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Suppose a diathermic movable piston separates an adiabatic container into $2$ parts. Thermodynamic parameters of gases in the two parts are $p_1,v_1,t_1$ and $p_2,v_2,t_2$. We assume quasi static processes.

If we apply first law of thermodynamics to these two parts separately and take into account that heat lost by one is gained by the other and change in internal energy of one gas is the negative of the change in internal energy of the other gas, the work done by one should be the negative of the work done by other.

  1. How are work done by the two gases equal in magnitude if the pressures of the two gases are unequal throughout till the end of the process?

  2. My teacher said that FLT can be applied to both the gases separately as well as together. I’m confused about how I can apply it to both gases simultaneously.

Any help in clearing my confusion will be very appreciated.

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    $\begingroup$ Nice question: the answer to your question is in the "macroscopic non-equilibrium" of the system: if pressure is different in the rooms at the 2 sides of the wall, and the wall is free to slide, the wall starts moving: if the wall has some mass, a contribution of the kinetic energy appears in the first principle of the thermodynamics. I have no time to go further in the details right now, but I leave you a link to hand-written notes about principles of TD for closed systems, basics.altervista.org/test/Physics/TD/td_principles.html $\endgroup$
    – basics
    Jan 6 at 10:11
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    $\begingroup$ You can easily remember the formulation and the meaning of the 1st principle of TD as: the variation of the total energy of a closed system equals the sum of the work done on it and the heat transferred to it. What's the total energy of a system? You can decompose it as the sum of the kinetic energy (due to MACROSCOPIC motion of bodies with mass) and all the remaining part, that we call internal energy (everything that is not related to macroscopic motion, but to other physical quantities, like temperature or strain and stress - for solid) $\endgroup$
    – basics
    Jan 6 at 10:15

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On your first question, the answer lies in the assumption that the process be quasistatic, which dictates that the system be in equilibrium at any time during the process. For the system to reach equilibrium quasistatically, there must be some kind of damping force, so as to render the two forces $P_1$ and $P_2$ infinitesmally different at any time. So the pressure on two sides of the piston indeed are unequal, but only infinitesmally so.

On your second question, I don't see why not. We can apply the first law to both chambers, and from this also justify the claim in the first paragraph: Denote the system with subscript $1$ as $\mathcal{S}_1$, that with subscript $2$ as $\mathcal{S}_2$, and the whole system $\mathcal{S} = \mathcal{S}_1+\mathcal{S}_2$. The first law of thermodynamics states that for a chosen system: $$ \delta Q +\delta W= \mathrm{d}U,$$ where we adopt the sign convention that work done on the system is positive.

Apply the first law to $\mathcal{S}_1$ and $\mathcal{S}_2$ separately at any given time, we get $$ \mathrm{d}U_1 = \delta Q_1 - P_1\mathrm{d}V_1, \quad \mathrm{d}U_2 = \delta Q_2 - P_1\mathrm{d}V_2. $$

But the outer walls are adiabatic and rigid, so applying the first law to $\mathcal{S}$ gives $\delta Q_1 = -\delta Q_2$, $\mathrm{d}V_1 = -\mathrm{d}V_2$, and $\mathrm{d}U_1 = -\mathrm{d}U_2$. This means $$ P_1\mathrm{d}V_1 = P_2\mathrm{d}V_1, $$ so $P_1$ and $P_2$ are identical up to a first order differential, i.e. $$ P_1 = P_2 + \delta P, $$ where $\delta$ is used here to imply a very small quantity, not an inexact differential.

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