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Starting from First Law of Thermodynamics, we have

ΔQ = ΔU + ΔW

Where, ΔW is the work done "by gas" on the surrounding.

Now, for an isobaric process, it is given by

ΔW = ∫PdV = PΔV

here P is the external pressure, which remains constant. Here is my doubt. Why do we take P as external pressure while calculating work done by "gas". Taking P as external pressure will calculated work done by surrounding, not by the gas.

Also, since internal pressure is changing and is not always equal to external pressure during the entire process, so both the work done are not same or equivalent.

I know that internal pressure is not same for the entire volume of gas and it is not easy to calculate work done by the gas. But this cannot be a reason why we take that P as external pressure because doing so we change the first law thermodynamics itself??

Please explain!!

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    $\begingroup$ who said that $P$ is the external pressure ? $\endgroup$
    – Ankit
    Commented Sep 25, 2020 at 10:01
  • $\begingroup$ Also suppose there was nothing to prevent the gas from expanding will it lose its energy while expanding ? $\endgroup$
    – Ankit
    Commented Sep 25, 2020 at 10:44
  • $\begingroup$ @Ankit ΔW = ∫PdV = PΔV. (I think u agree with this equation). Notice P is taken out of the integral sign, means P is constant. Since external pressure is constant in isobaric process and internal pressure varies. So I assumed (no one told me) that it is the external pressure. $\endgroup$
    – Kartikey
    Commented Sep 25, 2020 at 10:57
  • $\begingroup$ The gas expands when energy is given to it and this expansion occurs so that it's pressure remains the same as the earlier pressure. If it contracts it's the gases will apply more pressure than earlier. Also external pressure is always constant. Then why do you use different formula to find the work done by gas in different situations ? $\endgroup$
    – Ankit
    Commented Sep 25, 2020 at 11:10
  • $\begingroup$ Also isn't the isobaric process a quasi static process ? $\endgroup$
    – Ankit
    Commented Sep 25, 2020 at 11:23

3 Answers 3

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The ideal gas law describes the relationship between pressure, volume, and temperature of an ideal gas at thermodynamic equilibrium. It also describes the PVT relationship for a gas experiencing a very slow deformation process (quasi-static, reversible), since a reversible process is just a continuous sequence of closely neighboring thermodynamic equilibrium states. But for a rapid deformation of a gas, the ideal gas law no longer allows you to calculate the gas pressure correctly (especially at the moving boundary where the gas is doing work), because it applies only at thermodynamic equilibrium, and a rapid irreversible process passes through a sequence of non-equilibrium states. We know from fluid dynamics that, what is happening in an irreversible rapid-deformation process is that "viscous stresses" contribute to the pressure at the moving boundary. So the pressure must differ from the ideal gas law.

Now for $P_{ext}$ vs P: $P_{ext}$ is supposed to represent the pressure of the surroundings at the moving boundary where work is being done, and P is supposed to represent the pressure of the gas at this interface. Whether a process is reversible or irreversible, by Newton's law of action-reaction, we must always have that $P=P_{ext}$. And, for thermodynamic equilibrium or for a reversible process, P can be determined from the ideal gas law (or other real-gas equation of state). But, for an irreversible process, we can't use the ideal gas law, so we are more limited. To calculate the work done at the moving boundary, we must impose the external pressure manually or by an automatic control system to dictate the pressure for calculating the work done by the gas on its surroundings.

So, in summary, for all processes, both reversible or irreversible, the work done on the surroundings is $$W=\int{P_{ext}dV}=\int{PdV}$$However, for an irreversible process, we can not calculate P from the ideal gas law, so we are stuck using $P_{ext}$, which must be specified by other means.

ADDENDUM

The force balance on the piston (assumed frictionless) reads $$P_gA-mg - P_{atm}A=m\frac{dv}{dt}$$where, if our "system" is the gas, $P_g=P_{ext}$ is the action-reaction pair at the interface between the system and surroundings. $P_{ext}$ is the pressure exerted by the surroundings (in this case, the inside face of the piston) on the gas. So we have$$P_gA=P_{ext}A=mg + P_{atm}A+m\frac{dv}{dt}$$

If we now multiply this equation by the piston velocity, we get $$P_gA\frac{dx}{dt}=P_{ext}A\frac{dx}{dt}=mg\frac{dx}{dt} + P_{atm}A\frac{dx}{dt}+mv\frac{dv}{dt}$$or, equivalently, $$P_g\frac{dV}{dt}=P_{ext}\frac{dV}{dt}=mg\frac{dx}{dt} + P_{atm}\frac{dV}{dt}+mv\frac{dv}{dt}$$If we next integrate this equation between time zero and time t during the process, we obtain: $$W_g(t)=W_{ext}(t)=mgx(t)+P_{atm}[V(t)-V(0)]+m\frac{v^2(t)}{2}$$Note that the work done by the gas on its surroundings is exactly equal to the integral of $P_{ext}dV$ and that the kinetic energy of the piston is included in this work. The piston is frictionless, so it can't damp the motion of the piston. Even so, do you think the piston will continue oscillating forever, or do you think there is some other physical effect present that will eventually damp the motion of the piston (after a long time)?

