# Work

Work

Work is defined as the force (F) multiplied by the displacement(x).

−w = F. x

The negative sign (−) is introduced to indicate that the work has been done by the system by spending a part of its internal energy.

The work,

1. is a path function.
2. appears only at the boundary of the system.
3. appears during the change in the state of the system.
4. In thermodynamics, surroundings is so large that macroscopic changes to surroundings do not happen.

### Units of work:

The SI unit of work is joule (J), which   is   defined   as   the   work   done by a force of one Newton through a displacement of one meter ( J= Nm). We often use kilojoule (kJ) for large quantities of work. 1 kJ = 1000 J.

### Sign convention of work:

The symbol of work is 'w'.

If work is done by the system, the energy of the system decreases, hence by convention, work is taken to be negative (− w).

If work is done on the system, the energy of the system increases, hence by convention, the work is taken to be positive (+w).

work done for reversible process

For understanding pressure- volume work, let us consider a cylinder which contains 'n' moles of an ideal gas fitted with a frictionless piston of cross sectional area A. The total volume of thegas inside is Vi and pressure of the gas inside is Pint.

If the external pressure Pextis greater than Pint, the piston moves inward till the pressure inside becomes equal to Pext. Let this change be achieved in a singlestep and the final volume be Vf.

In this case, the work is done on the system (+w). It can be calculated as follows

w = –F .dx

where dx is the distance moved by the piston during the compression and F is the force acting on the gas.

F   =   PextA

On Substituting

=   – Pext. A. dx

A.dx = change in volume = Vf- Vi

w  =  – Pext. (Vf- Vi)

= Pext.dV

Since work is done on the system, it is a positive quantity.

If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. In such a case we can calculate the work done on the gas by the relation

In a compression process, Pextthe external pressure is always greater than the pressure of the system.

In an expansion process, the external pressure is always less than the pressure of the system

When pressure is not constant and changes in infinitesimally small steps (reversible conditions) during compression from Vi to Vf , the P-V plot looks like in fig Work done on the gas is represented by the shaded area.

In general case we can write, Pext = (Pint + dP). Such processes are called reversible processes. For a compression process work can be related to internal pressure of the system under reversible conditions by writing equation

If  Vf>Vi (expansion), the sign of work done by the process is negative.

If Vf<Vi (compression) the sign of work done on the process is positive.

work done for irreversible process

Where P is the pressure and ΔV is the change in volume.

In isothermal process, as the temperature remains constant, the internal energy exchange between medium and surroundings is 0

In irreversible process, we can not get back to the exact previous conditions from where we started the experiment.

It can be of two types:

1. Irreversible isothermal expansion
2. Irreversible isothermal contraction

### In case of expansion,

Suppose, a gas is expanded from P, V₁ to P, V₂, where V₂ > V₁

So, the work done is , W = - P (V₂ - V₁)

This will be '-ve', as V₂ > V₁

### In case of contraction,

Suppose, a gas is expanded from P, V₁ to P, V₂, where V₂ < V₁

So, the work done is , W = - P (V₂ - V₁)

or, W = + P(V₁ - V₂)

This will be '+ve', as V₂ < V₁

Free expansion

The expansion of a gas in a vacuum (pex = 0) is known as free expansion. During free expansion of an ideal gas, no work is done whether the process is reversible or irreversible. This is because there is no external pressure acting on the system, and the gas expands without any resistance.

The first law of thermodynamics, can be written as

∆U = q + w

If we substitute w = -pex∆V, we obtain:

∆U = q + (-pex∆V)

For a process carried out at constant volume (∆V = 0), the equation becomes:

∆U = qV

Here, the subscript V in qV indicates that heat is supplied at constant volume.

During isothermal (T = constant) expansion of an ideal gas into a vacuum, the work done is zero (w = 0) since there is no external pressure acting on the system. Additionally, it was experimentally determined by Joule that the heat supplied to the system is also zero (q = 0). Therefore, the change in internal energy (∆U) during isothermal and free expansion of an ideal gas is also zero.