 hence for two successive equilibrium states for an adiabetic process Pext should vary as internal gas pressure varies according to PVλ=constant.

So my question is does the external pressure vary during a reversible adiabetic expansion or compression of a gas?

14. Apr 26, 2014

### Staff: Mentor

I think I understand your confusion. When you write $P_1 = P_\mathrm{ext}$ and $P_2 = P_\mathrm{ext}$, you are assuming that $P_\mathrm{ext}$ is constant, but it can't be! If you have an adiabatic process for an ideal gas in a cylinder with a piston, the only external driving you have is the pressure. If the pressure doesn't change, then nothing happens.

In other words, at the beginning you have $P_1 = P_\mathrm{ext}$. At the end of the adiabatic process, $P_1 \neq P_2$, according to $PV^\gamma = \textrm{const.}$. Since the system is then again at equilibrium, $P_2 = P_\mathrm{ext}'$. Therefore, $P_\mathrm{ext} \neq P_\mathrm{ext}'$: the external pressure has changed.

15. Apr 26, 2014

### Soumalya

That's exactly what I meant!!!!!

Thanks DrClaude