Thermally insulated container with alternating series of walls

[Meden Agan]

Homework Statement

A thermally insulated container is divided into $2n+1$ equal sections by $2n$ walls, each containing monoatomic ideal gas with the same pressure and temperature. The walls with odd numbers (from $1$ to $2n-1$) are fixed and good heat conductors, the walls with even numbers (from $2$ to $2n$) are thermally insulating and freely moving like pistons. Similarly the $0^{\text{th}}$ wall is thermally insulating and is moveable. This gas in the first section of the container is slowly being compressed by moving the $0^{\text{th}}$ wall. How does the temperature of the gas in this section change as a function of its volume? What can we say about the displacements of the freely moving walls?

Relevant Equations

$TV^{\gamma -1}$, where $\gamma$ is heat capacity ratio.

IMO, this problem is really tricky. Here below is my attempt.

Let us take a perfectly insulated container and imagine dividing it with an alternating series of walls: those with an odd index are rigid and conduct heat, bringing the two adjacent compartments into thermal contact; those with an even index, including the $0^{\text{th}}$ we are pushing, are ideal adiabatic cylinders, i.e. they slide without friction maintaining the same pressure on both sides but not allowing heat to pass through.
The compartments are initially all identical, filled with the same monoatomic gas at temperature $T_0$ and volume $V_{\text{initial}}$.
The conductive walls weld the compartments $0$-$1, 2$-$3, \ldots , 2n-2$-$2n-1$ in pairs: this results in $n$ "double blocks", while compartment $2n$ remains the only "single block". Since each block is enclosed by adiabatic pistons, the gas it contains obeys the polytropic process equation $T V^{\gamma-1}= \text{constant}$ with $\gamma =5/3$; in addition, the pressure must remain the same in all compartments for the pistons not to gain acceleration. These two conditions force the volume of a double block to remain exactly twice that of the single block throughout the transformation: $V_{\text{double}} : V_{\text{single}} = 2 : 1$.

Let us now suppose we slowly push $0^{\text{th}}$ piston, reducing its volume from $V_{\text{initial}}$ to $V$ (if we are compressing it, $V < V_{\text{initial}}$). The overall volume of the gas changes from $(2n+1) V_{\text{initial}}$ to $V_{\text{tot}} = 2n V_{\text{initial}} + V$. The $0$-$1$ block must preserve the $2:1$ ratio and therefore takes $V_{01}=\dfrac{2(2n V_{\text{initial}} + V)}{2n+1}$ at each instant. Applying the polytropic process equation to that block, we immediately obtain the temperature of the gas in compartment $0$: $$T(V)=T_0 \left[\frac{(2n+1) V_{\text{initial}}}{2n V_{\text{initial}}+V}\right]^{2/3}.$$
This result shows that the longer the chain of pistons ($n$ large), the slower the temperature rises: the work we exert on the first piston is redistributed over many blocks. In the case $n \to \infty$, we would obtain $T(V) \to T_0$.

I am very suspicious of this result. It is expressed as a function of unknown variables from the text, and I don't know if the train of thought is correct.



As for the displacements of the internal pistons, we denote $\Delta V=V-V_{\text{initial}}$ ($< 0$ in compression) and let $S$ be the cross-section of the container. The piston of index $2k \,\, (k=0,1, \ldots, n)$ moves back by $$\Delta x_{2k}=\frac{V_{\text{initial}}-V}{S} \cdot \frac{2n+1-2k}{2n+1} \qquad \text{for} \quad k = 0, 1, \ldots, n.$$
Piston $0$ runs $(V_{\text{initial}}-V)/S$, piston $2n$ just runs $\dfrac{1}{2n+1} \cdot \dfrac{V_{\text{initial}}-V}{S}$, and between those extremes, the displacements decrease linearly: each internal piston always advances by a fixed amount less than the previous one.

Again, I am very suspicious of this result. It is expressed as a function of unknown variables from the text, and I don't know if the train of thought is correct.

 

[.Scott]

I believe you are calculating the conditions after the system has returned to equilibrium. So, it is not incredible that the temperature change will eventually become distributed over the entire system
After wall-0 is moved, the other even walls will begin moving as well. There is presumably some mechanism to keep them from oscillating.

 

[haruspex]

I don’t understand how you can apply that polytropic process equation to a double block. Don’t you need a region at uniform pressure for that?

 

[Meden Agan]

I don’t understand how you can apply that polytropic process equation to a double block. Don’t you need a region at uniform pressure for that?

You are quite right, but at first it struck me as the most appropriate approach to tackle the problem.
That said, I'm in over my head. I really don't know where we go from here.
Could you please give me a hint on how to start?

 

[haruspex]

You are quite right, but at first it struck me as the most appropriate approach to tackle the problem.
That said, I'm in over my head. I really don't know where we go from here.
Could you please give me a hint on how to start?

I would create unknowns for the temperatures, pressures and volumes of the 2n compartments and try to come up with recurrence relations for neighbouring compartments.

 

[Meden Agan]

I would create unknowns for the temperatures, pressures and volumes of the 2n compartments and try to come up with recurrence relations for neighbouring compartments.

Let's try to create unknowns for temperatures, pressures and volumes. Let me know if it's OK so far; if so, I'll go ahead with recurrence relations.

We would like to find the temperature $T_0$ of the gas in the first section $S_0$ as a function of its volume $V_0$, and the displacement of the freely moving walls. The system has $2n+1$ sections, each initially containing the same amount of monatomic ideal gas ($N_m$ moles) at pressure $P_{\text{initial}}$, temperature $T_{\text{initial}}$, and volume $V_{\text{initial}}$.
The gas constant is $R$, its molar heat capacity is $C_V = \dfrac{3}{2} R$, and the heat capacity ratio is $\gamma = 5/3$.
Let $P_i, V_i, T_i$ denote the pressure, volume, and temperature of section $S_i$. The processes are quasi-static.



1) Odd-numbered walls $(1, 3, \ldots, 2n-1)$ are fixed and conducting.
- Fixed position means $x_{2k-1}$ (with $k = 1, \ldots, n$) is constant.
- Conducting means $T_{2k-2}=T_{2k-1}$ for $k = 1, \ldots, n$. Let this common temperature be $\mathcal T_k$. So: $\mathcal T_1 = T_0 = T_1, \, \mathcal T_2 = T_2 = T_3, \, \ldots, \, \mathcal T_n = T_{2n-2} = T_{2n-1}$.

2) Even-numbered walls $(2, 4, \ldots, 2n)$ are insulating and freely moving.
- Insulating means that the process is adiabatic.
- Freely moving means $P_{2k-1}=P_{2k}$ for $k = 1, \ldots, n$. Using the relation $P = \dfrac{N_m R T}{V}$, this implies
$$\begin{aligned}
P_{2k-1}=P_{2k} &\implies \frac{N_m \, R \, T_{2k-1}}{V_{2k-1}} = \frac{N_m \, R \, T_{2k}}{V_{2k}} \\
&\iff \frac{\cancel{N_m \, R} \, T_{2k-1}}{V_{2k-1}} = \frac{\cancel{N_m \, R} \, T_{2k}}{V_{2k}} \\
&\iff \frac{T_{2k-1}}{V_{2k-1}} = \frac{T_{2k}}{V_{2k}}.
\end{aligned}$$
So, $\dfrac{T_{2k-1}}{V_{2k-1}} = \dfrac{T_{2k}}{V_{2k}}$.
Since $T_{2k-1} = \mathcal T_k$, we have $$\frac{\mathcal T_k}{V_{2k-1}} = \frac{T_{2k}}{V_{2k}}.$$

Wall $0$ is insulating and moveable, used to compress $S_0$; wall $2n+1$ (rightmost end) is fixed and insulating.



Let's consider the pair of sections $(S_{2k-2}, S_{2k-1})$ for $k = 1, \ldots, n$. This system is enclosed by wall $2k-2$ (moveable and insulating, or wall $0$ for $k=1$) and wall $2k$ (moveable and insulating). The internal wall $2k-1$ is fixed and conducting. The entire system $(S_{2k-2}, S_{2k-1})$ is adiabatically enclosed.

