I recently had the following homework problem (Knight 3e, exercise 16.72, page 467):

The cylinder in Figure CP16.72 has a moveable piston attached to a spring. The cylinder's cross-section area is $10 \, \mathrm{cm^2}$, it contains $0.0040 \, \mathrm{mol}$ of gas, and the spring constant is $1500 \, \mathrm{N/m}$. At $20 \, \mathrm{°C}$ the spring is neither compressed nor stretched. How far is the spring compressed if the gas temperature is raised to $100 \, \mathrm{°C}$?

Figure CP16.72, showing the piston setup described in the question (no additional information is discernible from the diagram)

I came up with the following solution:

There are three forces acting on the piston at any given time: $pA$ to the right, $p_{\mathrm{atmos}} A$ to the left, and $kx$ to the right. When the system is in equilibrium, $pA = p_{\mathrm{atmos}} A + kx$. At $T_1 = 293 \, \mathrm{K}$, the string is neither stretched nor compressed, so $x = 0$; further, given that the system is at equilibrium, we have $p_1 A = p_{\mathrm{atmos}} A$, so $p_1 = p_{\mathrm{atmos}} = 101.3 \, \mathrm{kPa}$. Therefore, the above invariant can be rearranged to \begin{align} p_2 &= \frac{1}{A} (p_1 A + kx). \tag{1} \end{align} Note also that \begin{align} V_2 = V_1 + Ax. \tag{2} \end{align} Then, from the ideal gas law, $p_2 V_2 = nRT_2$; substituting $(1)$ and $(2)$ gives \begin{align*} \frac{1}{A} (p_1 A + kx) (V_1 + Ax) &= nR T_2 \\ (p_1 A + kx) (V_1 + Ax) &= nR T_2 A \\ (Ak)x^2 + (V_1 k + p_1 A^2) x + A(p_1 V_1 - nR T_2) &= 0. \tag{3} \end{align*} This is a rather ugly but quite valid quadratic equation for $x$. Numerical evaluation yields a solution of $x = 1.0198 \, \mathrm{cm}$. Interestingly, there is also an equilibrium point at $x = -17.3927 \, \mathrm{cm}$, but this is longer than the depth of the cylinder and thus impossible.

I checked with my professor, and the solution manual uses a similar approach and ends up with the same equation in $(3)$.

I'm wondering what that extra, negative solution for $x$ represents. It doesn't make sense to me that there should be a negative solution: if $x$ is negative, then the spring is expanded and the gas is compressed, so there's nothing to counteract the piston's movement to the right and keep the system at equilibrium (except for the insufficiently strong atmospheric pressure).

My instructor suggested that deriving a potential energy formula $U(x)$ for the whole system might show that the desired solution is impossible to get to (i.e., would require infinite force or something because it's past a place where the required energy diverged), but I don't know how to go about doing that.

What's the significance or meaning of this solution? Is there any case where there could be a negative solution that's not longer than the length of the cylinder—and thus is a valid solution? If so, what would such a system look like?

  • $\begingroup$ As you note, the second solution gives a length longer than the cylinder depth. This would mean in the equation you wrote, the volume would be negative. If you just look at the ideal gas law, nothing mathematical prevents negative pressure or volume. So the extra solution is when the negative pressure due to negative volume balances the pull of the spring. In other words, it's an absurd solution. $\endgroup$ – MonkeysUncle Mar 4 '15 at 2:26

A negative solution like that is essentially just a sideresult from the mathematical method.

You could consider your physical model as just a piece of a more general mathematical model. Solving the problem on the mathematical model then gives you all solutions; but only the solutions lying within the physical part of this model are actually solving the original problem.

For a physical model you always start out with some boundaries. Something like the allowed range of a variable $x$ and so on. The other results outside of the boundaries are not usable for that physical model and do not have any real significance.

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    $\begingroup$ While I don't see a really useful interpretation this time, I would encourage you not to just dismiss all "sideresult"s as meaningless. My usual classroom example is a person thowing a ball from non-zero height and asking when does it get to the ground. Students get a positive and a negative answer where the postive answer is the desired one. But the negative answer is when the ball would need to have been projected from the ground to arrive at the trowing hand at the nonminal moment of release, which is a strange bu phyically meaningful quantity. $\endgroup$ – dmckee --- ex-moderator kitten Nov 4 '19 at 18:31

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