Concept of 'Work' in Thermodynamics

 

The concepts of heat and work are of fundamental importance in thermodynamics. Work and heat are algebraic quantities and can be positive or negative. They appear only during a change in the state of the system and appear only at the boundary of the system. They are manifestations of energy.

Work

Work is a basic mode of energy transfer. An object is displaced through a distance dx against a force F(x) then the amount of work which has to be done is defined as

W = -F(x)dx

In SI System of units, work is expressed in joule or kilojoule; 1J = 1 Nm. When work is done on the system W and when work is done by the system W is negative. There are many types of work and all of them could be expressed as the product of two factors; (i) intensity, and (ii) capacity. Some of these types are:

Gravitational work: Gravitational work is said to be done when a body is raised through a certain height against the gravitational field. If a body of mass m is raised through a height h against the gravitational field g, the magnitude of gravitational work is mgh. In this expression mg, the force required to overcome the gravity, is intensity factor and height h is the capacity factor.

Gravitational work

GW = mg x h =mgh

Electrical work: Electric work is said to be done when a charge body is moved from one potential region to another. If the charge is expressed in coulombs and the potential difference in volts, then the electrical work is given by QV. Here potential difference (V) is the intensity factor and the quantity of electricity Q is the capacity factor.

Mechanical work: Work associated with the change in volume of a system against an external pressure is referred to as the mechanical or pressure-volume work. The magnitude of this work in an isothermal expansion of a gas can be obtained as follows. Consider a certain quantity of a gas enclosed in a cylinder fitted with a weightless and frictionless piston. The gas is held in position by a constant external pressure, P_ext. The force acting on the piston is given by the product of the external pressure and the cross sectional area of the piston (P_ext.A), Where A is its area of cross section. The cylinder is immersed in a thermostat to ensure constant temperature during the expansion of the gas. The initial state of the system is described by P₁, V₁ and T. When external pressure is reduced, the gas tends to expand. During expansion, the Piston moves to a new position where the variables of the system are P₂, V₂ and T. Let the height up to which the Piston has moved be h. The work done by the gas during expansion is given by

W = -(Force x Distance)

Since the work is being done by the system on the surrounding, decreases the internal energy of the system and consequently, the work is given a negative sign. Similarly, if work is being done on the system, it increases the internal energy of the system and therefore, work is given a positive sign. The work of expansion is given by

W = -P_ext. A.h    where Force = P_ext/A and P_ext is the external pressure.

But (A.h) is the change in volume, ΔV during the expansion and is equal to (V₂-V₁). Hence the work of expansion is

W = P_ext. ΔV where P_ext is the external pressure

This work is a mechanical work and expressed as above and can be shown as

Pressure = force/area

P_ext = F_ext/A

Now consider a force, F applied on the piston and the piston move distance dx.

Mechanical work, dW = F_ext.dx………………………………………………(1)

Then the change of volume (V) dV = dl×A

                                                       = d(b-x)A

                                                       = -dxA [as b is fixed]

Therefore, dx = -dV/A ………………………………………………………...(2)

From (1) and (2) we get

dW = F_ext.dV/A

dW = -P_ext. A.dV/A

dW = -P_ext.dV

The sign of workdone (W) is determined by the sign of ΔV. Since P_ext is always a positive quantity. In expansion, ΔV is positive and w is negative. A negative value of work implies that work is being done by the system on the surroundings. In compression, ΔV is negative and therefore W will be positive meaning thereby that the work is done by the surroundings on the system. Therefore according to IUPAC convention

a) Work done by the system = (-)ve

b) Work done on the system = (+)ve

Work done in free expansion of the gas: if the external pressure is zero, i.e., the gas expands in vacuum, no work is done by the system

-δW = P_ext.dV = 0

Work involved in reversible expansion of gas

As seen above, in a two-step expansion the magnitude of the work produced is more than that involved in a single-step expansion. If the expansion were carried out in an infinite number of steps such that the external pressure at each stage of the expansion is only infinitesimally less than the pressure of the gas, the magnitude of work goes on increasing and attains a maximum value. This is given by integrating

The work given in this case is obtained only under reversible conditions such that the system does not deviate from its equilibrium state. This is the maximum amount of work that can be derived from the system. If the expansion is carried out rapidly the equilibrium of the system will be distributed and the process will be irreversible. Consequently, the magnitude of work would be less. This can be seen from the following comparison in which an ideal gas expands reversibly and irreversibly from state P₁, V₁ to P₂, V₂ under isothermal conditions.

