Equilibrium Conditions for a Smooth Adiabatic Piston

​ Feynman mentions in his lectures on physics Vol. I, §39-4 that if the pressures on both sides of a smooth adiabatic piston are equal but the temperatures are unequal, energy will transfer from the higher-temperature gas to the lower-temperature gas through microscopic vibrations of the piston, even when all heat conduction is neglected.

​ This article attempts to examine microscopically whether the piston can achieve equilibrium and the process of reaching the final equilibrium state.

We begin our discussion with a simple model: As shown in Fig. 1, a smooth adiabatic piston divides an adiabatic container into two compartments. For the piston, the cross-sectional area is denoted by \(A\) and its mass by \(M\). For the gas, the mass of a single molecule is denoted by \(m ({\ll} M)\), the constant-pressure molar heat capacity by \(C^{mol}_p\), and the gas can be treated as ideal.

In the initial state, the amount of substance of the left gas is \(\nu_1\), pressure \(p_0\), temperature \(T_{10}\); the right gas has amount of substance \(\nu_2\), pressure \(p_0\), and temperature \(T_{20} (> T_{10})\).

At some time \(t\) during the process, let the pressure of the left gas be \(p\), temperature \(T_{1}\); the right gas has pressure \(p\), temperature \(T_{2}\). The velocity of the piston is \(u\) (defined as positive to the right).

Collision Between Gas Molecules and the Piston #

At a moment when the piston’s velocity is \(u\), consider a gas molecule on the left with velocity component perpendicular to the piston \(v_x\).

After collision with the piston, its velocity becomes

$$ v_x' =\frac{\left ( m-M \right )v_x+2Mu }{M+m} \tag{1} $$

Energy Change #

The energy change of this gas molecule before and after collision is

$$ \begin{align} \triangle \varepsilon\left ( v_x \right )=&\frac{1}2{} m\left ( v_x'^2 -v_x^2 \right ) \notag \\ =& 2Mm\frac{Mu^2-\left ( M-m \right )uv_x-mv_x^2 }{\left ( M+m \right )^2 } \tag{2} \end{align} $$

Considering the randomness of \(u\). Taking the average, we obtain

$$ \left \langle \triangle \varepsilon \right \rangle _{(v_x)}=\frac{2Mm}{\left ( M+m \right )^2}\left ( M\left \langle u^2 \right \rangle - (M - m) v_x \left \langle u \right \rangle - mv_x^2 \right ) \tag{3} $$

Given that \(\sqrt{\frac{kT_1}{m} }\gg u\), we integrate over \(v_x\in[0,+\infty]\) to find the energy change per unit time for the left gas

$$ \begin{align} P_1 &=\int_{0}^{\infty }\triangle \varepsilon _{\left ( v_x \right )}v_xAn_1\sqrt{\frac{m}{2\pi kT_1} }e^{-\frac{m}{2kT_1}v_x^2}dv_x \notag \\ &=\frac{2Mm}{\left ( M+m \right )^2 }An_1\sqrt{\frac{m}{2\pi kT_1} }\frac{kT_1}{m}\left ( M\left \langle u^2 \right \rangle - \sqrt{\frac{\pi k T_1}{2 m}} (M-m) \left \langle u \right \rangle - 2kT_1 \right ) \notag \quad \text{(4)} \end{align} $$

Using \(p=n_1kT_1\), we get

$$ P_1=\frac{2Mm}{\left ( M+m \right )^2 }Ap \sqrt{\frac{m}{2\pi kT_1} } \frac{1}{m} \left ( M\left \langle u^2 \right \rangle - \sqrt{\frac{\pi k T_1}{2 m}} (M-m) \left \langle u \right \rangle - 2kT_1 \right ) \quad \text{(5)} $$

Effective Temperature of the Piston #

Define

$$ \left \langle u \right \rangle=0\quad,\quad\frac{1}2M\left \langle u^2 \right \rangle = \frac{1}2kT_0 \tag{6} $$

where \(T_0\) is the effective temperature of the piston. Note that \(\left \langle u \right \rangle \neq 0\) in reality, but it will be shown later that \(\left \langle u^2 \right \rangle \gg \sqrt{\frac{\pi k T_1}{2 m}} \left \langle u \right \rangle\), allowing \(\left \langle u \right \rangle\) to be neglected. Thus

$$ P_1=\frac{2Mm}{\left ( M+m \right )^2 }Ap \sqrt{\frac{m}{2\pi kT_1} } \frac{1}{m} \left ( kT_0-2kT_1 \right ) \tag{7} $$

Similarly, the energy change per unit time for the right gas is

$$ P_2=\frac{2Mm}{\left ( M+m \right )^2 }Ap \sqrt{\frac{m}{2\pi kT_2} } \frac{1}{m} \left ( kT_0-2kT_2 \right ) \tag{8} $$

Given \(\frac{1}{2}M\left \langle u^2 \right \rangle\ll \nu_{1,2}C^{mol}_V T_{1,2}\), energy conservation for the system can be written as

$$ P_1+P_2=0 \tag{9} $$

Therefore

$$ \frac{T_0}{\sqrt{T_1}}-2\sqrt{T_1}+\frac{T_0}{\sqrt{T_2} }-2\sqrt{T_2}=0 \tag{10} $$

$$ T_0=2\sqrt{T_1T_2} \tag{11} $$

Energy Flux #

Substituting (11) into (7), the energy flux from the right gas to the left gas is

$$ P=\frac{4Mm}{\left ( M+m \right )^2 }A\sqrt{\frac{k}{2\pi m} }p\left ( \sqrt{T_2}-\sqrt{T_1} \right ) \tag{12} $$