The internal energy of the wall $2k-2$ is $U_{2k-2} = N_m \, C_V \, T_{2k-2}$, so $\mathrm dU_{2k-2} = N_m \, C_V \, \mathrm dT_{2k-2}$. By the First Law of Thermodynamics,
$$\begin{aligned}
\mathrm dU_{2k-2} &= \cancelto{0}{\delta Q} - \delta W_{2k-2} \\
&= - \delta W_{2k-2} \\
&= - P_{2k-2} \, \mathrm dV_{2k-2} \\
&= -\frac{N_m \, R \, T_{2k-2}}{V_{2k-2}} \mathrm dV_{2k-2}.
\end{aligned}$$
So $$\begin{aligned}
N_m \, C_V \, \mathrm dT_{2k-2} = -\frac{N_m \, R \, T_{2k-2}}{V_{2k-2}} \mathrm dV_{2k-2} &\iff \cancel{N_m} \, C_V \, \mathrm dT_{2k-2} = -\frac{\cancel{N_m} \, R \, T_{2k-2}}{V_{2k-2}} \mathrm dV_{2k-2} \\
&\iff C_V \, \mathrm dT_{2k-2} = -\frac{R \, T_{2k-2}}{V_{2k-2}} \mathrm dV_{2k-2} \\
&\iff \frac{C_V}{R} \frac{\mathrm dT_{2k-2}}{T_{2k-2}} = - \frac{\mathrm dV_{2k-2}}{V_{2k-2}}\\
&\iff \frac{C_V}{R} \frac{\mathrm d \mathcal T_k}{\mathcal T_k} = - \frac{\mathrm dV_{2k-2}}{V_{2k-2}}.
\end{aligned}$$
Thus: $$\frac{C_V}{R} \frac{\mathrm d \mathcal T_k}{\mathcal T_k} = - \frac{\mathrm dV_{2k-2}}{V_{2k-2}}. \tag{1}$$

The internal energy of the wall $2k-1$ is $U_{2k-1} = N_m \, C_V \, T_{2k-1}$, so $\mathrm dU_{2k-1} = N_m \, C_V \, \mathrm dT_{2k-1}$. By the First Law of Thermodynamics,
$$\begin{aligned}
\mathrm dU_{2k-1} &= \cancelto{0}{\delta Q} - \delta W_{2k-1} \\
&= - \delta W_{2k-1} \\
&= - P_{2k-1} \, \mathrm dV_{2k-1} \\
&= -\frac{N_m \, R \, T_{2k-1}}{V_{2k-1}} \mathrm dV_{2k-1}.
\end{aligned}$$
So $$\begin{aligned}
N_m \, C_V \, \mathrm dT_{2k-1} = -\frac{N_m \, R \, T_{2k-1}}{V_{2k-1}} \mathrm dV_{2k-1} &\iff \cancel{N_m} \, C_V \, \mathrm dT_{2k-1} = -\frac{\cancel{N_m} \, R \, T_{2k-1}}{V_{2k-1}} \mathrm dV_{2k-1} \\
&\iff C_V \, \mathrm dT_{2k-1} = -\frac{R \, T_{2k-1}}{V_{2k-2}} \mathrm dV_{2k-1} \\
&\iff \frac{C_V}{R} \frac{\mathrm dT_{2k-1}}{T_{2k-1}} = - \frac{\mathrm dV_{2k-1}}{V_{2k-1}} \\
&\iff \frac{C_V}{R} \frac{\mathrm d \mathcal T_k}{\mathcal T_k} = - \frac{\mathrm dV_{2k-1}}{V_{2k-1}}.
\end{aligned}$$
Thus: $$\frac{C_V}{R} \frac{\mathrm d \mathcal T_k}{\mathcal T_k} = - \frac{\mathrm dV_{2k-1}}{V_{2k-1}}. \tag{2}$$

Adding $(2)$ to $(1)$:
$$2 \frac{C_V}{R} \frac{\mathrm d \mathcal T_k}{\mathcal T_k} = - \frac{\mathrm dV_{2k-2}}{V_{2k-2}} - \frac{\mathrm dV_{2k-1}}{V_{2k-1}}. \tag{3}$$
Initially, $\mathcal T_k = T_{\text{initial}}, \, V_{2k-2} = V_{\text{initial}}, \, V_{2k-1} = V_{\text{initial}}$. So, we integrate LHS of equation $(3)$ from $T_{\text{initial}}$ to $\mathcal T_k$ and RHS of equation $(3)$ from $V_{\text{initial}}$ to $V_{2k-2}$ (first term) and from $V_{\text{initial}}$ to $V_{2k-1}$ (second term). Then:
$$\begin{aligned}
2 \frac{C_V}{R} \int\limits_{T_{\text{initial}}}^{\mathcal T_k} \frac{\mathrm d \mathcal T_k}{\mathcal T_k} = - \int\limits_{V_{\text{initial}}}^{V_{2k-2}} \frac{\mathrm dV_{2k-2}}{V_{2k-2}} - \int\limits_{V_{\text{initial}}}^{V_{2k-1}} \frac{\mathrm dV_{2k-1}}{V_{2k-1}} &\implies 2 \frac{C_V}{R} \ln \left(\frac{\mathcal T_k}{T_{\text{initial}}}\right) = \ln \left(\frac{V_{\text{initial}}^2}{V_{2k-2} V_{2k-1}}\right) \\
&\iff \ln \left(\frac{\mathcal T_k}{T_{\text{initial}}}\right)^{2 \, C_V/R} = \ln \left(\frac{V_{\text{initial}}^2}{V_{2k-2} V_{2k-1}}\right) \\
&\iff \left(\frac{\mathcal T_k}{T_{\text{initial}}}\right)^{2 \, C_V/R} = \left(\frac{V_{\text{initial}}^2}{V_{2k-2} V_{2k-1}}\right) \\
&\iff \mathcal T_k^{2 \, C_V/R} \, V_{2k-2} V_{2k-1} = T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2.
\end{aligned}$$
Thus: $$\mathcal T_k^{2 \, C_V/R} \, V_{2k-2} V_{2k-1} = T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2 \equiv \verb|constant| \qquad \text{for} \quad k= 1, \ldots, n. \tag{A}$$

Now, let's consider volumes around a fixed wall $2k-1$. If $A$ is the area of the section, then $V_{2k-1} = A \left(x_{2k}-x_{2k-1}\right)$ and $V_{2k} = A \left(x_{2k+1}-x_{2k}\right)$ (assuming $x_{2k+1}$ is fixed, which is true if wall $2k+1$ is an odd-numbered fixed wall). If walls $2k-1$ and $2k+1$ are fixed, then $V_{2k-1} + V_{2k} = A \left(x_{2k+1}-x_{2k-1}\right) = \text{constant}$. This applies for $k= 0, \ldots, n-1$ (wall $2(0)+1 = 1$ is fixed, wall $2(1)+1 = 3$ is fixed, ..., wall $2(n-1)+1 = 2n-1$ is fixed). For $k=n$, wall $2n+1$ is the fixed end of the container.
Initially, $V_{2k-1} = V_{\text{initial}}, \, V_{2k} = V_{\text{initial}}$. Thus: $$V_{2k-1} + V_{2k} = 2 V_{\text{initial}} \equiv \text{constant} \qquad \text{for} \quad k= 1, \ldots, n. \tag{B}$$

 

[Chestermiller]

Let's consider the pair of sections $(S_{2k-2}, S_{2k-1})$ for $k = 1, \ldots, n$. This system is enclosed by wall $2k-2$ (moveable and insulating, or wall $0$ for $k=1$) and wall $2k$ (moveable and insulating). The internal wall $2k-1$ is fixed and conducting. The entire system $(S_{2k-2}, S_{2k-1})$ is adiabatically enclosed.

