# P-V Work – The First Law of Thermodynamics

####
**Work**

** Work in
thermodynamics is defined as in classical mechanics.

** When part of
the surroundings exerts a macroscopically measurable force F on
matter in the system while this matter moves a distance dx at the point of
application of F, then the surroundings has done work on the system:

**dw = F**

_{x}dx (1)
** where F

_{x}is the component of F in the direction of the displacement.
** F may be
a mechanical, electrical, or magnetic force and may act on and displace the
entire system or only a part of the system.

** When F

_{x }and the displacement dx are in the same direction, positive work is done on the system: dw > 0.
** When F

_{x}and dx are in opposite directions, dw is negative: dw < 0####
**Reversible P-V Work**

**(a)**The most common way work is done on a thermodynamic system is by a change in the system’s volume.

** Consider the
system of Fig(1):

** The system
consists of the matter contained within the piston and cylinder walls and has
pressure P.

** Let the
external pressure on the frictionless piston also be P. Equal opposing forces
act on the piston, and it is in mechanical equilibrium.

** Let x denote
the piston’s location.

** If the
external pressure on the piston is now increased by an infinitesimal amount,
this increase will produce an infinitesimal imbalance in forces on the piston.
The piston will move inward by an infinitesimal distance dx, thereby decreasing
the system’s volume and increasing its pressure until the system pressure again
balances the external pressure.

** During this
infinitesimal process, which occurs at an infinitesimal rate, the system will
be infinitesimally close to equilibrium. The piston, which is part of the
surroundings, exerted a force, which we denote by F

_{x}, on matter in the system at the system–piston boundary while this matter moved a distance dx.
** The
surroundings therefore did work dw = F

_{x}dx on the system.
** Let F be the
magnitude of the force exerted by the system on the piston. Newton’s third law
(action = reaction) gives F = F

_{x}.
** The
definition P = F/A of the system’s pressure P gives F

_{x}= F = PA, where A is the piston’s cross-sectional area.
** Therefore the
work dw = F

_{x}dx done on the system in Fig (1) is:

**dw = PA dx (2)**

** The system has
cross-sectional area A and length l = b - x (Fig 1), where x is the piston’s
position and b is the position of the fixed end of the system.

** The volume of this cylindrical system is V = Al = Ab - Ax. The change in system
volume when the piston moves by dx is dV = d(Ab - Ax) = - A dx. Equation (2)
becomes:

**dw**

_{rev}= - P dV (3)*closed system, reversible process*

The
subscript(rev) stands for reversible.

**(b)**The meaning of “reversible” will be discussed shortly.We implicitly assumed a closed system in deriving (equation 3). When matter is transported between system and surroundings, the meaning of work becomes ambiguous; we shall not consider this case. We derived (equation 3) for a particular shape of system, but it can be shown to be valid for every system shape .**(c)**We derived (3) by considering a contraction of the system’s volume (dV < 0). For an expansion (dV > 0), the piston moves outward (in the negative x direction), and the displacement dx of the matter at the system–piston boundary is negative (dx < 0). Since F

_{x}is positive (the force exerted by the piston on the system is in the positive x direction), the work dw = F

_{x}dx done on the system by the surroundings is negative when the system expands. For an expansion, the system’s volume change is still given by dV = - A dx (where dx < 0 and dV > 0), and ( equation 3) still holds.

**(d)**In a contraction, the work done on the system is positive (dw > 0). In an expansion, the work done on the system is negative (dw < 0). (In an expansion, the work done on the surroundings is positive.)

**(e)**So far we have considered only an infinitesimal volume change. Suppose we carry out an infinite number of successive infinitesimal changes in the external pressure. At each such change, the system’s volume changes by dV and work - P dV is done on the system, where P is the current value of the system’s pressure. The total work w done on the system is the sum of the infinitesimal amounts of work, and this sum of infinitesimal quantities is the following definite integral:

**(f)**The finite volume change to which (equation 4) applies consists of an infinite number of infinitesimal steps and takes an infinite amount of time to carry out. In this process, the difference between the pressures on the two sides of the piston is always infinitesimally small, so finite unbalanced forces never act and the system remains infinitesimally close to equilibrium throughout the process. Moreover, the process can be reversed at any stage by an infinitesimal change in conditions, namely, by infinitesimally changing the external pressure. Reversal of the process will restore both system and surroundings to their initial conditions.