Thermodynamic Quantities #

Initial state

$$ p_0V_{10}={\nu} _1RT_{10}, \quad p_0V_{20}=\nu_2RT_{20} \tag{13} $$

State at time \(t\)

$$ pV_1=\nu_1 RT_1, \quad pV_2=\nu_2 RT_2 \tag{14} $$

Again, given \(\frac{1}{2}M\left \langle u^2 \right \rangle\ll \nu_{1,2}C^{mol}_V T_{1,2}\), energy conservation for the system is

$$ \nu _1C^{mol}_V T_{10}+\nu _2C^{mol}_V T_{20}=\nu _1C^{mol}_V T_1+\nu _2C^{mol}_V T_2 \tag{15} $$

From

$$ V_{10}+V_{20}=V_1+V_2 \tag{16} $$

it follows that

$$ p=p_0 \tag{17} $$

Differential Equations for Thermodynamic Quantities #

For a time element \(t\) ~ \(t+dt\), suppose the piston moves right by \(dx\). The energy conservation equation for the left gas is

$$ \nu _1C_V^{mol}dT_1=Pdt-p_0Adx \tag{18} $$

Since pressure is constant by (17), we have

$$ \frac{dT_1}{T_1}-\frac{Adx}{V_1}=0 \tag{19} $$

i.e.,

$$ p_0Adx=\nu _1RdT_1 \tag{20} $$

Substituting into (18) gives

$$ \nu _1C_p^{mol}dT_1=Pdt \tag{21} $$

Substituting (12) and eliminating \(P\), we obtain

$$ \nu _1C_p^{mol}dT_1=\frac{4Mm}{\left ( M+m \right )^2 }A\sqrt{\frac{k}{2\pi m} }p_0\left ( \sqrt{T_2}-\sqrt{T_1} \right )dt \tag{22} $$

Combining with (15) yields the system of differential equations for \(T_1(t)\), \(T_2(t)\)

$$ \begin{cases} \ \nu _1T_{10}+\nu _2T_{20}=\nu _1T_1+\nu _2T_2 \\\\ \ \nu _1C_P^{mol}\frac{dT_1}{dt}=\frac{4Mm}{\left ( M+m \right )^2 }A\sqrt{\frac{k}{2\pi m} }P_0\left ( \sqrt{T_2}-\sqrt{T_1} \right ) \end{cases} \tag{23} $$

Approximate Solution #

(23) cannot be solved analytically. Consider the case where \(T_{10}-T_{20}=\delta T_0 \ll T_1, T_2\). The final equilibrium temperature is

$$ T_f=\frac{\nu _1T_{10}+\nu _2T_{20}}{\nu _1+\nu _2} \tag{24} $$

During the process, let \(\delta T=T_1-T_2\), then

$$ {\begin{cases}T_1=T_f+\frac{\nu _2}{\nu _1+\nu _2}\delta T \\T_2=T_f-\frac{\nu _1}{\nu _1+\nu _2}\delta T\end{cases}} \tag{25} $$

Substituting into (22) and retaining first-order small quantities gives

$$ \frac{d\left ( \delta T \right ) }{\delta T}=-\frac{2Mm}{\left ( M+m \right )^2 }A\sqrt{\frac{kT_f}{2\pi m} }p_0\frac{1}{T_f}\frac{1}{C_p^{mol}}\frac{\nu_1+\nu _2}{\nu _1\nu _2}dt \tag{26} $$

i.e.,

$$ \delta T=\delta T_0 e^{-\frac{t}{\tau }} \tag{27} $$

$$ T_1-T_2=(T_{10}-T_{20}) e^{-\frac{t}{\tau }} \tag{28} $$

where the characteristic time constant \(\tau\) is

$$ \begin{align} \tau &=\frac{\left ( M+m \right )^2 }{2Mm}\frac{T_f}{Ap_0}\sqrt{\frac{2\pi m}{kT_f} }C_p^{mol}\frac{\nu _1\nu _2}{\nu _1+\nu _2} \notag \\ &=\frac{\left ( M+m \right )^2 }{2Mm}\frac{1}{Ap_0}\sqrt{\frac{2\pi m}{k} \cdot \frac{\nu _1T_{10}+\nu _2T_{20}}{\nu _1+\nu _2}}C_p^{mol}\frac{\nu _1\nu _2}{\nu _1+\nu _2} \tag{29} \end{align} $$

As a reference, using values \(M=10^{-3} \text{kg}, m=5\times 10^{-26} \text{kg}, A=10^{-2} \text{m}^2, p_0=1 \text{atm}, T_{10}=310 \text{K}, T_{20}=300 \text{K}, \nu_1=\nu_2=10 \text{mol}, C^{mol}_p=\frac{7}{2}R\), we find

$$ \tau \approx 1.2\times10^{14} \text{a (years)} \tag{30} $$

Thus, this effect is negligible in macroscopic systems. (Additionally, \(\left \langle u \right \rangle \sim \frac{l}{\tau}\), where \(l \sim \frac{\nu R T}{p A}\) is the characteristic length of the container. This validates the earlier assumption \(\left \langle u^2 \right \rangle \gg \sqrt{\frac{\pi k T_1}{2 m}} \left \langle u \right \rangle\) )