The internal energy of the wall $2k-2$ is $U_{2k-2} = N_m \, C_V \, T_{2k-2}$, so $\mathrm dU_{2k-2} = N_m \, C_V \, \mathrm dT_{2k-2}$. By the First Law of Thermodynamics,
$$\begin{aligned}
\mathrm dU_{2k-2} &= \cancelto{0}{\delta Q} - \delta W_{2k-2} \\
&= - \delta W_{2k-2} \\
&= - P_{2k-2} \, \mathrm dV_{2k-2} \\
&= -\frac{N_m \, R \, T_{2k-2}}{V_{2k-2}} \mathrm dV_{2k-2}.
\end{aligned}$$two

This does not seem correct to me. Even though the combined two sections are adiabatic, each one individually is not adiabatic, and heat can flow across the intervening diathermic fixed partition.

 

[Meden Agan]

This does not seem correct to me. Even though the combined two sections are adiabatic, each one individually is not adiabatic, and heat can flow across the intervening diathermic fixed partition.

OK, then I'm in over my head. If you have detailed hints on how to do that, let's hear them.

 

[Chestermiller]

x−y

OK, then I'm in over my head. If you have detailed hints on how to do that, let's hear them.

After further consideration, I am withdrawing my incorrect objection to what you did in post #6. This analysis you presented was correct. Sorry. My bad.

 

[Meden Agan]

After further consideration, I am withdrawing my incorrect objection to what you did in post #6. This analysis you presented was correct. Sorry. My bad.

Mhm, then I'm confused. To me, your objection in post #7 seems OK. May I ask what leads you to think your objection is incorrect?

 

[Chestermiller]

I analyzed the system is a slightly different way, and came to the same result. The key is to recognize that the two cells between adiabatic barriers experience an adiabatic reversible deformation and that the conductive partition between these cells guarantees that the temperatures of the two adjacent cells is the same. This is basically what you did, and it was correct.

 

[Meden Agan]

I analyzed the system is a slightly different way, and came to the same result. The key is to recognize that the two cells between adiabatic barriers experience an adiabatic reversible deformation and that the conductive partition between these cells guarantees that the temperatures of the two adjacent cells is the same. This is basically what you did, and it was correct.

OK. May I ask you to write down the maths about that "slightly different way"?

 

[Chestermiller]

OK. May I ask you to write down the maths about that "slightly different way"?

I would like to (and tried to do it yesterday), but I'm having a problem entering and previewing equations. The problem lies with my computer, the Physics Forums app, or the Safari browser. I don't know how to even describe it.

 

[Meden Agan]

I would like to (and tried to do it yesterday), but I'm having a problem entering and previewing equations. The problem lies with my computer, the Physics Forums app, or the Safari browser. I don't know how to even describe it.

Mhm, sad. Of course, I don't know what to do to help you.
Perhaps you can send an image of your handwritten notes, and I can try writing in LaTeX?

 

[haruspex]

I would like to (and tried to do it yesterday), but I'm having a problem entering and previewing equations. The problem lies with my computer, the Physics Forums app, or the Safari browser. I don't know how to even describe it.

Hi Chet,

The problem I ran into is that if I use the latex preview it looks ok but actually messes up the latex, so the submitted post is screwed.
To get around it, I have been copying the unpreviewed text as typed into another window (email, say), using preview to see if it is right (may require browser screen refresh), correcting any errors in the saved copy, deleting the whole unsubmitted post and pasting it back in from the saved and edited copy.
When it previews ok, do the "delete and paste saved back in" one more time before submitting.

 

[Steve4Physics]

I would like to (and tried to do it yesterday), but I'm having a problem entering and previewing equations. The problem lies with my computer, the Physics Forums app, or the Safari browser. I don't know how to even describe it.

Preview doesn’t work for me at all. LaTeX is only rendered in messages that have actually been sent.

Here’s my work-around. It’s based on (but I think slightly different to) @haruspex ’s.

1. Open Physics Forum in two separate tabs (say A and B).

2. Write/edit the message using Tab-A. Take a copy.

3. Using Tab-B, create a PM to yourself (i.e. Direct Messages/Send direct message). Paste the message into the PM and post it.

4. Read the received PM (using Tab-B). For some reason, LaTeX in the PM is rendered.

5. If needed repeat steps 2-4.

6. When you’re happy, post the message from Tab-A.

Occasionally delete the PMs you've sent yourself.

 

[Chestermiller]

I did something similar by going to a different forum that also uses LaTex, Ann editing the post there. Here is what I got:

If we apply the first law of thermodynamic to the 0 and 1 cells of the sequence between the 0 and 1 adiabatic barriers, we obtain $$dU=nC_vdT_0+nC_vdT_1=P_0A dx_0-P_1Adx_2\tag{1}$$with $$P_0=\frac{nRT_0}{A(x_1-x_0)}$$and $$P_1=\frac{nRT_1}{A(x_2-x_1)}$$ If we substitute these into Eqn. 1, we obtain:$$dU=nC_vdT_0+nC_vdT_1=nRT_0\frac{dx_0}{x_1-x_0}-nRT_1\frac{dx_2}{x_2-x_1}\tag{2}$$Next, let $$x_1-x_0=(x_1-x_{0,init})-(x_{0}-x_{0,init})=\delta(1-\xi_0)\tag{3}$$with $\delta$ representing the initial spacing between adjacent partitions $$\delta=(x_1-x_{0,init})=(x_{2,init}-x_1)\ \ etc$$ and $$\xi_0=\frac{x_0-x_{0,init}}{\delta}$$Similarly, let $$x_2-x_1=(z_{2,init}-x_1)+(x_2-x_{2,init})=\delta(1+\xi_2)\tag{4}$$If we substitute Eons, 3 and 4 into Eqn. 2, we obtain $$nC_v(dT_0+dT_1)=nRT_0\frac{d\xi_0}{1-\xi_0}-nRT_1\frac{d\xi_2}{1+\xi_2}\tag{5}$$But, since the barrier is conductive, we have $T_1=T_0$, and Eqn. 5 reduces to $$2nC_v dT_0=nRT_0\left[\frac{d\xi_0}{1-\xi_0}-\frac{d\xi_2}{1+\xi_2}\right]$$Integrating this equation between the initial and final states gives: $$2\frac{C_v}{R}\ln{\left(\frac{T_0}{T_{0,init}}\right)}=-\ln[{(1-\xi_0)(1+\xi_2)}]$$or$$\frac{T_0}{T_{init}}=\frac{T_1}{T_{init}}=[{(1-\xi_0)(1+\xi_2)}]^{-\frac{\gamma-1}{2}}$$

 

[Meden Agan]

I did something similar by going to a different forum that also uses LaTex, Ann editing the post there. Here is what I got:

If we apply the first law of thermodynamic to the 0 and 1 cells of the sequence between the 0 and 1 adiabatic barriers, we obtain $$dU=nC_vdT_0+nC_vdT_1=P_0A dx_0-P_1Adx_2\tag{1}$$with $$P_0=\frac{nRT_0}{A(x_1-x_0)}$$and $$P_1=\frac{nRT_1}{A(x_2-x_1)}$$ If we substitute these into Eqn. 1, we obtain:$$dU=nC_vdT_0+nC_vdT_1=nRT_0\frac{dx_0}{x_1-x_0}-nRT_1\frac{dx_2}{x_2-x_1}\tag{2}$$Next, let $$x_1-x_0=(x_1-x_{0,init})-(x_{0}-x_{0,init})=\delta(1-\xi_0)\tag{3}$$with $\delt$ representing the initial spacing between adjacent partitions $$\delta=(x_1-x_{0,init})=(x_{2,init}-x_1)\ \ etc$$ and $$\xi_0=\frac{x_0-x_{0,init}}{\delta}$$Similarly, let $$x_2-x_1=(z_{2,init}-x_1)+(x_2-x_{2,init})=\delta(1+\xi_2)\tag{4}$$If we substitute Eons, 3 and 4 into Eqn. 2, we obtain $$nC_v(dT_0+dT_1)=nRT_0\frac{d\xi_0}{1-\xi_0}-nRT_1\frac{d\xi_2}{1+\xi_2}\tag{5}$$But, since the barrier is conductive, we have $T_1=T_0$, and Eqn. 5 reduces to $$2nC_v dT_0=nRT_0\left[\frac{d\xi_0}{1-\xi_0}-\frac{d\xi_2}{1+\xi_2}\right]$$Integrating this equation between the initial and final states gives: $$2\frac{C_v}{R}\ln{\left(\frac{T_0}{T_{0,init}}\right)}=-\ln[{(1-\xi_0)(1+\xi_2)}]$$or$$\frac{T_0}{T_{init}}=\frac{T_1}{T_{init}}=[{(1-\xi_0)(1+\xi_2)}]^{-\frac{\gamma-1}{2}}$$

Lovely approach! Your analysis agrees with mine, just set
$$\begin{aligned}
V_{\text{initial}} &= A \, \underbrace{(x_1 - x_{0, \, \text{initial}})}_{\delta} = A \, \underbrace{(x_{2, \, \text{initial}} - x_1)}_{\delta} \\
V_0 &= A(x_1-x_0) \\
V_1 &= A (x_2 - x_1).
\end{aligned}$$
Now I have to try to come up with recurrence relations for neighbouring compartments, don't I?

