# Work done in expanding a gas reversibly and irreversibly

So, my chemistry teacher gave the class following $$P_{external}$$ versus $$Volume$$ diagrams for reversible and irreversible expansion of a gas which are as follows.

(Reversible expansion)

(Irreversible expansion)

Then he gave us following results to memorize: $$\mid \left(W_{reversible}\right)_{expansion}\mid \gt\mid \left(W_{irreversible}\right)_{expansion}\mid$$ Where W is for work. According to him, these results are always true. But consider the following graph

If I'm not wrong, this can also be considered an irreversible expansion. But as we can see clearly, $$\mid \left(W_{reversible}\right)_{expansion}\mid \lt\mid \left(W_{irreversible}\right)_{expansion}\mid$$ This thought is really bothering me. Where did I go wrong?

• The upper case looks like reversible isothermic expansion, the middle case like irreversible adiabatic expansion, the bottom case like reversible isobaric expansion and reversible isochoric compression.( All implies the system contains a constant amount of an ideal gas ). Jan 6 at 7:43
• @Poutnik - Is it right to call it a compression while already considering it "isochoric"? And why not call the middle case isochoric then isobaric expansion instead of irreversible adiabatic expansion? Jan 6 at 7:58
• Hehe, you are right, I revoke it. It is an isochoric cooling down. Jan 6 at 8:00
• The middle case can have 2 possible explanations. It is either irreversible adiabatic expension after sudden decrease of external pressure, either it is reversible isochoric cooling down and then reversible isobaric expansion. Jan 6 at 8:03
• Is the eternal force per unit area exerted by the inside face of the piston on the gas equal to the internal force per unit area exerted by the gas on the inside face of the piston, or is Newton's 3rd law of action-reaction no longer valid? Do you think that, in an irreversible rapid expansion, the "pressure" of the gas (force per unit area) is still determined by the ideal gas law (or other equation of state), or does the equation of state only apply to thermodynamic equilibrium states and reversible processes (comprised of a continuous sequence of thermodynamic equilibrium states)? Jan 6 at 13:58

I will call your processes (1), (2), and (3). I assume (1) is reversible isothermal expansion, (2) is irreversible isothermal expansion, and (3) is unspecified.

The key idea is that the inequality $$w_\text{rev} \geq w_\text{irrev}$$ can only be applied to reversible and irreversible analogues of the same process. (1) and (2) satisfy this requirement because (1) is the reversible analogue of (2), but (1) is not the reversible analogue of (3).

Physically, (1) is like having a pile of sand on top of a container (containing an ideal gas) whose lid is a piston, and then slowly blowing away the sand. (2) is the same scenario, except you sweep off the sand in one grand gesture. But (3) is a different scenario entirely.

If we refer to the first law, $$\Delta U = q + w$$, it's clear that, given an initial and a final state, we can make $$w$$ as large as we want as long as we have a commensurate value of $$q$$. Just because processes (1), (2) and (3) all have the same initial and final states does not mean that it makes sense to apply the inequality $$w_\text{rev} \geq w_\text{irrev}$$ to any two of them.

• I think you're right. I didn't think about the situation this way. Jan 13 at 16:02

The ideal gas model represents the limit of the behavior of real gases in the limit of low pressures. Real gases exhibit a non-zero viscosity, even in the ideal gas limit. So ideal gases must be considered to exhibit viscous behavior.

When a fluid (liquid or gas) is deformed very slowly, the state of stress within the fluid is isotropic (acting equally at all directions at each given location) and the force per unit area on any surface is equal to the pressure determined from the equation of state (e.g., the ideal gas law). However, as the rate of deformation becomes more rapid, the state of stress is no longer isotropic, and the viscous contributions to the stresses are proportional to the rate of deformation. For a gas in a cylinder experiencing a rapid (irreversible) deformation, this translates into the force per unit area exerted by the gas on the piston face varying not only with the volume of the gas, but also with the rate of change of volume. A crude approximation to the compressive stress on the piston face in a rapidly deforming gas can be expressed as: $$\sigma=\frac{nRT}{V}-\eta\frac{1}{V}\frac{dV}{dt}$$The first term represents the ideal gas behavior, and the second term represents the viscous contribution; the parameter $$\eta$$ is proportional to the gas viscosity.

In a rapid irreversible expansion of compression at constant externally applied pressure, we can set $$P_{ext}=\sigma=\frac{nRT}{V}-\eta\frac{1}{V}\frac{dV}{dt}$$ Note that, even if this is different from the original pressure and from the pressure calculated from the ideal gas law, the force per unit area exerted by the gas on the piston face can still match the external pressure if the gas deforms rapidly enough, such that the viscous contribution makes up the difference. In fact, this is what actually takes place as the gas response automatically adjusts to the externally applied pressure in a rapid (irreversible) process.

You are misunderstanding how pressure is changing.

Work is given by $$\delta W = -p_{ext}dV$$

Work is performed when the external pressure on the system is smaller than the internal pressure of your gas. $$p(A)$$ and $$p(B)$$ are the final and initial external pressures of your system. If you are expanding isobarically (as in your second image), $$W = -p(B) \Delta V$$ in the reversible and non-reversible case. Therefore, work done in an isobaric expansion is the same for reversible and non-reversible process. It may happen than once the pressure drops, the system might expand again, but we are not interested in that regime right now. No work will be done in an infinitesimal amount of time after a sudden instantaneous drop in pressure.

If the pressure change is adiabatic from $$p(B)$$ to $$p(A)$$, you will have the curve in image 1.

If the pressure change is irreversible i.e. you have a sudden change in pressure (in image 2, the pressure drops suddenly, not in a smooth, differentiable manner), you can apply $$W_{irrev} = -\int _{V_A} ^{V_B} p_{ext} dV = p(A)\Delta V$$ If it were reversible and adiabatic, $$W_{rev} = -\int p_{ext}dV = -nRT\ln \left( \frac{V(B)}{V(A)}\right)$$ assuming an ideal gas. You can check that $$W_{rev} \geq W_{irrev}$$

Note the $$\geq$$ rather than $$>$$ in the above expression. That was a slight error on the part of your teacher.

• Could you please explain these lines. Sorry, but I think, I don't clear understand it.- "Work is performed when the external pressure on the system is smaller than the external pressure of your gas" Jan 6 at 9:31
• Oops, sorry I made a mistake. It should be internal pressure of your gas. When your gas is pushing the boundary with a certain pressure $P$, the system will only expand if the outside pressure is less than $P$. Jan 6 at 9:46
• One more thing please. Should the third diagram be considered irreversible or not? Jan 6 at 11:51
• Since there is a discontinuity at $V=V(B)$, it is irreversible. Generally, any smooth curve on a $P-V$ plot is a reversible process, unless explicitly stated otherwise. Jan 6 at 12:12
• If you have a massless frictionless piston,, how can you say that the external pressure is different from the internal pressure. Or, if you regard the boundary of your system as the inside face of a piston, irrespective of whether the piston is massless or frictionless, how can you say that the external pressure is different from the internal pressure. Jan 6 at 12:23