User:PAR/Work7

The essential point that Smith makes is the condition on which the mass added to a system does no work and adds no heat to that system.

The no-work no-heat (NWNH) scenario

This section gives a scenario where the only increase in internal energy of a system is due to the addition of an infinitesimal amount of mass.

Consider an infinitesimally small system (the small system) connected by a mass-permeable wall to a macroscopic system (the large system), the composite system surrounded by an isolating enclosure. Both systems are composed of the same single component, both are at the same temperature, pressure, density, etc.

The small system has infinitesimal extensive variables of energy, entropy, volume, and mole fraction $\Delta U,\Delta S,\Delta V,\Delta N$. (Again, the $\Delta 's$ are infinitesimals, not finite differences.) The large system has extensive variables $U,S,V,N$. Both systems have the same intensive variables $u=\Delta U/\Delta N=U/N,\,s=\Delta S/\Delta N=S/N,\,v=\Delta V/\Delta N=V/N$ and the same temperature T and pressure P.

The small system, is reversibly compressed until its volume is zero and all of its mass has been injected into the large system. This work adds energy to the large system in the amount of $-P\,\Delta V$. Now extract work energy by allowing the system to reversibly expand, regaining the lost volume volume $\Delta V$. Energy in the amount of $P\,\Delta V$ is extracted, the net work done is now zero. The net effect is the same as if the mass-permeable wall were simply removed and the large system was redefined to include the small system, obviously requiring no work and adding no heat. It is also equivalent to reversibly compressing and expanding the composite system by volume $\Delta V$, again requiring no work and adding no heat.

My analysis of the NWNW scenario

Both Smith and Alvarez agree that $dU=dQ+dW+dR$ where $dR$ is the energy added due to the addition of mass. The Euler equation for the small system is $\Delta U=T\Delta S-P\Delta V+\mu \Delta N$ and for the large system $U=TS-PV+\mu N$. In terms of mass specific quantities, the Euler equation is the same for both systems $u=Ts-Pv+\mu$. In terms of densities ($u'=U/V$, etc.), the Euler equation is the same for both systems $u'=Ts'-P+\mu n'$. The fundamental equation for the large system in terms of mass specific quantities is $du=Tds-Pdv$ and in terms of densities, $du=Tds+\mu dn$. Changes in the large system are denoted by $dU,\,dS,\,dV,\,dN$, etc.

In general, when both systems have the same intensive parameters, and since the total change in mass of the large system is $dN=\Delta N$

$dR=\Delta U=u\,dN$

$dW=-P(dV-\Delta V)=-P(dV-v\,dN)$

In other words, the reversible mechanical work done on the large system is the negative of the pressure times that portion of the increase in volume not associated with the added mass $dV-\Delta V$. The heat is therefore

$dQ=dU-dW-dR=(dU-u\,dN)+P(dV-v\,dN)$

In the NWNH scenario, the total changes are only those due to the added mass. $dU$ is just the energy added by the mass and is equal to $u\Delta N=u\,dN$ so the first term on the right is zero. Likewise, the change in volume $dV$ is just the volume of the added mass and is equal to $dV=v\,\Delta N=v\,dN$ so the second term on the right (the work) is zero. It follows that the energy added as heat is zero. This definition of the energy added by the mass $dR=\Delta U=u\,\Delta N$ seems intuitively obvious, and the definition of work as the negative of pressure times that portion of the increase in volume not associated with the added mass $dV-\Delta V$ seems also intuitively obvious.

Using the Euler equation for the large system, the heat may also be expressed as:

$dQ=dU+PdV-(Ts+\mu )dN$

Alvarez' analysis of the NWNW scenario

According to Alvarez

$dR=(\mu +Ts)dN\,\,\,\,\,\mathrm {(eq.23)}$

Using the Euler equation this becomes:

$dR=(u+Pv)dN$

rather than $dR=u\,dN$. The work is:

$dW=-P\,dV$

rather than $dW=-P(dV-v\,dN)$ and therefore

$dQ=dU+PdV-(\mu +Ts)dN$

which agrees with the above result for heat, but the work (dW) and the energy of the added mass (dR) do not agree. Alvarez states that the work done in the NWNH scenario is not zero but rather $-P\,dV_{m}$, which does not seem correct.