The effect I'm talking about is viscous stresses within the gas which allow the time variations in $P_g(t)$ (and, thus, $P_{ext}(t)$ to adjust in such a way that they act to slow the movement of the piston, and eventually bring it to a stop. So, in the end, V=$$W_g(\infty)=W_{ext}(\infty)=mgx(\infty)+P_{atm}[V(\infty)-V(0)]$$

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  • $\begingroup$ "by Newton's law of action-reaction, we must always have that P=Pext. "-How Newton's third law guarantees that? In fact, they don't even comprise an action-reaction pair. The pairs are Fext(by surrounding on the piston) and the piston exerting an equal and opposite -Fext(on the surrounding. Second pair is the gas pressure(or force) pussing the piston outwards and in reaction, the piston pussing the gas inwards. $\endgroup$
    – Kartikey
    Commented Sep 26, 2020 at 13:05
  • $\begingroup$ Also if P=Pext and piston was initially at rest then it should not move at all. This is guaranteed by Newton's law of inertia(first law). Net force is zero and body was initially at rest so, no movement afterwards. No work done in any case!!! How? Please explain! $\endgroup$
    – Kartikey
    Commented Sep 26, 2020 at 13:07
  • $\begingroup$ @Kartikey If I may add to Chet's comment, the smaller the difference in pressure the slower the process and the lower the rate at which work is done, all other things being equal. That's one of the things that makes a reversible process so impractical. $\endgroup$
    – Bob D
    Commented Sep 26, 2020 at 16:34
  • $\begingroup$ @Kartikey Would you like me to provide a detailed analysis on how this plays out for a piston that has finite mass. I can do that if you feel it would help. $\endgroup$ Commented Sep 26, 2020 at 17:17
  • $\begingroup$ If the gas is the system, then the inside face of the piston is its interface with the surroundings, and $P_{ext}$ represents the pressure that the inside face of the piston exerts on the system. By Newton's 3rd law, this is equal to the pressure that the gas exerts on the inside face of the piston. So, yes, it is an action-reaction pair. The net force on the gas is not zero, and the net force on the surroundings is not zero. When we apply Newton's laws, and we are considering the forces acting on body A, we include the force that body B exerts on A, not the force that A exerts on B. $\endgroup$ Commented Sep 27, 2020 at 1:56
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In this answer, I am assuming an ideal gas in a cylinder fitted with a frictionless, leak-free and movable piston.

When the gas expands isobarically it does work against its surroundings. The work done by the gas is

$$ W = P(V_f - V_i) $$

Here $P$ is the pressure of the system, which for the most part is the gas.

The process by which the gas expands occurs quasi-statically. Heat is supplied quasi-statically to the gas. If it wasn't, I don't see how the process could be isobaric.

Since the process is quasi-static each set of values of the state variables $(P,V,T)$ between the final and initial state are possible equilibrium values. Thus, $P$ is equal to the external pressure.

If the gas pressure is not constant, the work done by the gas is

$$ W = \int_{V_{i}}^{V_{f}} P\,dV\,. $$

Edit: note that the initial and final state of the gas are equilibrium states.

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Here is my doubt. Why do we take P as external pressure while calculating work done by "gas". Taking P as external pressure will calculated work done by surrounding, not by the gas.

When calculating the work done by a gas on its surroundings you always use the external pressure. The reason is it is the external pressure that the gas must overcome in order to expand and do work. The internal gas pressure will be the same as the external pressure if the process is carried out very slowly (reversibly). But it is not necessary to carry out the process slowly because the work done by the gas only manifests itself with respect to the energy transferred to the surroundings, and that only depends on the external pressure.

For example, take a vertically oriented cylinder fitted with a massless piston and enclosing an ideal gas. A weight has been placed on top of the piston. The gas is initially in equilibrium both internally and with the surroundings which consists of the pressure of the weight and atmospheric pressure. The gas can be made to do work in the following two ways.

  1. We can very slowly (reversibly) add heat to the gas so that the gas expands very slowly doing work against the pressure of the weight and atmospheric pressure raising the weight a height $h$, while always being in equilibrium with the surroundings. The work done by the gas where $m$ is the mass of the weight and $A$ is the piston surface area, is then

$$W=P_{ext}\Delta V$$

$$W=\biggr(\frac{mg}{A}+1atm\biggl)Ah$$

  1. We can rapidly (irreversibly) heat the gas so that it expands more quickly and raises the weight to the height $h$ at equilibrium. In this case the gas is not in pressure equilibrium with the surroundings. Only at the boundary between the gas and surroundings is the pressure the same. Within the gas pressure gradients exist. The end result (i.e. work done) however is the same as when the process was carried out reversibly.

Bottom line: The work done by the gas depends only on the external pressure.

Hope this helps.

p.s.

Your first law equation $\Delta Q=\Delta U+\Delta W$ should be written as $Q=\Delta U+W$. The $\Delta$ symbol is reserved for changes in thermodynamic properties between equilibrium states such as $\Delta P$, $\Delta V$, $\Delta T$, $\Delta U$, and $\Delta S$. Heat $Q$ and work $W$ are not thermodynamic properties, but energy transfers. It makes no sense to speak of a change in heat or a change in work.

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