 

[Chestermiller]

Lovely approach! Your analysis agrees with mine, just set
$$\begin{aligned}
V_{\text{initial}} &= A \, \underbrace{(x_1 - x_{0, \, \text{initial}})}_{\delta} = A \, \underbrace{(x_{2, \, \text{initial}} - x_1)}_{\delta} \\
V_0 &= A(x_1-x_0) \\
V_1 &= A (x_2 - x_1).
\end{aligned}$$
Now I have to try to come up with recurrence relations for neighbouring compartments, don't I?

Yes. It looks like there are more unknowns than variables, but I think closure will be provided at the far end.

 

[Meden Agan]

Yes. It looks like there are more unknowns than variables, but I think closure will be provided at the far end.

Any hint on obtaining recurrence relations? I only manage to come up with messy maths and incomplete expressions.

 

[Chestermiller]

Any hint on obtaining recurrence relations? I only manage to come up with messy maths and incomplete expressions.

Well, another relation is the matching of pressures on the two sides of the insulated partitions, and you can also make us of the matching of temperature again on the two sides of the conductive partitions. I haven't tried to develop the recurrence relationships but, if run into trouble, I can see what can do.

 

[Meden Agan]

Well, another relation is the matching of pressures on the two sides of the insulated partitions, and you can also make us of the matching of temperature again on the two sides of the conductive partitions. I haven't tried to develop the recurrence relationships but, if run into trouble, I can see what can do.

Well, the maths I come up with is long and very, very tedious. To avoid overloading you, I post one chunk at a time. Let me know if it is OK each time.

In post #6, I come up with:
$$\mathcal T_k^{2 \, C_V/R} \, V_{2k-2} V_{2k-1} = T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2 \equiv \verb|constant| \qquad \text{for} \quad k= 1, \ldots, n\tag{A}$$
$$V_{2k-1} + V_{2k} = 2 V_{\text{initial}} \equiv \text{constant} \qquad \text{for} \quad k= 1, \ldots, n. \tag{B}$$
$$\frac{\mathcal T_k}{V_{2k-1}} = \frac{T_{2k}}{V_{2k}}. \tag{C}$$
Incidentally, Equation $\text{(C)}$ stems from the matching of pressures on the two sides of the insulated, freely moving partitions.

Let $X_k = \mathcal T_k^{2 \, C_V/R} \, V_{2k-2}$ and $T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2 = \mathcal C_1$.
From Equation $\text{(A)}$:
$$\begin{aligned}
V_{2k-1} &= \frac{T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2}{\mathcal T_k^{2 \, C_V/R} \, V_{2k-2}} \\
&= \frac{\mathcal C_1}{X_k},
\end{aligned}$$
thus $$V_{2k-1} = \frac{\mathcal C_1}{X_k}. \tag{A1}$$
Let $2 \, V_{\text{initial}} = \mathcal C_2$.
From Equation $\text{(B)}$:
$$\begin{aligned}
V_{2k} &= 2 \, V_{\text{initial}} - V_{2k-1} \\
&= \mathcal C_2 - \frac{\mathcal C_1}{X_k} \\
&= \frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k},
\end{aligned}$$
thus
$$V_{2k} = \frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k}. \tag{B1}$$
From Equation $\text{(C)}$:
$$\begin{aligned}
T_{2k} &= \mathcal T_k \frac{V_{2k}}{V_{2k-1}} \\
&= \mathcal T_k \frac{\dfrac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k}}{\dfrac{\mathcal C_1}{X_k}} \\
&= \mathcal T_k \frac{\dfrac{\mathcal C_2 \, X_k - \mathcal C_1}{\cancel{X_k}}}{\dfrac{\mathcal C_1}{\cancel{X_k}}} \\
&= \mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1},
\end{aligned}$$
thus
$$T_{2k} = \mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1}. \tag{*}$$
We make use of the matching of temperatures on the two sides of the conducting partitions: $T_{2k-2} = T_{2k-1} = \mathcal T_k$ for $k = 1, \ldots, n$. Symmetrically, $T_{2 \, (k+1)-2} = T_{2 \, (k+1)-1} = \mathcal T_{k+1}$. Since $T_{2 \, (k+1)-2}=T_{2k +2-2} = T_{2k}$ and $T_{2 \, (k+1)-2} = \mathcal T_{k+1}$, we also have $T_{2k} = \mathcal T_{k+1}$.
Equation $(*)$ becomes
$$\mathcal T_{k+1} = \mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1}. \tag{C1}$$

Now, we have
$$\begin{aligned}
X_{k+1} &= \mathcal T_{k+1}^{2 \, C_V/R} \, V_{2 \, (k+1)-2} \\
&= \mathcal T_{k+1}^{2 \, C_V/R} \, V_{2k+2-2} \\
&= \mathcal T_{k+1}^{2 \, C_V/R} \, V_{2k} \\
&= \underbrace{\left(\mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1} \right)^{2 \, C_V/R}}_{\text{(C1)}} \, \underbrace{\frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k}}_{ \text{(B1)}} \\
&= \mathcal T_k^{2 \, C_V/R} \, \left(\frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1}\right)^{2 \, C_V/R} \, \frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k} \\
&= \mathcal T_k^{2 \, C_V/R} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R}}{C_1^{2 \, C_V/R}} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)}{X_k} \\
&= \mathcal T_k^{2 \, C_V/R} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \ + \, 1}}{C_1^{2 \, C_V/R} \, X_k} \\
&= \mathcal T_k^{2 \, C_V/R} \cdot \frac{V_{2k-2}}{V_{2k-2}} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \, 1}}{C_1^{2 \, C_V/R} \, X_k} \\
&= \underbrace{\mathcal T_k^{2 \, C_V/R} \, V_{2k-2}}_{\displaystyle X_k} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \, 1}}{C_1^{2 \, C_V/R} \, X_k \, V_{2k-2}} \\
&= X_k \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \, 1}}{C_1^{2 \, C_V/R} \, X_k \, V_{2k-2}}.
\end{aligned}$$
Thus: $$X_{k+1} =X_k \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \ + \, 1}}{C_1^{2 \, C_V/R} \, X_k \, V_{2k-2}} \qquad \text{for} \quad k = 1, \ldots, n-1.$$

For $k=1$, we have $X_k = X_1$, $X_{k+1} = X_2$ and $V_{2k-2} = V_0$. Then:
$$X_{2} =X_1 \frac{\left(\mathcal C_2 \, X_1 - \mathcal C_1\right)^{2 \, C_V/R + 1}}{C_1^{2 \, C_V/R} \, X_1 \, V_{0}}.$$
For $k \geqslant 2$, we have $V_{2k-2} = V_{2(k-1)} = \frac{\mathcal C_2 \, X_{k-1} - \mathcal C_1}{X_{k-1}}$ according to $\text{(B1)}$. Then:
$$X_{k+1} = X_k \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \ 1} \, X_{k-1}}{C_1^{2 \, C_V/R} \, X_k \, \left(\mathcal C_2 \, X_{k-1} - \mathcal C_1\right)} \qquad \text{for} \quad k = 2, \ldots, n-1.$$

OK so far?

