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We know from the First Principle that:

$$W = Q - \Delta U$$

Suppose that we know the initial and final temperature of an isobaric irreversible process; then we can deduct many things:

$$\Delta U = n c_V \Delta T$$

and for an isobaric process: $$Q = n c_p \Delta T$$

and can't we always deduct the work as: $$W = Q - \Delta U = n \left( c_p - c_V \right) \Delta T = n R \Delta T$$

?

Is my thought process correct? If indeed it is, then what's the difference in terms of efficiency between a reversible and an irreversible isobaric process that operate in the same range of temperatures? I thought that at least the work should have been different...

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An "irreversible isobaric process" is the term we use to describe a situation in which a gas is at an initial pressure $P_0$ and we suddenly drop (or raise) the force per unit area exerted on the gas by the inner face of the piston to $P_1\neq P_0$, and hold it at this value while the piston moves and the gas re-equilibrates. Under these circumstances, the work done is $$W=P_1(V_1-V_0)=P_1\left(\frac{nRT_1}{P_1}-\frac{nRT_0}{P_0}\right)=nR\left(T_1-\frac{P_1}{P_0}T_0\right)$$Note that, only at a single instant of time during this irreversible change in state was the gas really at its initial pressure $P_0$. Therefore, grammatically, it is not quite really correct to call this an isobaric irreversible process. However, we still do, and we know what we mean.

The change in internal energy, which depends only on the two end points is still $$\Delta U=nC_v(T_1-T_0)$$But, the heat added is now:$$Q=\Delta U+W=nC_p(T_1-T_0)+nR\left(1-\frac{P_1}{P_0}\right)T_0$$

If $P_1$ were equal to $P_0$ (and the system were merely receiving heating or cooling by the surroundings at the constant external force), then, even for irreversible heating or cooling, the equations you wrote for the work, heat, and change in internal energy would still be correct. This is one of those interesting cases where, irrespective of the path reversibility (provided the external force is constant), the heat, work, and changes in entropy and internal energy of the system would be the same. However, any entropy generated within the system in the process would be transferred from the system to the reservoirs that are being used to bring about the heating or cooling. So, the entropy of the combination of system and surroundings would increase for an irreversible heating or cooling path.

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  • $\begingroup$ I came across this when researching irreversible isobaric processes. While I agree with your answer, I wonder if the only irreversible isobaric process involves a sudden pressure drop. That example involves mechanical disequilibrium. How about thermal disequilibrium, such as due to a finite temperature difference between the surroundings and the system during the expansion? $\endgroup$ – Bob D Nov 29 '18 at 22:51
  • $\begingroup$ @BobD Yes, you can certainly do that too, plus, in that case, if the rate of heat transfer is high enough, the gas will deform very rapidly and create mechanical disequilibrium (even if the external pressure is held constant at the original gas pressure). In such a case, viscous stresses within the gas would make up the difference. $\endgroup$ – Chet Miller Nov 29 '18 at 23:39

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