Pressure–volume work is the work performed by a thermodynamic system when it undergoes a change in volume against an external pressure. It is the most common and important form of work encountered in thermodynamics, particularly in the study of gaseous systems.
Thermodynamic Notation (Very Important)
In thermodynamics, internal energy U is a state function and its differential is exact, written as dU. Heat q and work w are path functions and their differentials are inexact, written as δq and δw. This distinction must be strictly followed in all derivations and numerical problems.
Physical Origin of Pressure–Volume Work
Consider a gas enclosed in a cylinder fitted with a frictionless movable piston. When the gas expands, it pushes the piston outward against the external pressure and performs work on the surroundings. During compression, the surroundings push the piston inward and do work on the gas.
Thus, pressure–volume work arises whenever there is a change in volume of a system under the action of an opposing pressure.
Mechanical Derivation of Pressure–Volume Work
Let the cross-sectional area of the piston be A and the external pressure opposing expansion be Pext. If the piston moves outward by an infinitesimal distance dx, the force acting on the piston is given by
\[F = P_{ext} \times A\]
By the mechanical definition of work, the infinitesimal work done is
\[\delta w = F\,dx\]
Substituting the value of F, we obtain
\[\delta w = P_{ext}A\,dx\]
Since the infinitesimal change in volume is dV = A dx, the above expression becomes
\[\boxed{\delta w = P_{ext}\,dV}\]
This is the fundamental expression for pressure–volume work and clearly shows that work is a path function.
Finite Pressure–Volume Work
For a finite change in volume from V1 to V2, the total work done is obtained by integration
\[\boxed{w = \int_{V_1}^{V_2} P_{ext}\,dV}\]
Irreversible Pressure–Volume Work
In an irreversible expansion, the system expands against a constant external pressure. Since Pext remains constant during the process, it can be taken outside the integral.
\[\boxed{w_{irrev} = P_{ext}(V_2 – V_1)}\]
This expression shows that irreversible work depends only on the external pressure and the change in volume.
Reversible Pressure–Volume Work
In a reversible process, the system remains in mechanical equilibrium with the surroundings at every stage. The external pressure differs infinitesimally from the internal pressure of the gas and may be replaced by P.
\[\boxed{w_{rev} = \int_{V_1}^{V_2} P\,dV}\]
Derivation of Reversible Isothermal Work (Ideal Gas)
For an ideal gas undergoing reversible isothermal expansion, the equation of state is
\[P = \frac{nRT}{V}\]
Substituting this expression into the work equation gives
\[w_{rev} = \int_{V_1}^{V_2} \frac{nRT}{V}\,dV\]
Since n, R, and T are constants for an isothermal process,
\[\boxed{w_{rev} = nRT \ln\frac{V_2}{V_1}}\]
Conceptual Explanation Using PV Diagram
In a pressure–volume diagram, the work done during a process is represented by the area under the curve. For the same initial and final states, the area under a reversible curve is greater than the area under an irreversible path, proving that reversible work is maximum.
Solved Numerical Problems (Full Book Style)
Example 1: Irreversible Expansion
Question. A gas expands irreversibly from 2 L to 6 L against a constant external pressure of 2 atm. Calculate the work done.
Given:
Pext = 2 atm
V1 = 2 L
V2 = 6 L
Formula:
\[w_{irrev} = P_{ext}(V_2 – V_1)\]
Substitution:
\[w = 2 \times (6 – 2) = 8\ \text{L atm}\]
Conversion:
\[1\ \text{L atm} = 101.3\ \text{J}\]
\[w = 8 \times 101.3 = 810.4\ \text{J}\]
Answer:
\[\boxed{w = +810.4\ \text{J}}\]
Example 2: Reversible Isothermal Expansion
Question. One mole of an ideal gas expands reversibly and isothermally at 300 K from 5 L to 15 L. Calculate the work done.
Given:
n = 1 mol
T = 300 K
V1 = 5 L
V2 = 15 L
R = 8.314 J mol⁻¹ K⁻¹
Formula:
\[w_{rev} = nRT \ln\frac{V_2}{V_1}\]
Substitution:
\[w = 1 \times 8.314 \times 300 \times \ln\left(\frac{15}{5}\right)\]
\[\ln 3 = 1.0986\]
Calculation:
\[w = 2494.2 \times 1.0986 = 2740\ \text{J (approx.)}\]
Answer:
\[\boxed{w_{rev} = +2740\ \text{J}}\]
Book-Style Practice Questions
1. Derive the expression for pressure–volume work starting from mechanical principles.
2. Explain why pressure–volume work is a path function.
3. Derive the expression for reversible isothermal work of an ideal gas.
4. Using PV diagrams, explain why reversible work is maximum.