Ideal Gas in Cylinder with Piston (Conceptual)

[fiestyman001]

You have a cylinder with an ideal gas inside, enclosed by a piston. There is a valve on the side of the cylinder could let gas through if opened. The valve is then opened and some of the gas escapes slowly while maintaining thermal equilibrium. Once it is closed, the piston is a little lower than it was before. Is the final pressure of the gas inside the cylinder greater, less than, or equal to the initial pressure?

I know n, and V change, but I'm not sure if P changes..

 

[h2oski1326]

Let me ask you this, what is causing the pressure in the cylinder?

Try a force balance on the piston, see if that answers your question.

 

[perucho]

You have a cylinder with an ideal gas inside, enclosed by a piston. There is a valve on the side of the cylinder could let gas through if opened. The valve is then opened and some of the gas escapes slowly while maintaining thermal equilibrium. Once it is closed, the piston is a little lower than it was before. Is the final pressure of the gas inside the cylinder greater, less than, or equal to the initial pressure?

I know n, and V change, but I'm not sure if P changes..

h2oski1326 observation is essential. Notice that the external forces exerted over the piston are its proper weight and the force due to atmospheric pressure. So if we assume a quasistatic process, then the internal pressure is equilibrated by the external one all the time during the process. Also to mantain thermal equilibrium (I assume, in advance, that the process is non-adiabatic as you are talking about of thermal equilibrium, but if the process were adiabatic then it is easy to show that it must be isentropic) some heat, in principle, should be transferred between the open system (control volume) and its surroundings. Furthermore, since T=const. and we are dealing with an ideal gas, the exit enthalpy may be asummed constant and equals to the system, that is h_e=C_p.T=const. But what n is? as in an isothermal process the polytropic constant is equal to 1. But I think n is the mole's number. So PV_1=m_1.R.T, and PV_2=m_2.R.T, where m_1 and m_2 are the initial and final mass resp. inside the control volume. Also by continuity eq. m_e=m_1-m_2. So by the first law of thermodynamics applied to a 'transitory' control volume
Q=m_e.h_e+m_2.u2-m_1.u_1+W,
where u_1=u_2=C_v.T, h_e=C_p.T. and W=-P(V_1-V_2), negative because work is efectued on the system by the piston.
Therefore,
Q=(m_1-m_2)C_p.T-(m_1-m_2)C_v.T-P(V_1-V_2)=(m_1-m_2)(C_p-C_v)T-P(V_1-V_2),
and we get
Q=(m_1-m_2)RT-P(V_1-V_2)=P(V_1-V_2)-P(V_1-V_2),
that is,
Q=0.
So, according to this result (and if I didn't commit mistakes), the process is also adiabatic in this very particular case. How much it is the entropy change?