recall and use W = p∆V for the work done when the volume of a gas changes at constant pressure and understand the difference between the work done by the gas and the work done on the gas

The First Law of Thermodynamics

Learning Objective

Recall and use the expression

\$W = p\Delta V\$

for the work done when the volume of a gas changes at constant pressure, and understand the difference between the work done by the gas and the work done on the gas.

Key Concepts

• Internal energy (U) – the total kinetic and potential energy of the particles in a system.
• First law of thermodynamics – the change in internal energy of a closed system equals the net heat supplied to the system plus the net work done on the system:

  \$\Delta U = Q + W_{\text{on}}\$

• Work in a gas – when a gas expands or contracts against a constant external pressure, the work done is

  \$W = p\Delta V\$

  where \$p\$ is the external pressure and \$\Delta V = V{\text{final}}-V{\text{initial}}\$.

Sign Conventions

It is essential to keep track of the sign of work depending on the perspective:

Derivation of \$W = p\Delta V\$ (Constant Pressure)

1. Consider a piston of cross‑sectional area \$A\$ moving a distance \$d\$.
2. The change in volume is \$\Delta V = A d\$.
3. The force exerted by the external pressure is \$F = pA\$.
4. Work done by the gas on the surroundings is \$W = F d = pA d = p\Delta V\$.

Worked Example

Problem: A gas expands at a constant external pressure of \$1.0\times10^5\ \text{Pa}\$ from \$2.0\ \text{L}\$ to \$5.0\ \text{L}\$. Calculate the work done by the gas.

Solution:

Convert volumes to cubic metres: \$2.0\ \text{L}=2.0\times10^{-3}\ \text{m}^3\$, \$5.0\ \text{L}=5.0\times10^{-3}\ \text{m}^3\$.

\$\Delta V = (5.0-2.0)\times10^{-3}\ \text{m}^3 = 3.0\times10^{-3}\ \text{m}^3\$

\$W = p\Delta V = (1.0\times10^5\ \text{Pa})(3.0\times10^{-3}\ \text{m}^3) = 300\ \text{J}\$

Since the gas expands, the work is done by the gas, so \$W{\text{by}} = +300\ \text{J}\$ and \$W{\text{on}} = -300\ \text{J}\$.

Common Misconceptions

• Confusing the sign of work: remember that expansion does positive work on the surroundings, but negative work on the system.
• Assuming \$W = p\Delta V\$ applies when pressure varies; in that case the integral \$W = \int p\,dV\$ must be used.
• Neglecting the units: pressure in pascals (Pa) and volume in cubic metres (m³) give work in joules (J).

Summary

The first law links heat, work and internal energy. For processes at constant external pressure the mechanical work term simplifies to \$W = p\Delta V\$. Correctly assigning the sign of \$W\$ distinguishes between work done by the gas (energy leaving the system) and work done on the gas (energy entering the system).