 

[Chestermiller]

Well, the maths I come up with is long and very, very tedious. To avoid overloading you, I post one chunk at a time. Let me know if it is OK each time.

In post #6, I come up with:
$$\mathcal T_k^{2 \, C_V/R} \, V_{2k-2} V_{2k-1} = T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2 \equiv \verb|constant| \qquad \text{for} \quad k= 1, \ldots, n\tag{A}$$
$$V_{2k-1} + V_{2k} = 2 V_{\text{initial}} \equiv \text{constant} \qquad \text{for} \quad k= 1, \ldots, n. \tag{B}$$
$$\frac{\mathcal T_k}{V_{2k-1}} = \frac{T_{2k}}{V_{2k}}. \tag{C}$$
Incidentally, Equation $\text{(C)}$ stems from the matching of pressures on the two sides of the insulated, freely moving partitions.

Let $X_k = \mathcal T_k^{2 \, C_V/R} \, V_{2k-2}$ and $T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2 = \mathcal C_1$.
From Equation $\text{(A)}$:
$$\begin{aligned}
V_{2k-1} &= \frac{T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2}{\mathcal T_k^{2 \, C_V/R} \, V_{2k-2}} \\
&= \frac{\mathcal C_1}{X_k},
\end{aligned}$$
thus $$V_{2k-1} = \frac{\mathcal C_1}{X_k}. \tag{A1}$$
Let $2 \, V_{\text{initial}} = \mathcal C_2$.
From Equation $\text{(B)}$:
$$\begin{aligned}
V_{2k} &= 2 \, V_{\text{initial}} - V_{2k-1} \\
&= \mathcal C_2 - \frac{\mathcal C_1}{X_k} \\
&= \frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k},
\end{aligned}$$
thus
$$V_{2k} = \frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k}. \tag{B1}$$
From Equation $\text{(C)}$:
$$\begin{aligned}
T_{2k} &= \mathcal T_k \frac{V_{2k}}{V_{2k-1}} \\
&= \mathcal T_k \frac{\dfrac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k}}{\dfrac{\mathcal C_1}{X_k}} \\
&= \mathcal T_k \frac{\dfrac{\mathcal C_2 \, X_k - \mathcal C_1}{\cancel{X_k}}}{\dfrac{\mathcal C_1}{\cancel{X_k}}} \\
&= \mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1},
\end{aligned}$$
thus
$$T_{2k} = \mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1}. \tag{*}$$
We make use of the matching of temperatures on the two sides of the conducting partitions: $T_{2k-2} = T_{2k-1} = \mathcal T_k$ for $k = 1, \ldots, n$. Symmetrically, $T_{2 \, (k+1)-2} = T_{2 \, (k+1)-1} = \mathcal T_{k+1}$. Since $T_{2 \, (k+1)-2}=T_{2k +2-2} = T_{2k}$ and $T_{2 \, (k+1)-2} = \mathcal T_{k+1}$, we also have $T_{2k} = \mathcal T_{k+1}$.
Equation $(*)$ becomes
$$\mathcal T_{k+1} = \mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1}. \tag{C1}$$

Now, we have
$$\begin{aligned}
X_{k+1} &= \mathcal T_{k+1}^{2 \, C_V/R} \, V_{2 \, (k+1)-2} \\
&= \mathcal T_{k+1}^{2 \, C_V/R} \, V_{2k+2-2} \\
&= \mathcal T_{k+1}^{2 \, C_V/R} \, V_{2k} \\
&= \underbrace{\left(\mathcal T_k \frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1} \right)^{2 \, C_V/R}}_{\text{(C1)}} \, \underbrace{\frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k}}_{ \text{(B1)}} \\
&= \mathcal T_k^{2 \, C_V/R} \, \left(\frac{\mathcal C_2 \, X_k - \mathcal C_1}{\mathcal C_1}\right)^{2 \, C_V/R} \, \frac{\mathcal C_2 \, X_k - \mathcal C_1}{X_k} \\
&= \mathcal T_k^{2 \, C_V/R} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R}}{C_1^{2 \, C_V/R}} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)}{X_k} \\
&= \mathcal T_k^{2 \, C_V/R} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \ + \, 1}}{C_1^{2 \, C_V/R} \, X_k} \\
&= \mathcal T_k^{2 \, C_V/R} \cdot \frac{V_{2k-2}}{V_{2k-2}} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \, 1}}{C_1^{2 \, C_V/R} \, X_k} \\
&= \underbrace{\mathcal T_k^{2 \, C_V/R} \, V_{2k-2}}_{\displaystyle X_k} \, \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \, 1}}{C_1^{2 \, C_V/R} \, X_k \, V_{2k-2}} \\
&= X_k \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \, 1}}{C_1^{2 \, C_V/R} \, X_k \, V_{2k-2}}.
\end{aligned}$$
Thus: $$X_{k+1} =X_k \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \ + \, 1}}{C_1^{2 \, C_V/R} \, X_k \, V_{2k-2}} \qquad \text{for} \quad k = 1, \ldots, n-1.$$

For $k=1$, we have $X_k = X_1$, $X_{k+1} = X_2$ and $V_{2k-2} = V_0$. Then:
$$X_{2} =X_1 \frac{\left(\mathcal C_2 \, X_1 - \mathcal C_1\right)^{2 \, C_V/R + 1}}{C_1^{2 \, C_V/R} \, X_1 \, V_{0}}.$$
For $k \geqslant 2$, we have $V_{2k-2} = V_{2(k-1)} = \frac{\mathcal C_2 \, X_{k-1} - \mathcal C_1}{X_{k-1}}$ according to $\text{(B1)}$. Then:
$$X_{k+1} = X_k \frac{\left(\mathcal C_2 \, X_k - \mathcal C_1\right)^{2 \, C_V/R \, + \ 1} \, X_{k-1}}{C_1^{2 \, C_V/R} \, X_k \, \left(\mathcal C_2 \, X_{k-1} - \mathcal C_1\right)} \qquad \text{for} \quad k = 2, \ldots, n-1.$$

OK so far?

I'm having trouble seeing through all the indexing. If you can please do me a favor and write this down (1) for the case of 2 cells total and (2) for the case of 4 cells total, that would be much appreciated. Also, instead of 2Cv/R, please write $1/(\gamma-1)=3$.

 

[Chestermiller]

For the case of 2 cells, my result in post #7 gives us enough to determine the final temperature in cell 0. $$\frac{T_0}{T_{init}}=\frac{T_1}{T_{init}}=[{(1-\xi_0)(1+\xi_2)}]^{-\frac{\gamma-1}{2}}$$ Taking $\xi_2=0$ for the insulated barrier on the right end, we have $$\frac{T_0}{T_{init}}=\frac{T_1}{T_{init}}=[{(1-\xi_0)}]^{-\frac{1}{3}}$$

 

[Meden Agan]

I'm having trouble seeing through all the indexing. If you can please do me a favor and write this down (1) for the case of 2 cells total and (2) for the case of 4 cells total, that would be much appreciated. Also, instead of 2Cv/R, please write 1/(γ−1)=3.

Sure.

