B Adiabatic or Isobaric process?

[Chestermiller]

To address this, I'm going to take off my Member hat and put on my Physics Forums Mentor hat. Part of my responsibility as a Physics Forums Mentor is to make sure that the correct information is made available to the members, so that neophytes to a subject are not led astray. I don't mean to be rude or harsh, but I have to be frank with you. You seem very very off base on your understanding of thermodynamics, and I recommend for you a thorough review of the subject (at least the first and 2nd laws). The gaps in your understanding are too broad to cover within the format of a Physics Forums thread. So, after I respond to this post, I'm going to close the present thread.

I agree again Sir, But Cp(T2-T1), is it not a reversible heat transfer?

Generally, this equation does not describe heat transfer at all. It describes the change in enthalpy per mole $\Delta h$ of an ideal gas between two thermodynamic equilibrium states. For an irreversible process, it applies only to the two end states. For a reversible process, since all the intermediate states are also thermodynamic equilibrium states, it applies to all intermediate states as well

Or you say it is irreversible heat transfer that will take place in a real process?

Irreversible heat transfer will take place in a real process.

It is reversible heat transfer through reservoirs of infinite thermal capacity at temperatures increasing steadily from T1 to T2 in infinite infinitesimal steps.Then using the fact that internal energy is a state function this heat transfer must be equal to the work done in the actual irreversible process. What I wish to state Sir is whenever we use integration to calculate either work done by the system or heat supplied to the system we make use of the concept of a reversible process only.

In the process being discussed in this thread, the system is adiabatic, so there is no heat transfer taking place at all (that is the definition of adiabatic). And there are no reservoirs, and the zero heat transfer is not equal to the finite amount of work done.

 

[Chestermiller]

I might have a way of solving the original problem via the entropy, but I'm not sure it's right. Anyway, the idea is this: The entropy of the surroundings is constant. If we imagine a reversible process between the same initial and final state, the entropy of the system is also constant (This might be be the problem.), since the total entropy has to be constant. Now we can use the Sackur-Tetrode equation to write down the entropy of the system for initial and final state and equate them. Then we also have the ideal gas law to eliminate the temperature in initial and final state. The result I get is V₂ p₂³ = V₁ p₁³. Does that make sense?

This is an adiabatic irreversible process, so entropy is generated within the system, and the change in entropy of the system is positive. Therefore, any reversible process between the same two end states must involve heat transfer to the system. It is therefore impossible to devise an adiabatic reversible path between the two end states of an adiabatic irreversible process.