Confusion in the validity of First Law of Thermodynamics

Question

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.

Answer 1

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.