Why do you think it's not possible to calculate the work done on or by the gas? Yes, it's rapid, but it sounds like you're theorizing that some physical law will not be applicable because the rate of change is large. What would the rate of change have to do with the applicability of thermodynamic laws?my problem is that here in this situation since the changes taking place inside the cylinder are very rapid... so we cannot calculate the work done by the gas inside the cylinder onto the piston..
The cylinder does necessarily oscillate for a while - but it is dampened oscillation. The energy of the compression wave in the cylinder gas ends up as heat in the gas. So you don't have to worry about the work done by the gas on the piston during the oscillations. You only have to determine the net work done by the piston on the gas. This is just a matter of applying the ideal gas law (and the first law of thermodynamics).i_island0 said:The actual question was:
Consider ‘n’ moles of an ideal monatomic gas placed in a vertical cylinder. The top of the cylinder is closed by a piston of mass M and cross section A. Initially the piston is fixed, and the gas has volume Vo and temperature To. Next, the piston is released, and after several oscillations comes to a stop. Disregarding friction and the heat capacity of the piston and cylinder, find the temperature and volume of the gas at equilibrium. The system is thermally isolated, and the pressure outside the cylinder is P.
I am not interested in the solution to the question... my problem is that here in this situation since the changes taking place inside the cylinder are very rapid... so we cannot calculate the work done by the gas inside the cylinder onto the piston.. but we can definitely calculate the work done by Mg and PA on the gas inside the cylinder.
So can we conclude that the work done by Mg and PA is numerically equal to work done by gas inside the cylinder.
You can figure it out yourself. W = PdV. For the work done on the atmosphere, you use the pressure of the atmosphere x volume change. For work done by the cylinder gas, use the pressure of the cylinder x volume change.i_island0 said:I want to know if in rapid adiabatic compression of a gas, is work done by gas on surrounding (atmosphere) numerically same as work done by surrounding on the gas.
Please tell if there exists some solid proof of the answers you guys are giving.
There some important subtlties here.russ_watters said:Yes...so I'm not seeing the problem either. If instead of the atmosphere, you just put a big weight on top of the piston, the resulting scenario is the same.
Could you explain why you think the word "rapid" makes a difference here? I'm guessing you think it has something to do with kinetic energy, but as you and others pointed out, the initial KE is zero and the final KE is zero and you aren't given any information about viscous damping (if a weight is placed on the piston, there need not even be damping), so there is no calculable effect.
If the question you are asking is about the effect of that damping, then it still depends on things like insulation, which will determine if the heat eventually goes into the cylinder (keeping it closer to adiabatic) or into the atmosphere (making it eventually isothermal).
Andrew Mason said:There some important subtlties here.
If you expand the gas rapidly more energy goes into the kinetic energy of the gas and piston (kinetic energy is proportional to v^2). That kinetic energy is reduced if the mass of the piston is increased. But it does not disappear. The piston's kinetic energy allows the gas to expand past the point at which its pressure is equal to the outside pressure (when the pressures are equal, the piston is still moving). The piston stops and starts going back the other way because the outside pressure is now greater than the inside gas pressure. It gains kinetic energy in the other direction. Repeat compressions and expansions (oscillations) occur like this. In each oscillation expansion and compression, the gas heats up. Eventually all the kinetic energy is converted to thermal energy of the gas (assuming no friction).
If you stop the piston system at its initial maximum extension (no kinetic energy), the change in internal energy of the gas is equal to -the work done by the gas ([itex]\Delta U = - P\Delta V[/itex] - first law). The cylinder is thermally insulated so no heat has flowed out of the gas. The result is that more work has been done in the rapid non-reversible expansion than would have been done in the reversible expansion.
If the process is adiabatic, it means that heat does not flow into or out of the cylinder. But energy can flow into or out of the gas and that energy can be converted to or draw out thermal energy from the gas. This is why I have maintained that we really should not talk about an adiabatic non-reversible process (but I appear to be in a minority on that).
AM
i_island0 said:I want to know if in rapid adiabatic compression of a gas, is work done by gas on surrounding (atmosphere) numerically same as work done by surrounding on the gas.
Please tell if there exists some solid proof of the answers you guys are giving.