# Adiabatic reversible and irreversible process

If I have 2 gases $$A$$ and $$B$$ which are at an initial state ($$P_1$$,$$V_1$$,$$T_1$$), and one gas say $$A$$ is taken through a reversible adiabatic process and the other gas $$B$$ through an irreversible adibatic process, such that their volume changes are the same - then how do I compare their final pressures, i.e., will the pressure of gas $$A$$ be more than that of $$B$$ or $$B$$ more than $$A$$?

The irreversible externally applied pressure does not have to be below the pressure for the reversible path over the entire process in order for the irreversible work to be less. The pressure for the irreversible process will typically start out lower (for example, you suddenly drop the externally imposed pressure to a new lower value and then hold it at that value), but the reversible process pressure later crosses over the irreversible process pressure at a certain volume and ends up below that of the irreversible process. So, the final temperature and pressure for the irreversible process will be higher. And the work done on the surroundings by the irreversible process will be lower.

In the irreversible process, it is typically not mechanical friction that causes the difference relative to the reversible process; it is viscous friction within the gas itself. This causes mechanical energy to dissipate to internal energy within the gas.

• could you please elaborate as to why the 2 curves will have to intersect each other May 18, 2018 at 13:16
• @physics123 Just do a sample problem in which, for the irreversible process, you suddenly drop $P_{ext}$ to a lower value and hold it constant at that value until the system re-equilibrates. Then compare this with what you get with a corresponding reversible process in which the final volume is the same as for the irreversible process. You will be able to see in detail how this plays out. May 18, 2018 at 13:40

In a reversible adiabatic expansion the gas does work against a piston, loses internal energy and therefore experiences a decrease in temperature. Therefore its pressure falls more than in an isothermal expansion between the same initial and final volumes. ($p=\frac{nRT}{V}$ in which not only $V$ increases, but $T$ decreases.)

One case of irreversible adiabatic expansion is when the piston moves out at a speed which is not negligible compared with the rms speeds of the molecules. In that case the local pressure next to the piston is less than in the bulk of the gas and less work will be done than in a reversible adiabatic expansion between the same two volumes. So the pressure won't drop as much as in the reversible case, but more than it does in the isothermal case (unless the piston moves so quickly as to leave a complete local vacuum behind it).

Note that the extreme case of an irreversible adiabatic expansion is when the gas expands into an additional insulated container, initially empty (evacuated), through a hole in a partition between the original container and the additional container. In that case no work at all is done and there is no change in internal energy. This is called 'Joule expansion'. For an ideal gas there would be no temperature change.

What's to be said about friction? Friction between the piston and the cylinder doesn't affect the amount of work (${p dV}$) that the gas itself does, (for a given gas pressure p and a given small volume increase, $dV$) though it reduces the amount of useful work available outside the cylinder. If gas and cylinder are insulated and the thermal capacity of the cylinder is assumed small, the work done against friction goes to increasing the internal energy of the gas (or, rather preventing it from falling as much as in the reversible case). In a nutshell, the friction prevents as much internal energy being lost from the system! So again the pressure won't fall as much as in the reversible case.

You can specify the state of the system by any pair of state variables, such as $$(p,V)$$ or $$(p,T)$$ but also other pairs such as $$(p,S)$$ and $$(U,S)$$ and things like that.

For present purposes let's specify the state by the variables $$(V,S)$$.

Gas A goes from $$(V_1, S_1)$$ to $$(V_2, S_1)$$. Here we argue that the final state has the same entropy as the initial state since the process is reversible and with no heat exchange.

Gas B goes from $$(V_1, S_1)$$ to $$(V_2, S)$$ where $$S$$ is not known. However we know that there is no heat exchange, so the only possibility is $$S > S_1.$$ Thus we have that gas B ends in a state with the same volume but higher entropy than gas A. It follows that the temperature of gas B will be higher, and therefore so will the pressure. To see why the temperature is higher, you must think about how the entropy and temperature are related. Consider the heat capacity at constant volume: $$C_V = T \left. \frac{\partial S}{\partial T} \right|_V$$ For a stable equilibrium this quantity must be positive, and we have positive temperature, so we have that $$S$$ is an increasing function of $$T$$ at constant volume, so higher $$S$$ does indeed imply higher $$T$$ (at given $$V$$).

Finally, we want to argue that higher temperature at a given volume must imply higher pressure. For an ideal gas this follows from the equation of state, but it is instructive to note that this result does not require that we assume the gas to be ideal. It would also be true for many other systems (though I guess it is not universally true).