# Isobaric process with non-boundary work?

I came up with a doubt on isobaric processes.

Firstly, in any isobaric process, (irreversible or not) the following hold $$Q=n c_p (T_B-T_A)\tag{1}$$ $$W=p (V_B-V_A)=n R(T_B-T_A)\tag{2}$$ Is that correct? If so consider the following situation.

A gas in a tank can expand at constant pressure $p$. When it expand it does an amount of work $p \Delta V$. Besides it, in the tank there is a fan that does work on the gas delivering a power $P$.

In total the amount of work is $$W=p\Delta V-P \cdot t$$

So in this case $(2)$ is not complete, but what about $(1)$? Is $(1)$ still valid?

• If the irreversible process involves a sudden drop or increase in external pressure to a new constant value at time zero (according to customary convention, for an irreversible process, this is regarded as isobaric), Eqn. 1 that you wrote for the work is incorrect. – Chet Miller Jul 11 '16 at 19:09
• @ChesterMiller Thanks for the reply! I found $(1)$ and $(2)$ on textbook as valid also for any isobaric process. Are you saying that $(1)$ and $(2)$ are valid only in the case of reversible isobaric process? – Sørën Jul 12 '16 at 7:39
• Eqn. 2 is valid only if the p in the equation is the external pressure $p_{ext}$ applied to the gas (which in both reversible and irreversible processes is equal to the force per unit area that the gas applies at the boundary). But in an irreversible expansion or compression, p cannot be calculated from the equation of state (e.g., the ideal gas law) because (1) the pressure and/or the temperature are not uniform in the system and (2) viscous stresses contribute to the force per unit area applied at the boundary. (Continued in next comment). – Chet Miller Jul 12 '16 at 13:33
• Also, by an irreversible isobaric expansion, we mean that, at time zero, the external force per unit area is suddenly dropped to a new value and subsequently held at that value. So there is a jump change in pressure at time zero. In the irreversible case, there is no jump change at time zero. – Chet Miller Jul 12 '16 at 13:37

if there is a fan inside the tank and if the tank's volume is kept constant, what will be happen? The air will circulate and viscosity will convert kinetic energy to thermal energy, i.e. temperature will increase. The electrical energy input will be eventually converted to thermal energy which is $\delta Q$. So $W=pdV-Pt$ is not correct. The equations are, $$Q+Pt=nc_p(T_B-T_A)$$ $$W=p(V_B-V_A)$$
• Thanks for the answer, the volume is not kept constant (otherwise it would be both isochoric and isobaric, which means no transformation at all). Anyway the work of the gas is against the fan so I think there should be a minus in front of $P t$ if that is the work done by the gas. – Sørën Jul 12 '16 at 7:42