###
**Summary**

** A reversible process is one where the system is
always infinitesimally close to equilibrium, and an infinitesimal change in
conditions can reverse the process to restore both system and surroundings to
their initial states. A reversible process is obviously an idealization.

** Equations (3) and (4) apply only to reversible
expansions and contractions. More precisely, they apply to mechanically
reversible volume changes. There could be a chemically irreversible process,
such as a chemical reaction, occurring in the system during the expansion, but
so long as the mechanical forces are only infinitesimally unbalanced, (3) and
(4) apply.

** The work (equation 4) done in a volume change
is called P-V work. Later on, we shall deal with electrical work and work of
changing the system’s surface area, but for now, only systems with P-V work
will be considered.

** We have defined the symbol w to stand for work
done

*on*the system by the surroundings. Some texts use w to mean work done*by*the system on its surroundings. Their w is the negative of ours.**Example: Find the work w**

_{rev}for processes (a) and (b) of Fig. blow if P_{1}= 3.00 atm, V_{1}= 500 cm^{3}, P_{2}= 1.00 atm, and V_{2}= 2000 cm^{3}. Also, find wrev for the reverse of process (a).

**Solution:**

####
**Irreversible P-V Work**

** The work

*w*in a mechanically irreversible volume change sometimes cannot be calculated with thermodynamics.
** For example, suppose the external pressure on
the piston in Fig.(1) is suddenly reduced by a finite amount and is held fixed
thereafter. The inner pressure on the piston is then greater than the outer
pressure by a finite amount, and the piston is accelerated outward. This
initial acceleration of the piston away from the system will destroy the
uniform pressure in the enclosed gas. The system’s pressure will be lower near
the piston than farther away from it. Moreover, the piston’s acceleration
produces turbulence in the gas. Thus we cannot give a thermodynamic description
of the state of the system.

** We have dw = F

_{x}dx. For P-V work, F_{x}is the force at the system–surroundings boundary, which is where the displacement dx is occurring. This boundary is the inner face of the piston, so dw_{irrev }= - P_{surf}dV, where P_{surf}is the pressure the system exerts on the inner face of the piston. (By Newton’s third law, P_{surf}is also the pressure the piston’s inner face exerts on the system.)
** Because we cannot use thermodynamics to
calculate P

_{surf}during the turbulent, irreversible expansion, we cannot find dw_{irrev}from thermodynamics. The law of conservation of energy can be used to show that, for a frictionless piston ,

**dw**

_{irrev}= - P_{ext}dV - dK_{pist}(5)P

_{ext}is the external pressure on the outer face of the piston

dK

_{pist}is the infinitesimal change in piston kinetic energy.
** The integrated form of (equation 5) is :

** If we wait long enough, the piston’s kinetic
energy will be dissipated by the internal friction (viscosity) in the gas. The
gas will be heated, and the piston will eventually come to rest (perhaps after
undergoing oscillations). Once the piston has come to rest, we have ΔK

_{pist}= 0 - 0 = 0, since the piston started and ended at rest. We then have:
** Hence we can find w

_{irrev}after the piston has come to rest. If, however, part of the piston’s kinetic energy is transferred to some other body in the surroundings before the piston comes to rest, then thermodynamics cannot calculate the work exchanged between system and surroundings.###
**Summary**

** For now, we
shall deal only with work done due to a volume change.

** The work done
on a closed system in an infinitesimal mechanically reversible process is:

**dw**

_{rev}= -P dV
** The work:

depends on
the path (the process) used to go from the initial state 1 to the final state
2.

**Reference:***physical Chemistry /Ira N. Levine / University of New York /6th .ed./ 2002 .*
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