Work in irreversible expansion of gas

In this case the system does not main thermodynamic equilibrium and the expansion was not done in infinitesimally small steps rather larger steps. 

Here P_ext is constant

dW_irrev = -P_ext(V₂-V₁)

This is general expansion for any gas

If, P_ext = P₂

-W_irrev = P₂(V₂-V₁)

For ideal gases

V = nRT/P

-W_{irrev, iso} = P₂(nRT/P₂ –nRT/P₁)

              = nRT(1-P₂/P₁)

The difference between work done in reversible and irreversible process for ideal gas can be seen as follows

W_{iso, rev, ideal} = nRTln(V₂/V₁) = nRTln(P₁/P₂)

So, the difference in the magnitude of W_rev and W_irrev is

W_{iso, rev, ideal} - W_{iso, irrev, ideal}

= nRTln(P₁/P₂) – nRT(1-P₂/P₁)

= nRT(P₁/P₂ – 1) – nRT(1-P₂/P₁)

= nRT(P₁-P₂)/P₂ – nRT(1-P₂/P₁)

= nRT(P₁-P₂)[1/P₂ -1/P₁]

= nRT(P₁-P₂)²/P₂P₁

Since (P₁-P₂)² is always positive irrespective of whether P₁ is less than, equal to or greater than P₂, hence

|W_rev| - |W_irrev| > 0

|W_rev| > |W_irrev|

This shows that the magnitude of work in a reversible process is more than that in an irreversible process.

Nature of work: W is not a thermodynamic quantity. Its value as seen from the previous values for the same change in state depends on the path followed for isothermal transformation of the system. Mathematically one can prove that W is not an exact differential. If W is an exact differential for mathematical work, one may write

-dW = PdV

Considering V as a function of P and T, i.e.,

V = f(P, T)

dV = (dV/dP)_TdP + (dV/dT)_PdT

-dW = P(dV/dP)_TdP + P(dV/dT)_PdT

Hence

If dP = 0, then

-(dW/dT)_P = P(dV/dT)_P

-(d²W/dPdT) = P(d²V/dPdT) + (dV/dT)_P

And at constant temperature dT = 0 and

-(dW/dP)T = P(dV/dP)T

-(d²W/dTdP) = P(d²V/dPdT)

If dW is an exact differential

(d²W/dTdP) = (d²W/dPdT)

P(d²V/dPdT) + (dV/dT)_P = P(d²V/dPdT)

(dV/dT)_P = 0

The conclusion is obviously incorrect which means dW is not an exact differential.

For a cyclic transformation carried out reversibly and irreversibly, the value of W for two different processes would be different. This can be shown as follows:

If the system contains n moles of a real gas obeying van der Waals equation

(P + an²/V²)(V-nb) = nRT

Then the work (-W_vd) in isothermal reversible expansion is given by

-W_vd = ∫PdV = ∫[nRT/(V-nb) – an²/V²]dV

-W_vd = [nRTln(V-nb) - an²/V²]_V1^V2

= 2.303nRTlog(V₂-nb)/(V₁-nb) + an²[ 1/V₁ -1/ V₂]

The difference in the magnitude of work of expansion for an ideal and van der Waals gas is given by,

|W_id| -|W_vd| = nRTln (V2/V1) - nRTln(V₂-nb)/(V₁-nb) + an²[ 1/V₁ -1/ V₂]

If nb << V, so that V-nb =V, then

|W_id| -|W_vd| = an²[ 1/V₁ -1/ V₂] = an²[ V₂-V₁/V₁V₂] = an²ΔV/V₁V₂ = a positive quantity

The magnitude of reversible work expansion for an ideal gas is always greater than that for a van der Waals gas as work has to be done in overcoming the attractive forces between the molecules of a van der Waals gas. Therefore one can say that understanding and quantifying work in thermodynamics is essential for analyzing various processes such as expansion, compression, heat transfer, and energy conversion etc.

 Reference

1) Peter Atkins, Julio de Paula, Atkins’ physical chemistry book, eighth edition.