I am interested in answering the question which has nothing to do with a "homework". The question has also no aim to "check my work" because I have no doubts in my calculations and I have added them only because some people in the comments found the result "not trivial". I have however still doubts in the basic concept.

In view of this I ask for reopening the question because there are obviously different opinions on the subject and people who do not agree with the accepted answer have right to tell their opinion.

The problem is following:

The volume of a cylinder is separated by a piston. There is some ideal gas at temperature $T_0$ and volume $V_0$ at one side of the piston and a spring at the other side. The piston is initially fixed so that the spring has the equilibrium length. Being relaxed the piston comes to rest at some position such that the gas volume becomes $V_1$ (obviously $V_1>V_0$). Find the new temperature of the ideal gas. Both the cylinder walls and piston are thermal insulators.

This assumed to be a school problem and it is easily solved assuming that the internal energy of the gas is given by $U=n c_V T$ and that the "lost" portion of the internal gas energy transfers to the potential energy of the spring.

Following the line one obtains: $$ \frac{T_0}{T_1}=1+\frac{\gamma-1}2\left(1-\frac{V_0}{V_1}\right)\tag1 $$ where $\gamma=\frac{c_p}{c_V}=1+\frac R{C_V}$ is the adiabatic index. Note that neither the mass of the piston nor the stiffness of the spring enters the solution.

However somebody with a high-school knowledge can consider this as an adiabatic process and use simply the equation $TV^{\gamma-1}=\text{const}$ so that $$ \frac{T_0}{T_1}=\left(\frac{V_1}{V_0}\right)^{\gamma-1}.\tag2 $$

Obviously the two solutions are different and therefore at least one of them is wrong. It seems Eq.(2) cannot be applied because the piston coming to rest implies the process is irreversible.

My question is therefore: is it conceptually wrong to consider an adiabatic process or can this be used in conjunction with other assumptions to provide a correct solution?

The derivation of the formula $(1)$

The energy conservation law reads: $$\begin{align} nc_V(T_0-T_1)&=\frac{k(x_1-x_0)^2}2\\ &=\frac{p_1A(x_1-x_0)}2\\& =\frac{p_1V_1(1-V_0/V_1)}2\\ &=\frac{nRT_1(1-V_0/V_1)}2.\tag{*} \end{align}$$ where we used $$\begin{align} &p_1A=k(x_1-x_0),&\text{net force is equal to $0$ in the final state}, \\ &A(x_1-x_0)=V_1-V_0,& \text{cylinder geometry},\\ &p_1V_1=nRT_1 & \text{ideal gas law}.\\ \end{align}$$

Resolving $(\text*)$ one obtains $(1)$.

  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Jul 19 at 1:30

2 Answers 2


The analysis of interest starts with a force balance on the piston: $$F_g-kx=M\frac{dv}{dt}$$where M is the mass of the piston and $F_g$ is the force that the gas exerts on the piston. If we multiply this equation by the piston velocity v and integrate with respect to t, we obtain $$W_g=\int_0^x{F_gdx}=k\frac{x^2}{2}+M\frac{v^2}{2}.$$ This is the mechanical energy balance on the piston.

When viscous stresses within the gas have caused the motion of the piston to be damped out, the work done by the gas up to the final state ($x=x_1$) reduces to $$W_{g,1}=k\frac{x_1^2}{2}$$ What will happen physically is that, in the initial stroke of the piston, the piston will overshoot the final equilibrium displacement $x_1$, and subsequently execute a damped oscillation until the system comes to equilibrium and the piston ultimately comes to rest at $x=x_1$. The cause of the piston damping is viscous stresses in the gas at the inside piston face.


The question is "why is $k\frac{(x_1-x_0)^2}{2}$ equal to $P_1\frac{(V_1-V_0)}{2}$?

