Clarifying free adiabatic expansion

Question

The free adiabatic expansion is described in my notes like this:

A monoatomic perfect gas is in equilibrium in half of the gas tank, which is thermally isolated (i.e. it does not exchange heat with the outside). We also assume that the gas tank is in the void, so that we can consider the external pressure to be equal to zero. The piston is released, and the expansion begins. The transformation is irreversible, but we can compute the new equilibrium state assuming that initially the gas was characterized by the thermodynamic state of n moles of monoatomic gas $P_0, V_0/2,T_0$.

Let’s analyze the different terms of the first law of thermodyammics in our free expansion. The expansion is adiabatic, no heat exchange with the outside is possible, and $Q = 0$. Moreover, as the external pressure is zero, the piston performs zero mechanical work ($Pext = 0$ implies that the force acting on the piston is zero), and thus $W = 0$. If both$ Q$ and $W$ are zero, we have that the variation of the internal energy $∆U = 0$ . $U = U(T )$ is a function of the temperature alone (indeed the Joule expansion was the crux experiment performed by Joule to show that the internal energy depends only on the temperature). This in turn implies that $T0 = T1$, and the new state of the system is$ P0=2$; $V0$; $T0$. As nothing from the thermodynamic point of view seems to have changed, what happens to the entropy? As the thermodynamic entropy is a state function, we can compute the entropy variation along any reversible transformation connecting the initial and the final state. The temperature is constant, therefore the isothermal transform is a good candidate, which yields $\Delta S= \int\frac{dQ}{T}=nRln2$

There are a few things I don't get

1. It says that the work is zero because the external pressure is zero.... But the internal pressure is not, doesn't the gas push the piston as it expands, which is work, even if no external pressure force is present? Perhaps I am not understanding what "the piston is released" means. Does it mean something was holding it and then they just leave it, so that the gas can push it? or does it mean that someone quickly pulls it all the way , before the gas can react (in that case the person would be doing work, wouldn't he?)

2. Since the heat is zero, because the system is adiabatic, shouldn't the entropy be zero, like this: $\Delta S= \int\frac{dQ}{T}=\int\frac{0}{T}=0$ ? Instead they use an isothermic path, but that would be changing the problem, wouldn't it? I know I can use any path to find the entropy, but in doing so, shouldn't I respect the previous conclusions( such as that the heat is zero, because the system remains isolated)

3. Why does this experiment shows internal energy depends only on temperature, how do I rule out any other dependence, provided I am able to have a list of possible dependences?

Can somebody shed some light?

Answer 1

Answer to Q1: If the piston is massless and frictionless, a force balance on the piston using Newton's 2nd law tells us that the force exerted by the gas on the piston must be essentially zero. But how can that be, since the gas must have "internal pressure." Well, in an irreversible expansion or compression, even an "ideal gas" does not satisfy the ideal gas law. The ideal gas law only applies at thermodynamic equilibrium. In an irreversible expansion, there are also viscous tensile stresses present which offset the "internal pressure" of the gas, and result in a net force of the gas on the piston of zero for free expansion. So the gas doesn't do any work.

Answer to Q2: You are very confused about how to determine the entropy change for an irreversible process. In determining the change in entropy for a system that has experienced an irreversible process, the first step is to totally forget about the irreversible process (entirely), and to focus, instead, on the initial and final thermodynamic equilibrium states only. The next step is to devise (i.e., dream up) an alternative reversible process between these exact same two end states. There are an infinite number of reversible process paths between these same two end states, and they all have the same entropy change. What you do is, for the specific reversible path that you have devised, calculate the integral of dq/T for that path. This will be the entropy change for both the reversible path and the irreversible path. When the write $dq_{rev}/T$, this is what they mean. For a cookbook primer on all this, including several worked examples, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Answer to Q3: The first law of thermodynamics tells us that $$\Delta U=Q-W$$If Q and W are both zero, then $\Delta U=0$. Generally, U is a function of T and V. But here, V has changed, and there is no change in U. So, for an ideal gas, U must be a function of T only.

ADDENDUM

And Why should I assume a force balance ? can't the piston accelerate as result of an internal unbalanced force and then stop , of course, when it's fully extended?

Just because you have completed a course in Freshman Physics does not mean that you are no longer allowed to apply what you have learned there, particularly when it can help you understand a new concept that you are struggling with.

Let's consider the case in which the piston has mass. If we do a force balance on the piston for this situation, we find that $$M_P\frac{d^2x}{dt^2}=F_g(t)$$where x is the displacement of the piston at time t after it is released and $F_G(t)$ is the force that the gas exerts on the piston at time t (so, in this case, the gas actually does exert a force on the piston). If we multiply this force balance by the velocity of the piston v = dx/dt, and integrate with respect to time, we obtain: $$W_G(t)=\int_0^t{F\frac{dx}{dt}dt}=M\frac{v^2}{2}=KE(t)$$where $W_G(t)$ is the work done by the gas on the piston up to time t and KE(t) is the kinetic energy of the piston at time t. The piston may bounce back off the closed end of the cylinder elastically, and partially recompress the gas as time progresses. But eventually, the kinetic energy of the piston will be dissipated by viscous forces in the gas. So, as you indicated, eventually the piston will come to rest at the closed end of the cylinder. So, at very long times, the net work that the gas does on the piston will be zero: $$W_G(\infty)=0$$

An alternate approach would be to treat the combination of piston plus gas as your system. This system certainly does no work on the vacuum, so, in this case, W = 0. Then, applying the first law of thermodynamics to this combined system gives: $$\Delta U_G+\Delta U_P=0$$where $U_P$ is the internal energy of the piston. Here again, we would have that, in the end, the piston is at rest. So, we obtain $$nC_v(T_f-T_i)+M_PC_P(T_f-T_i)=0$$or$$T_f=T_i$$$$\Delta U_G=0$$ and $$\Delta U_P=0$$

In Q1, what how does "not satisfying the ideal has law"(ie PV=nRT does not hold) traslates into not exerting a force on the piston?

If and when you study fluid mechanics, you will learn how viscous forces contribute to the force the gas exerts and how, for the case of a massless piston, the force the gas exerts on the piston can be zero. And, even for the case of a piston with mass, you will understand quantitatively on how the force the deforming gas exerts can differ from that predicted by the ideal gas law.

when they say the "piston is released", does it move while the gas expands or as, the picture suggests, someone pulls it all the way before the gas starts expanding(if this is possible at all)?

The picture doesn't accurately describe what happens. Nobody pulls the piston. With a massless piston, the vanishingly small force imbalance between the gas and the vacuum causes the piston to move. For a piston with mass, the finite force imbalance between the gas and the vacuum causes the piston to move (although, because of viscous effects, the force of the gas is less than predicted by the ideal gas law).