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?
Best Answer
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: $$\frac{T_f}{T_0}=\frac{1}{\gamma}+\frac{P_2}{P_0}\frac{(\gamma-1)}{\gamma}$$