Non Quasi-static expansion in a piston cylinder

Question

Consider two identical piston cylinder arrangements, inside which we have the same gas. Sand grains are present over the piston in both the cases, and the initial state of the system (the gas) is same in both the cases.

I remove the sand grains in both the cases to obtain an expansion process. However, in the first arrangement the grains are removed slowly, one by one, and sufficient time is given for the system to adjust to uniform properties. In the second, a chunk of sand is removed instantly.

In both the cases the total amount of sand removed is same (so that what remains is also the same). If in the first arrangement the final pressure is $P_2$, temperature $T_2$ and volume $V_2$, will the state be the same even in the second arrangement, after infinite time? Even though intuitively I feel the final states after infinite time will be same, I don't have an explanation for it as to why?

Furthermore, how will the intermediate configurations in the second arrangement look like? In the sense that, will there be any oscillation of the piston?

Answer 1

If you do a force balance on the piston plus sand for the 2nd case (assuming vacuum pressure outside the cylinder), you get $$F_G(t)-mg=m\frac{dv}{dt}$$ where v is the velocity of the piston, m is the mass of piston plus sand throughout case 2, and $F_G(t)$ is the force exerted by the gas on the inside face of the piston at time t. At infinite time, the piston has stopped moving and the force the gas exerts on the piston face is equal to $$F_G(\infty)=mg=P_2A$$ Therefore, our force balance equation on the piston becomes: $$F_G(t)-P_2A=m\frac{dv}{dt}$$If we multiply this equation by the piston velocity $v=\frac{dx}{dt}$ (where x is the upward displacement of the piston up to time t), we obtain: $$F_G(t)\frac{dx}{dt}-P_2A\frac{dx}{dt}+mv\frac{dv}{dt}$$Integrating this equation between time zero and time t yields: $$W_G(t)=\int_0^t{F_G(t)\frac{dx}{dt}dt}=P_2(V(t)-V_0)+K(t)$$were K(t) is the kinetic energy of the piston at time t ($K(t)=m\frac{v^2(t)}{2}$) and $W_G(t)$ is the work done by the gas on the piston up to time t.

After an infinite amount of time, when the system has equilibrated and the oscillations of the piston have damped out, the total amount of work done by the gas on the piston will be $$W_{G}(\infty)=P_2(V_f-V_0)$$where $V_f(\gt V_2)$ is the final volume of gas in case 2 and $V_2$ is the final volume of gas in case 1.

From the first law of thermodynamics, it follows that $$\Delta U=nC_v(T_f-T_0)=-W_G(\infty)=-P_2(V_f-V_0)=-P_2\left(\frac{nRT_f}{P_2}-\frac{nRT_0}{P_0}\right)$$Solving this for the final temperature then gives: $$\frac{T_f}{T_0}=\frac{1}{\gamma}+\frac{P_2}{P_0}\frac{(\gamma-1)}{\gamma}$$

Answer 2

If in the first arrangement the final pressure is $P_2$, temperature $T_2$ and volume $V_2$, will the state be the same even in the second arrangement, after infinite time?

No.

You have indicated that the piston/cylinder is thermally insulated. As such, and assuming no friction between the piston and cylinder, the first process (one grain of sand at a time) is a reversible adiabatic process and the second is an irreversible adiabatic process. A reversible and irreversible adiabatic process cannot connect the same two equilibrium states. Fig 1 below assumes they do connect the same states 1 and 2.

The process in black is the reversible adiabatic expansion where the sand is removed one grain at a time. The process in red is the irreversible adiabatic expansion where the external pressure suddenly drops (removal of a chunk of sand equal to the amount removed grain by grain).

Both processes are governed by the first law

$$\Delta U=Q-W$$

Since both processes are adiabatic, $Q=0$ and

$$\Delta U=-W$$

Clearly, the work done by the reversible process (area under the black curve) is greater than the work done by the irreversible process (area under the red curve) and therefore the change in internal energy is different for the two processes (less for the irreversible process). But the change in internal energy has to be the same between the same two equilibrium states since it is a state function independent of the path. Therefore, the two processes cannot connect the same two equilibrium states as shown in the Fig 1.

So how does the final equilibrium state for the irreversible process differ?

Let's assume we are dealing with an ideal gas. For an ideal gas, any process,

$$\Delta U=C_{V}(T_{f}-T_{i})$$

Where $f$ and $i$ are the final and initial states. Therefore, for an ideal gas

$$C_{V}(T_{f}-T_{i})=-W$$

Since $W$ is less for the irreversible process, its decrease in internal energy is less making its final temperature greater than that of the reversible process, i.e., $T_{f}>T_{2}$. From the ideal gas law, for the irreversible process.

$$\frac{P_{1}V_{1}}{T_1}=\frac{P_{2}V_{f}}{T_{f}}$$

For the reversible process it is

$$\frac{P_{1}V_{1}}{T_1}=\frac{P_{2}V_{2}}{T_{2}}$$

Therefore

$$\frac{P_{2}V_{f}}{T_{f}}=\frac{P_{2}V_{2}}{T_{2}}$$

Which tells us that the volume to temperature ratio for the final equilibrium states is the same for the two processes. Given that $T_{f}>T_2$, then $V_{f}>V_2$.

The final equilibrium volume and temperature will be greater for the irreversible process. See FIG 2. @Chet Miller rigorous treatment in his answer gives the actual final volume and temperature for the irreversible process.

Hope this helps.