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Suppose we have a heat insulated and rigid container with a frictionless piston dividing the container into two parts, the left part being filled with some ideal gas(say mono-atomic), and the right one is vacuum with a spring connected between the right wall and the piston. How do we decide what process the system undergoes now? Will it be a polytropic process? A quastistatic one? Both? Neither?

I think it'll be quasi-static but not polytropic, and as a consequence, the piston's net force will always be zero, but I don't have any explanation for it. How do I explain that to someone?

Also, is it possible to find the heat capacity of this system, considering the heat capacities of the piston, spring etc. negligible for these n moles of ideal gas? How?

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It won't be polytropic or even quasi-static. The gas will suffer an irreversible expansion, and the final state of the system can be determined.

Assuming that the spring is initially at its unextended length, the force exerted by the spring on the piston during the process will be $-k(x-x_0)$, where x is the location relative to the left wall of the cylinder, and $x_0$ is the its original location. But, if A represents the area of the piston, then $-k(x-x_0)=-\frac{k}{A}(V-V_0)$ where V is the gas volume at time t, and $V_0$ is the initial gas volume. So, if we apply Newton's 2nd law of motion to derive a force balance on the piston at any time t, we obtain: $$F_g-\frac{k}{A}(V-V_0)=m\frac{d^2x}{dt^2}=\frac{m}{A}\frac{d^2V}{dt^2}$$where $F_g$ is the force exerted by the gas on the piston at time t and m is the mass of the piston. If we multiply this equation by dV/dt and integrate with respect to time, we can obtain the work done by the gas on the piston (its surroundings) up to time t: $$W_g=\frac{k}{2A}(V-V_0)^2+\frac{m}{2A}\left(\frac{dV}{dt}\right)^2$$In this equation, the first term on the right hand side represents the elastic energy stored in the spring, and the second term represents the kinetic energy of the piston. Even though the piston may oscillate back and forth during this process, eventually, because of viscous damping forces within the gas, the piston will come to rest, and the system will be in its final equilibrium state. At that point, the work done by the gas will be given by:$$W_g=\frac{k}{2A}(V_f-V_0)^2$$where $V_f$ is the final equilibrium gas volume. The final pressure of the gas will then be given by: $$P_f=\frac{F_{gf}}{A}=\frac{k}{A^2}(V_f-V_0)\tag{1}$$ Next, applying the first law of thermodynamics to the system, we have $$\Delta U=nC_v(T_f-T_0)=-W_{gf}=-\frac{k}{2A}(V_f-V_0)^2$$But, from the ideal gas law, it follows that $$nC_v(T_f-T_0)=\frac{C_v}{R}(P_fV_f-P_0V_0)$$Therefore, we have: $$\frac{C_v}{R}(P_fV_f-P_0V_0)=-\frac{k}{2A}(V_f-V_0)^2\tag{2}$$Eqn. 2 can be combined with Eqn. 1 to provide a single quadratic equation for calculating the final volume $V_f$ (and then the final temperature and final pressure).

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  • $\begingroup$ Are you sure? What if the total length of the container is the initial (and the relaxed) length of the spring? I think the net force on piston is always zero. In your answer, can you also show how to compute the specific heat of the gas in the process it undergoes? $\endgroup$
    – kushal
    Commented Jan 8, 2020 at 14:36
  • $\begingroup$ Of course I’m sure. The specific heat of the gas is assumed to be known, either from knowing the degrees of freedom of the gas molecules or from a separate experiment. Specific heat is a physical property of the gas, independent if any process. $\endgroup$ Commented Jan 8, 2020 at 15:21
  • $\begingroup$ I'm not talking about finding the specific heat of the gas at constant volume, which is $\frac{dQ}{ndT} = \frac{dU}{ndT} = C_v$. I'm talking about the specific heat of his particular process. $\endgroup$
    – kushal
    Commented Jan 9, 2020 at 12:12
  • $\begingroup$ If the container is insulated (as you specified), Q=0. $\endgroup$ Commented Jan 9, 2020 at 12:32

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