The spring force is $$F=k(x-x_0)=\frac{k}{A}(V-V_0)$$At $x=x_1$ and $V=V_1$, $F=P_1A$. So, $$\frac{k}{A}(V_1-V_0)=AP_1$$So, $$k=A^2\frac{P_1}{(V_1-V_0)}$$The spring work is $$W=k\frac{(x_1-x_0)^2}{2}=\frac{k}{A^2}\frac{(V_1-V_0)^2}{2}$$If we eliminate k between the previous two equations, the work becomes $$W=\frac{P_1}{2}(V_1-V_0)$$

  • $\begingroup$ Can you describe how this gets one to the solution given in the OP's eq 1? (the derivation of which the OP states is "trivial") $\endgroup$
    – Bob D
    Jul 17 at 16:20
  • $\begingroup$ @BobD I added the derivation. $\endgroup$
    – drer
    Jul 17 at 17:28
  • $\begingroup$ @drer When you write :"where we used $p_1A=k(x_1-x_0)"$ it looks you are assuming the expansion is a constant pressure process at the final pressure. On what do you base that assumption? $\endgroup$
    – Bob D
    Jul 17 at 17:31
  • $\begingroup$ I see. It is wrong of course because there is other work done either on friction or as kinetic energy. The $PV^\gamma$ route is correct becaase friction does not affect the reversibility of the thermodynamic processes in the gas. $\endgroup$
    – mike stone
    Jul 17 at 17:43
  • $\begingroup$ See the addendum to my answer. $\endgroup$ Jul 17 at 19:33

Your earlier version of this question got closed faster than I could have had the time to type up an answer.

It is good to see that you have made some improvements, but it seems like your reasoning is still somewhat flawed.

Initially, the gas obeys the ideal gas equation $$\tag1p_0V_0=nRT_0$$ for some unspecified $p_0$ and $N$ that must be thus absent from the final solution.

The spring is initially relaxed and at equilibrium. This means that any motion $x$ is already the amount of compression of the spring. There is no need to be so complicated by dealing with $x_1-x_0$

Now, the moment you let go of the restriction, there is a pressure and thus a force from the gas, and there is nothing balancing it. It is not possible for this system to be evolving quasistatically. Since a process is only reversible if it is quasistatic, it is thus also not reversible. Any argument using $p V^\gamma$ or $TV^{\gamma-1}$ is not acceptable because this assumes a reversible process by definition.

Instead, you may only appeal to ideal gas law of the final equilibrium state. That is, after waiting forever for the system to come back to equilibrium, we should find that the ideal gas law is obeyed, and also $$\tag2p_1A=kx$$ Now, let $\ell=V_0/A$, and this means $V_1=A(\ell+x)$, which implies $$\tag3p_1V_1=nRT_1=p_1A(\ell+x)=kx(\ell+x)$$ Then we note that the system was isolated and so energy can only be exchanged between the spring and the gas, so that \begin{align} \tag4\text{CoE}:\qquad\sum E_i+W&=\sum E_f\\ \tag5nC_VT_0+0+0&=nC_VT_1+\frac12kx^2\\ \tag62nC_V(T_0-T_1)&=kx^2\\ \tag7\therefore\qquad p_1V_1-p_1V_0&=2nC_V(T_0-T_1)\\ \tag8\frac{nRT_1}{V_1}(V_1-V_0)&=2nC_V(T_0-T_1)\\ \tag9T_1\left(1-\frac{V_0}{V_1}+2\frac{C_V}R\right)&=2\frac{C_V}RT_0\\ \tag{10}\frac{T_0}{T_1}&=1+\frac R{2C_V}\left(1-\frac{V_0}{V_1}\right) \end{align} which is what you obtained in your Equation (1), and we have combined Equations (2), (3) and (6) to get Equation (7)

If the piston is doing work against friction, even the above arguments will be wrong, because then the energy is being lost to friction.

If the piston is incredibly heavy, that does not change anything, because the spring will just be taking a much longer time to vibrate around this equilibrium position due to the huge mass of the piston.

Thus, given this question, you cannot be arriving at any other answer and be expected to be given credit.

  • $\begingroup$ I don't quite understand why is my reasoning "still somewhat flowed" taking into account that you arrived at the same answer. $\endgroup$
    – drer
    Jul 17 at 19:34
  • $\begingroup$ Excuse me my ignorance, what is a credit? $\endgroup$
    – drer
    Jul 17 at 19:34
  • $\begingroup$ @drer I think naturallyinconsistent is referring to your overcomplicating the problem by dealing with $x_{1}-x_{0}$ $\endgroup$
    – Bob D
    Jul 17 at 20:14
  • $\begingroup$ @BobD Dealing with $x_1-x_0$ simplifies the problem. $\endgroup$
    – drer
    Jul 17 at 20:22
  • $\begingroup$ @drer $kx_0$ implies a spring force initially exists at the initial position $x_0$, though it is specified that the spring is initially relaxed. $\endgroup$
    – Bob D
    Jul 17 at 20:29

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