The container has $2 \, k$ cells, for $k = 1, \ldots, n$, plus the moving piston $0$.
The conducting partition between two moveable partitions guarantees that the temperatures of the two adjacent cells is the same: $T_{2k-2}=T_{2k-1}$ for $k = 1, \ldots, n$. Let this common temperature be $\mathcal T_k$.
Thus: $$\mathcal T_k = T_{2k-2}=T_{2k-1} \qquad \text{for} \quad k = 1, \ldots, n.$$
Let's consider the pair of sections $(S_{2k-2}, S_{2k-1})$ for $k = 1, \ldots, n$. Section $S_{2k-2}$ is composed by wall $2k-2$ and wall $2k-1$; section $S_{2k-1}$ is composed by wall $2k-1$ and wall $2k$. This system is enclosed by wall $2k-2$ (moveable and insulating, or wall $0$ for $k=1$) and wall $2k$ (moveable and insulating). The internal wall $2k-1$ is fixed and conducting.
Let $V_{2k-2}$ and $V_{2k-1}$ be the volumes of the sections $S_{2k-2}$ and $S_{2k-1}$, respectively. Note that, in the present analysis, we are working with the volume of sections. Not with the volume of each cell.
In post #6, I found $$\mathcal T_k^{2/(\gamma -1)} \, V_{2k-2} \, V_{2k-1} = T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \equiv \verb|constant| \qquad \text{for} \quad k= 1, \ldots, n. \tag{A}$$


(1) Case of $2$ cells. For $k= 1$, the container has $2$ cells (cell $1$ and cell $2$), plus the moving piston $0$.
The conducting partition $1$ between piston $0$ and cell $2$ guarantees that the temperatures of the two adjacent cells is the same: $T_{0}=T_{1}$. Let this common temperature be $\mathcal T_1$.
Thus: $$\mathcal T_1 = T_{0}=T_{1}.$$
Let's consider the pair of sections $(S_{0}, S_{1})$. Section $S_0$ is composed by wall $0$ and wall $1$; section $S_1$ is composed by wall $1$ and wall $2$. This system is enclosed by piston $0$ and wall $2$ (moveable and insulating). The internal wall $1$ is fixed and conducting.
Let $V_{0}$ and $V_{1}$ be the volumes of the sections $S_0$ and $S_1$, respectively. Note again that, in the present analysis, we are working with the volume of sections. Not with the volume of each cell.
Substituting $k = 1$ in $\text{(A)}$:
$$\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, V_{1} = T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2. \tag{1}$$
We know that $V_1 = A \, (x_2 - x_1)$.
Plugging that equation into $(1)$:
$$\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, A \, (x_2 - x_1) = T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2.$$
As a boundary condition, barrier $2$ must be fixed and insulating, since it is the closed end of the container. Thus, $$x_2 = x_{2, \, \text{initial}} \iff x_2 - x_{2, \, \text{initial}} = 0. \tag{BC}$$
Moreover, we know that $$V_{\text{initial}} = A \, (x_1 - x_{0, \, \text{initial}}) = A \, (x_{2, \, \text{initial}} - x_1).\tag{2}$$
Then:
$$\begin{aligned}
\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, A \, (x_2 - x_1) &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, A \, (\overbrace{x_2 - x_{2, \, \text{initial}}}^{\displaystyle 0 \,\, \text{as per (BC)}} + x_{2, \, \text{initial}} - x_1) &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, \underbrace{A \, (x_{2, \, \text{initial}} - x_1)}_{\displaystyle := V_{\text{initial}} \,\, \text{as per (2)}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, V_{\text{initial}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
\mathcal T_1^{2/(\gamma -1)} \, V_{0} \, \cancel{V_{\text{initial}}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^{\cancel{2}} \\
\mathcal T_1^{2/(\gamma -1)} \, V_{0} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\left(\frac{T_1}{T_{\text{initial}}}\right)^{2/(\gamma -1)} &= \frac{V_{\text{initial}}}{V_{0}} \\
\frac{T_1}{T_{\text{initial}}} &= \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{\frac{\gamma-1}{2}} \\
T_1 &= T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{\frac{\gamma-1}{2}}.
\end{aligned}$$
So: $$T_1 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{\frac{\gamma-1}{2}},$$
which agrees with your result.


If it is OK so far, I'll go ahead with the case of $4$ cells, but I'm not able to achieve a closure.

 

[Chestermiller]

For 4 cells, if I did the algebra right, I get $$\frac{T_0}{T_{init}}=\left[\frac{1+(1-\xi_0)^{1/4}}{2(1-\xi_0)}\right]^{1/3}$$

 

[Meden Agan]

Again, from post #6, I have
$$\mathcal T_k^{2 \, C_V/R} \, V_{2k-2} V_{2k-1} = T_{\text{initial}}^{2 \, C_V/R} \, V_{\text{initial}}^2 \equiv \verb|constant| \qquad \text{for} \quad k= 1, \ldots, n\tag{A}$$
$$V_{2k-1} + V_{2k} = 2 V_{\text{initial}} \equiv \text{constant} \qquad \text{for} \quad k= 1, \ldots, n \tag{B}$$
$$\frac{\mathcal T_k}{V_{2k-1}} = \frac{T_{2k}}{V_{2k}}. \tag{C}$$

