Problem (inspired by this question): a bubble of ideal gas at pressure $P_0$ and volume $V_0$ is adiabatically compressed to conditions $P_1, V_1$ such that its radius $r$ is reduced by $a$, where $a\ll r$. Find the work done on the bubble.
(Note that because $a$ is small, we can Taylor series expand expressions such as $(r-a)^3$ and retain only the initial terms to obtain approximations such as $V_1-V_0\approx\frac{3a}{r}V_0$ and $V_0/V_1\approx 1+\frac{3a}{r}$.)
Person 1: Everyone knows that for adiabatic ideal gas processes, $PV^\gamma=\mathrm{constant}$, where $\gamma$ is the heat capacity ratio. From the integral of work $W=\int_{V_0}^{V_1}-P\,dV=\int_{V_0}^{V_1}-P_0V_0^\gamma V^{-\gamma}\,dV$, we derive the expression $W=\frac{P_1V_1-P_0V_0}{\gamma-1}$. With $P_1=P_0(V_0/V_1)^\gamma\approx P_0\left(1+\frac{3a\gamma}{r}\right)$ and $V_1\approx V_0\left(1- \frac{3a}{r}\right)$, we obtain $$W\approx\boldsymbol{3P_0V_0a/r}(=\boldsymbol{4\pi ar^2P_0}).$$
Person 2: With such slight compression, it doesn't matter if the process is adiabatic or isothermal, as the heating is so slight. I'll just write the work integral using the ideal gas law: $$W=\int_{V_0}^{V_1}-P\,dV=-nRT\int_{V_0}^{V_1}dV/V=-P_0V_0\ln(V_1/V_0)\approx\boldsymbol{3P_0V_0a/r},$$ where I Taylor series expanded the logarithm as $\ln\left(\frac{V_1}{V_0}\right)\approx\ln\left(1-\frac{3a}{r}\right)\approx-\frac{3a}{r}$ for small $a$. See, the answers match.
Person 3: With such slight compression, the pressure won't change much. I'll just calculate the work as $$W=P\Delta V\approx \boldsymbol{3P_0V_0a/r}.$$ Same answer. You two are working too hard, no pun intended.
Person 4: I'll treat the gas as an elastic object with stiffness (i.e., bulk modulus) $K\equiv-V\left(\frac{\partial P}{\partial V}\right)=-V\left(\frac{\partial (P_0V_0^\gamma V^{-\gamma})}{\partial V}\right)=\gamma P_0\left(\frac{V_0}{V}\right)^\gamma\approx\gamma P_0\left(1+\frac{3x\gamma}{r}\right)$ for radial contraction $x$ (from 0 to $a$). This is approximately $\gamma P_0$ for minimal contraction. (Indeed, it's widely known that the adiabatic bulk modulus of a gas is $\gamma P$.) The volumetric energy gained by an elastic object under strain $\varepsilon$ is $K\varepsilon^2/2$, so the work done is $$W=KV\varepsilon^2/2\approx\boldsymbol{\frac{9}{2}\gamma P_0V_0a^2/r^2}(=\boldsymbol{6\pi \gamma a^2rP_0}).$$
Various others: This result is much smaller than ours. You've probably made too many assumptions, or your energy expression is invalid.
Person 4: Fine; I'll go back to the conjugate variables of stress $\sigma$ and strain $\varepsilon$; in differential terms, the energy increases by $V\sigma \,d\varepsilon=VE\varepsilon\,d\varepsilon$, where $E$ is a stiffness, a modulus of elasticity (this appears in myriad textbooks of elasticity). The volumetric strain $\varepsilon(x)\approx-\frac{3x}{r},$ so $d\varepsilon=-\frac{3}{r}dx$. The volume is $V\approx V_0\left(1-\frac{3x}{r}\right)$. And I showed above that the stiffness or bulk modulus is $K\approx\gamma P_0\left(1+\frac{3x\gamma}{r}\right).$ Now all I've done is make the same small-$a$ approximations that you have, correct?
Everyone else: Yes.
Person 4: Now I'll integrate:
$$W=\int_{0}^a\left[V_0\left(1-\frac{3x}{r}\right)\right]\left[\gamma P_0\left(1+\frac{3x\gamma}{r}\right)\right]\left(-\frac{3x}{r}\right)\left(-\frac{3}{r}\right)dx=\frac{9 a^2 \gamma [2 r^2+ 4 a (\gamma – 1) r-9 a^2 \gamma ]}{4 r^4}.$$
Ignoring higher-order terms of $a$, I again obtain $$W\approx\boldsymbol{\frac{9}{2}P_0V_0\gamma a^2/r^2},$$
which differs from the other answers. Why?
Various others: ???
Best Answer
There is indeed a problem with the energy expression $KV\varepsilon^2/2$. The problem lies in the derivation step $V\sigma\,d\varepsilon=VE\varepsilon\,d\varepsilon$, which relies on the Hooke's Law substitution $\sigma=E\varepsilon$. Person 4 is conflating the original configuration of bubble radius $r$ (i.e., $\varepsilon=0$) with a stress-free state $\sigma=0$, but this isn't the case; the original configuration involves a stress of $\sigma=-P_0$.
Put another way, we don't approximate a small amount of work done by multiplying a stress increase by a strain and dividing by two, as described in the linked question. That would essentially be finding the work from $-dP\,dV$ instead of the correct $-P\,dV$, with the resulting estimate being far too small ($\frac{9}{2}P_0V_0\gamma a^2/r^2\ll3P_0V_0a/r$).
This point isn't specific to the ideal gas; for any material, the strain energy is $EV\varepsilon^2/2$ only if $\varepsilon=0$ corresponds to the stress-free or unconfined state. If you hand me a block of steel clamped in a vise and instruct me to add 1 MPa to the stress, I can do it with no problem just by tightening the vise a tad. But if you ask me how much strain energy I added, I'm clueless without knowing the starting stress.
For condensed matter, we sometimes cheat a bit and use 1 atm as a reference state, as the volume doesn't change much from 0 atm to 1 atm. We may even conflate changes in the internal energy $\Delta U$ with changes in the enthalpy $\Delta H=\Delta U+P\Delta V$, and the associated error is often minimal. Condensed matter is stiff compared to the stresses we typically apply to it. But gases are extremely compliant; their isothermal stiffness essentially equals their pressure (exactly so for the ideal gas). We thus can't consider the pressurization to $P_0$ in this problem to be ignorable.
Person 4 could have instead used $V(\sigma-P_0)\,d\varepsilon=V(K\varepsilon-P_0)\,d\varepsilon$, where $\sigma$ is the additional stress, to obtain $$W\approx\frac{3 a P_0V_0 (27 a^3 \gamma^2 - 12 a^2 (\gamma - 1) \gamma r - 6 a (\gamma + 1) r^2 + 4 r^3)}{4 r^4},$$
which after dropping small terms clearly reduces to $W\approx\boldsymbol{3P_0V_0a/r}$, matching the other answers.