
#1
Mar3108, 02:52 PM

P: 23

work done in an reversible process is easy to find we have the function of P internal bt when it comes to irreversible process then how do we find work done by the gas or on it and if ur answer is (sry i am new to forum i dont know how to use the features) [ integration p external dv] then plz explain why is it so?




#2
Nov108, 11:09 AM

P: 4

the same question from me.... :(




#3
Nov108, 01:06 PM

P: 1,672

The work done by a reversible process and irreversible process are the same. Work done is just the work done, entropy isn't going to change that.




#4
Nov108, 06:52 PM

P: 2,159

work in irreversible process?
What matters is if the process is quasistatic or not. Quasistatic means that the process proceeds slow enough for internal thermodynamic equilibrium to be maintained.
If the irreversible process is quasistatic then the work done is given by the integral of P dV. Consider e.g. an ideal gas that is expanding slowly while kept at constant temperature. The gas will do an equal amount of work as the heat it absorbs. Quasistatic irreversible processes can always be realized in a completely reversible way. E.g. in the above case where the entropy of the gas increases due to absorbing heat, the heat can be supplied to the gas by a heat bath and the temperature difference between the heat bath and the gas can be made arbitrarily small. The entropy increase of the heat bath plus the entropy increase of the heat bath is then zero. The entire process is then reversible. In case of nonquasistatic processes you can no longer apply the usual thermodynamic equations. Work done is then no longer given by the integral of P dV. In general you cannot even define the pressure for nonequilibrium processes in unambiguous way. But even if the pressure is well defined and the integral of P dV is well defined, it is not the same as the work done. To see why this is the case, consider the first law of thermodynamics: dE = dQ  dW (1) which expresses conservation of energy: The change in internal energy is the absorbed heat minus the work done. In case of a quasistatic change of the system, we know that: dQ = T dS (2) and dW = P dV (3) Inserting (2) and (3) in (1) gives: dE = T dS  P dV (4) which is known as the fundamental thermodynamic relation. Now, even though this was derived assuming the change in the system was quasistatic, it is also valid for nonquasistatic changes. The reason is that the internal energy E is a thermodynamic state variable which can be uniquely specified by the entropy and volume. Then, if you change the volume and entropy, the internal energy change is some fixed quantity, independent of that change is realized (quasistatic or not). But for nonquasistatic changes we know that (2) is not valid. In general we have: T dS >= dQ (5) Then since (4) must be valid no matter what, we have: P dV = T dS  dE Using (1) we get: P dV = T dS  dQ + dW Eq. (5) then implies: P dV >= dW (6) So, the work done in irreversible processes will, in general, be less than the integral of P dV. You have equality in the quasistatic case, even if the process is irreversible (but such processes can always be realized in a reversible way). In some books reversible volume pressure work is defined to be any process in which the equality (3) holds. This can be justified because if (3) holds and you plug that in Eq. (4) , then you obtain (using Eq.(1)) Eq (2), which can always be realized in a reversible way. Loosely speaking you can say that P dV is the work done but part of it can get lost due to internal friction. That internal friction can be considered to be heat (it will certainly lead to entropy increase). But because we define heat as energy that flows to the system from outside the system boundary, it is not the accounted for by the term dQ in the equations. So, what you have is that part of the P dV is dissipated in the form of heat which then casuses the entropy change to be more than dQ/T, while the work that the system performs is less than P dV, because work done is energy that flows out of the system in the form of work performed on an external system and the part of the P dV that has been dissipated is no longer available for this. An exteme example is free expansion of a gas in avacuum. In that case no work at all is performed, but this process is so violent that you can object by saying that the pressure of the gas is not well defined. But we can perform a free expansion in small steps by moving a piston very fast from one position to a slighly different position, faster than the gas can expand and then fixing the piston in that position. Then what happens is that the gas expands and bumps into the piston in that new position. The P dV term is then equal to the increase in kinetic energy of the gas which comes at the expense of the internal energy, but this kinetic energy gets dissipated after a while, so the internal energy stays the same. 


Register to reply 
Related Discussions  
work done in irreversible adiabatic process  Classical Physics  2  
Entropy of an irreversible process  General Physics  11  
Irreversible process  Introductory Physics Homework  5  
Irreversible process  Introductory Physics Homework  0  
delta G of irreversible process  General Physics  1 