(2) Case of $4$ cells. For $k= 2$, the container has $4$ cells (cell $1$, cell $2$, cell $3$, cell $4$), plus the moving piston $0$.
The conducting partition $3$ between cell $2$ and cell $4$ guarantees that the temperatures of the two adjacent cells is the same: $T_{2}=T_{3}$. Let this common temperature be $\mathcal T_2$.
Thus: $$\mathcal T_2 = T_{2}=T_{3}.$$
Let's consider the pair of sections $(S_{2}, S_{3})$. Section $S_2$ is composed by wall $2$ and wall $3$; section $S_3$ is composed by wall $3$ and wall $4$. This system is enclosed by wall $2$ (moveable and insulating) and wall $4$ (moveable and insulating). The internal wall $3$ is fixed and conducting.
Let $V_{2}$ and $V_{3}$ be the volumes of the sections $S_2$ and $S_3$, respectively. Note that, in the present analysis, we are working with the volume of sections. Not with the volume of each cell.
Substituting $k = 2$ in $\text{(A)}$:
$$\mathcal T_2^{2/(\gamma -1)} \, V_{2} \, V_{3} = T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2. \tag{1.1}$$
Substituting $k = 2$ in $\text{(B)}$:
$$V_3+V_4 = 2 \, V_{\text{initial}}. \tag{2.1}$$
As a boundary condition, barrier $4$ must be fixed and insulating, since it is the closed end of the container. Thus, $$x_4 = x_{4, \, \text{initial}} \iff x_4 - x_{4, \, \text{initial}} = 0. \tag{BC1}$$
We know that $V_4 = A \, (x_4 - x_3)$ and $V_{\text{initial}} = A \, (x_{4, \, \text{initial}} - x_3)$. Then we can write
$$\begin{aligned}
V_4 &= A \, (x_4 - x_3) \\
&= A \, (\underbrace{x_4 - x_{4, \, \text{initial}}}_{\displaystyle :=0} + x_{4, \, \text{initial}} - x_3) \\
&= \underbrace{A \, (x_{4, \, \text{initial}} - x_3)}_{\displaystyle := V_{\text{initial}}} \\
&= V_{\text{initial}}.
\end{aligned}$$
Thus $V_4 = V_{\text{initial}}$. Plugging that expression into $(2.1)$:
$$V_3 + V_{\text{initial}} = 2 \, V_{\text{initial}} \iff V_3 = V_{\text{initial}}. \tag{3.1}$$
Substituting $(3.1)$ in $(1.1)$:
$$\begin{aligned}
\mathcal T_2^{2/(\gamma -1)} \, V_{2} \, V_{\text{initial}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
\mathcal T_2^{2/(\gamma -1)} \, V_{2} \, \cancel{V_{\text{initial}}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^{\cancel{2}} \\
\mathcal T_2^{2/(\gamma -1)} \, V_{2} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}.
\end{aligned}$$
So $$\mathcal T_2^{2/(\gamma -1)} \, V_{2} = T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}. \tag{E1}$$
From Equation $\text{(A)}$ $$V_{2k-1} = \frac{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_k^{2/(\gamma -1)} \, V_{2k-2}}. \tag{A1}$$
From Equation $\text{(B)}$
$$\begin{aligned}
V_{2k} &= 2 \, V_{\text{initial}} - V_{2k-1} \\
&= 2 \, V_{\text{initial}} - \frac{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_k^{2/(\gamma -1)} \, V_{2k-2}} \\
&= \frac{2 \, V_{\text{initial}} \, \mathcal T_k^{2/(\gamma -1)} \, V_{2k-2} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_k^{2/(\gamma -1)} \, V_{2k-2}},
\end{aligned}$$
so $$V_{2k} = \frac{2 \, V_{\text{initial}} \, \mathcal T_k^{2/(\gamma -1)} \, V_{2k-2} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_k^{2/(\gamma -1)} \, V_{2k-2}}. \tag{B1}$$
Substituting $k=1$ in $\text{(A1)}$, we obtain an expression for $V_1$:
$$V_1 = \frac{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_1^{2/(\gamma -1)} \, V_{0}}. \tag{E2}$$
Substituting $k=1$ in $\text{(B1)}$, we obtain an expression for $V_2$:
$$V_2 = \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_1^{2/(\gamma -1)} \, V_{0}}. \tag{E3}$$
From Equation $\text{(C)}$ $$T_{2k} = \mathcal T_k \frac{V_{2k}}{V_{2k-1}},$$ and for $k=1$ we obtain an expression for $\mathcal T_2$:
$$\begin{aligned}
T_2 &= \mathcal T_1 \frac{V_{2}}{V_{1}} \\
&= \mathcal T_1 \frac{\dfrac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_1^{2/(\gamma -1)} \, V_{0}}}{\dfrac{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_1^{2/(\gamma -1)} \, V_{0}}} \\
&= \mathcal T_1 \frac{\dfrac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\cancel{\mathcal T_1^{2/(\gamma -1)} \, V_{0}}}}{\dfrac{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\cancel{\mathcal T_1^{2/(\gamma -1)} \, V_{0}}}} \\
&= \mathcal T_1 \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}.
\end{aligned}$$
Thus: $$T_2 = \mathcal T_1 \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}. \tag{E4}$$
Plugging the expressions $(\text{E3})$ and $(\text{E4})$ for $V_2$ and $T_2$, respectively, into the expression $(\text{E1})$, we obtain
$$\begin{aligned}
\mathcal T_2^{2/(\gamma -1)} \, V_{2} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\left(\mathcal T_1 \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}\right)^{2/(\gamma -1)} \, \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_1^{2/(\gamma -1)} \, V_{0}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\mathcal T_1^{2/(\gamma -1)} \frac{\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}}{\left(T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}} \, \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\mathcal T_1^{2/(\gamma -1)} \, V_{0}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\cancel{\mathcal T_1^{2/(\gamma -1)}} \frac{\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}}{\left(T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}} \, \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{\cancel{\mathcal T_1^{2/(\gamma -1)}} \, V_{0}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\frac{\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}}{\left(T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}} \, \frac{2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2}{V_{0}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\frac{\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1) \, + \, 1}}{V_0 \, \left(T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)}} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \\
\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1) \, + \, 1} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \, V_0 \,\left(T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1)} \\
\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1) \, + \, 1} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}} \, V_0 \,T_{\text{initial}}^{4/(\gamma -1)^2} \, V_{\text{initial}}^{4/(\gamma -1)} \\
\left(2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2\right)^{2/(\gamma -1) \, + \, 1} &= T_{\text{initial}}^{2/(\gamma -1) \, + \, 4/(\gamma -1)^2} \, V_{\text{initial}}^{1 \, + \, 4/(\gamma -1)} \, V_0 \\
2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} - T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 &= T_{\text{initial}}^{\frac{2/(\gamma -1) \, + \, 4/(\gamma -1)^2}{2/(\gamma -1) \, + \, 1}} \, V_{\text{initial}}^{\frac{1 \, + \, 4/(\gamma -1)}{2/(\gamma -1) \, + \, 1}} \, V_0^{\frac{1}{2/(\gamma -1) \, + \, 1}} \\
2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} &= T_{\text{initial}}^{\frac{2/(\gamma -1) \, + \, 4/(\gamma -1)^2}{2/(\gamma -1) \, + \, 1}} \, V_{\text{initial}}^{\frac{1 \, + \, 4/(\gamma -1)}{2/(\gamma -1) \, + \, 1}} \, V_0^{\frac{1}{2/(\gamma -1) \, + \, 1}} + T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
2 \, V_{\text{initial}} \, \mathcal T_1^{2/(\gamma -1)} \, V_{0} &= T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^{\frac{1 \, + \, 4/(\gamma -1)}{2/(\gamma -1) \, + \, 1}} \, V_0^{\frac{1}{2/(\gamma -1) \, + \, 1}} + T_{\text{initial}}^{2/(\gamma -1)} \, V_{\text{initial}}^2 \\
\mathcal T_1^{2/(\gamma-1)} &= T_{\text{initial}}^{2/(\gamma -1)} \left(\frac{V_{\text{initial}}^{\frac{2}{\gamma+1}} \, V_0^{\frac{\gamma-1}{\gamma+1}} + V_{\text{initial}}}{2 \, V_0}\right) \\
\mathcal T_1 &= T_{\text{initial}} \left(\frac{V_{\text{initial}}^{\frac{2}{\gamma+1}} \, V_0^{\frac{\gamma-1}{\gamma+1}} + V_{\text{initial}}}{2 \, V_0}\right)^{\frac{\gamma-1}{2}} \\
\mathcal T_1 &= T_{\text{initial}} \left(\frac{V_{\text{initial}}^{\frac{2}{\gamma+1}} \left(V_0^{\frac{\gamma-1}{\gamma+1}}+ V_{\text{initial}}^{\frac{\gamma-1}{\gamma+1}}\right)}{2 \, V_0}\right)^{\frac{\gamma-1}{2}} \\
\mathcal T_1 &= T_{\text{initial}} \, V_{\text{initial}}^{\frac{\gamma-1}{\gamma+1}} \left(\frac{V_0^{\frac{\gamma-1}{\gamma+1}}+ V_{\text{initial}}^{\frac{\gamma-1}{\gamma+1}}}{2 \, V_0}\right)^{\frac{\gamma-1}{2}}.
\end{aligned}$$
Then $$\mathcal T_1 = T_0 = T_1 = T_{\text{initial}} \, V_{\text{initial}}^{\frac{\gamma-1}{\gamma+1}} \left(\frac{V_0^{\frac{\gamma-1}{\gamma+1}}+ V_{\text{initial}}^{\frac{\gamma-1}{\gamma+1}}}{2 \, V_0}\right)^{\frac{\gamma-1}{2}}.$$
For $\gamma = 5/3$ I get $$\mathcal T_1 = T_{\text{initial}} \, V_{\text{initial}}^{1/4} \left(\frac{V_0^{1/4} + V_{\text{initial}}^{1/4}}{2 \, V_0}\right)^{1/3},$$
which agrees with your result.

May I ask you to write down your algebra?

 

[Chestermiller]

FOUR CELL CASE:

In the previous development, I showed for cells 0 and 1 that $$\frac{T_0}{T_{init}}=\frac{T_1}{T_{init}}=\frac{1}{[(1-\xi_0)(1+\xi_2)]^{1/3}}\tag{1}$$By extension, for cells 2 and 3, we have $$\frac{T_2}{T_{init}}=\frac{T_3}{T_{init}}=\frac{1}{[(1-\xi_2)(1+\xi_4)]^{1/3}}$$But, from the boundary condition at the closed end of the cylinder we have, $$\xi_4=0$$Therefore, in the case of 4 cells, we have:$$\frac{T_2}{T_{init}}=\frac{T_3}{T_{init}}=\frac{1}{(1-\xi_2)^{1/3}}\tag{2}$$From the condition of equal pressures on the two sides of an floating insulated partition, we have $$P_2=P_3$$or, from the ideal gas law, $$\frac{T_2}{1-\xi_2}=\frac{T_1}{1+\xi_2}\tag{3}$$
Eqns. 1-3 are sufficient to solve the 4 cell case. The number of unknowns matches the number of equations.

 

[Meden Agan]

FOUR CELL CASE:

In the previous development, I showed for cells 0 and 1 that $$\frac{T_0}{T_{init}}=\frac{T_1}{T_{init}}=\frac{1}{[(1-\xi_0)(1+\xi_2)]^{1/3}}\tag{1}$$By extension, for cells 2 and 3, we have $$\frac{T_2}{T_{init}}=\frac{T_3}{T_{init}}=\frac{1}{[(1-\xi_2)(1+\xi_4)]^{1/3}}$$But, from the boundary condition at the closed end of the cylinder we have, $$\xi_4=0$$Therefore, in the case of 4 cells, we have:$$\frac{T_2}{T_{init}}=\frac{T_3}{T_{init}}=\frac{1}{(1-\xi_2)^{1/3}}\tag{2}$$From the condition of equal pressures on the two sides of an floating insulated partition, we have $$P_2=P_3$$or, from the ideal gas law, $$\frac{T_2}{1-\xi_2}=\frac{T_1}{1+\xi_2}\tag{3}$$
Eqns. 1-3 are sufficient to solve the 4 cell case. The number of unknowns matches the number of equations.

