Work done in irreversible adiabatic process

In summary, the conversation discusses a question about an ideal monoatomic gas in a vertical cylinder with a piston on top. The piston is initially fixed and the gas has a specific volume and temperature. After being released, the piston undergoes oscillations and eventually reaches equilibrium. The question asks about the work done by an external agent, and the answer is that it can be considered the same as the work done by the gas inside the cylinder. The conversation also mentions doubts about how the piston can come to a stop without any friction.
  • #1
i_island0
123
0
This question came some time back in Stony Brooks exam. I have the solution also. But i am having some conceptual doubts. Can someone help me please.

The question is:
Consider ‘n’ moles of an ideal monoatomic 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 (due o gravity) 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.

Here i just want to know if the work done by the external agent.. (i.e. Mg & P.A).. can be taken same as work done by the gas inside the cylinder.
 
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  • #2
as this is reversible adiabatic process the work done on the gas by the surrounding is the work done by Mg and by atmospheric pressure.
 
  • #3
yes i think tht bt wht i did not get is how does the piston stop if there is no friction?
 

1. What is an irreversible adiabatic process?

An irreversible adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings and the system cannot return to its initial state. This means that the process is both non-adiabatic and non-reversible, resulting in a decrease in the system's entropy.

2. How is work done in an irreversible adiabatic process?

In an irreversible adiabatic process, work is done by the system on its surroundings. This is due to the fact that the system is unable to exchange heat with its surroundings, so all the energy must be converted into work. This work is typically in the form of expansion or compression of a gas.

3. What is the equation for calculating work done in an irreversible adiabatic process?

The equation for work done in an irreversible adiabatic process is W = PΔV, where W is the work done, P is the pressure, and ΔV is the change in volume of the system. This equation assumes that the process is a constant pressure process and the gas behaves ideally.

4. How does work done in an irreversible adiabatic process differ from that in a reversible adiabatic process?

In a reversible adiabatic process, the system is able to return to its initial state and no energy is lost due to friction or other non-adiabatic effects. Therefore, the work done in a reversible adiabatic process is equal to the change in internal energy (ΔU) of the system. In an irreversible adiabatic process, however, some energy is lost as the system cannot return to its initial state, resulting in a decrease in the system's entropy.

5. What are some real-life examples of irreversible adiabatic processes?

One example of an irreversible adiabatic process is the expansion of a gas through a nozzle. The gas expands rapidly and does work on its surroundings, but due to the high speed of the expansion, the process is irreversible. Another example is the compression of a gas in a diesel engine, where the compression is fast and adiabatic, resulting in a significant increase in the temperature of the gas.

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