Your approach is way, way nicer than mine.
We analyzed the cases of $2$ cells and $4$ cells. But now, we have the general case of $2k$ cells, plus the piston $0$. I outlined an approach in post #22. Any hints?

 

[Chestermiller]

Start at the closed end and work backwards.

 

[Meden Agan]

Start at the closed end and work backwards.

Mhm, I'm not sure I understand that hint. If you could be more specific, please?

 

[Chestermiller]

Mhm, I'm not sure I understand that hint. If you could be more specific, please?

In my analysis in post #28, I started out with the equation for the final pair of cells, and set the dimensionless displacement of the insulated end $\xi_4$ equal to zero. Then I took the equations for the previous set of cells, and solved for the dimensional displacement $\xi_2$ in terms of the displacement in the one pair of cells further towards the open end. Would it help if I showed the algebra I used for solving the equations in post #28?

 

[Meden Agan]

In my analysis in post #28, I started out with the equation for the final pair of cells, and set the dimensionless displacement of the insulated end $\xi_4$ equal to zero. Then I took the equations for the previous set of cells, and solved for the dimensional displacement $\xi_2$ in terms of the displacement in the one pair of cells further towards the open end. Would it help if I showed the algebra I used for solving the equations in post #28?

Sure it would help. But I did something similar in post #27. I got the same result as you.
Post #27 is a special case of the generalization of post #22, case of $2k$ cells. I cannot get a closure, though. Were you able to get a closed form for the general case of $2k$ cells?

 

[Chestermiller]

Sure it would help. But I did something similar in post #27. I got the same result as you.
Post #27 is a special case of the generalization of post #22, case of $2k$ cells. I cannot get a closure, though. Were you able to get a closed form for the general case of $2k$ cells?

I'll provide my algebraic solution to the equations in post #28 later today.

But, in the meantime, remember that the problem statement calls for the results only for a monatomic gas with Cv=3R/2 and $\gamma=5/3$. I suggest you use this in your analysis to simplify the equations and to simplify the job of documenting the derivations.

 

[Meden Agan]

I'll provide my algebraic solution to the equations in post #28 later today.

But, in the meantime, remember that the problem statement calls for the results only for a monatomic gas with Cv=3R/2 and $\gamma=5/3$. I suggest you use this in your analysis to simplify the equations and to simplify the job of documenting the derivations.

Sure, I'll simplify my algebra. I'm quite aware as to how to obtain $T_0$ for the cases of $2$ cells and $4$ cells. But I'm not about how to get $T_0$ in the case of $2k$ cells.
May I ask you to post your final result for $2k$ cells, if you got it?

 

[Chestermiller]

Sure, I'll simplify my algebra. I'm quite aware as to how to obtain $T_0$ for the cases of $2$ cells and $4$ cells. But I'm not about how to get $T_0$ in the case of $2k$ cells.
May I ask you to post your final result for $2k$ cells, if you got it?

I haven’t done it for the general case yet. I thought they doing it for four shells would be sufficient to give us an idea of how to do that. I’ll take a shot of doing it for the general case in a little while.

 

[Chestermiller]

When I went to 2k cells, or even only 2k=6 cells, the math got very unwieldy and there was not a simple algebraic solution as for the case of 2 or 4 cells. This nonlinear problem would, in my opinion, have to be solved numerically. Of course, some software automatically handles recursive implicit relations even if the math is complicated.

 

[Meden Agan]

When I went to 2k cells, or even only 2k=6 cells, the math got very unwieldy and there was not a simple algebraic solution as for the case of 2 or 4 cells. This nonlinear problem would, in my opinion, have to be solved numerically. Of course, some software automatically handles recursive implicit relations even if the math is complicated.

Mhm, I had this bad feeling. Perhaps if we apply the polytropic process equation to the last section of partitions? How about it?

 

[Chestermiller]

Mhm, I had this bad feeling. Perhaps if we apply the polytropic process equation to the last section of partitions? How about it?

This is going to boil down to a set of non-linear algebraic equations describing a split boundary value problem. It will be solved using a non-linear equation solver. The unknowns are the displacements $x_i$ and the Temperatures of the cells.

 

[Meden Agan]

This is going to boil down to a set of non-linear algebraic equations describing a split boundary value problem. It will be solved using a non-linear equation solver. The unknowns are the displacements $x_i$ and the Temperatures of the cells.

May I ask you to use such a non-linear equation solver and tell me what result do you get?

 

[Chestermiller]

May I ask you to use such a non-linear equation solver and tell me what result do you get?

I no longer have access to software of this type, because I have been retired 22 years now. I'm not willing to subscribe to such software out of my own pocket. Do you have access to the IMSL library or even less extensive software?

 

[Meden Agan]

I no longer have access to software of this type, because I have been retired 22 years now. I'm not willing to subscribe to such software out of my own pocket. Do you have access to the IMSL library or even less extensive software?

Sadly, no. I think I'll give up this problem for a while.

 

[Chestermiller]

Sadly, no. I think I'll give up this problem for a while.

https://www.mathworks.com/help/optim/ug/fsolve.html

 

[Meden Agan]

https://www.mathworks.com/help/optim/ug/fsolve.html

Mhm, I have no idea how to use it.
According to the official solution, the final result is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_0}\right)^{f_n},$$ where $f_n =\dfrac{2}{7-\dfrac{(4+\sqrt{15})^2+(4+\sqrt{15})^{2n}}{(4+\sqrt{15})+(4+\sqrt{15})^{2n \, + \,1}}}$.
It sounds too nice to be true.

 

[Chestermiller]

Mhm, I have no idea how to use it.
According to the official solution, the final result is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_0}\right)^{f_n},$$ where $f_n =\dfrac{2}{7-\dfrac{(4+\sqrt{15})^2+(4+\sqrt{15})^{2n}}{(4+\sqrt{15})+(4+\sqrt{15})^{2n \, + \,1}}}$.
It sounds too nice to be true.

I have no idea where they got that result. It looks like some cancelation is possible between the numerator and denominator. How does this result compare with our results for n=1 and n=2? Have you checked?

 

[Meden Agan]

I have no idea where they got that result. It looks like some cancelation is possible between the numerator and denominator. How does this result compare with our results for n=1 and n=2? Have you checked?

Our result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{1/3}.$$
Their result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{2/3}.$$

For $n=1$, the expressions are very different.

 

[Chestermiller]

Our result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{1/3}.$$
Their result for $n=0$ is $$T_0 = T_{\text{initial}} \left(\frac{V_{\text{initial}}}{V_{0}}\right)^{2/3}.$$

For $n=1$, the expressions are very different.

I meant for two shells and four shells

 

[Meden Agan]

$$4+\sqrt{15}=(\sqrt{5}+\sqrt{3})^2=3(\sqrt{\gamma }+1)^2$$

OK...? How is that useful?

Minor typo: $${\color{red}{2}} \, \left(4+\sqrt{15}\right)= \left(\sqrt{5}+\sqrt{3}\right)^2=3 \, \left(\sqrt{\gamma}+1 \right)^2.$$

 

[Chestermiller]

OK...? How is that useful?

Minor typo: $${\color{red}{2}} \, \left(4+\sqrt{15}\right)= \left(\sqrt{5}+\sqrt{3}\right)^2=3 \, \left(\sqrt{\gamma}+1 \right)^2.$$

I can't use this equation editor. They have now changed it into something that now sucks.

 

[Chestermiller]

$$4+\sqrt{15}=\frac{8+2\sqrt{15}}{2}=\frac{(\sqrt{5}+\sqrt{3})^2}{2}$$