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 (Cambridge 9702)

Learning Objective

Students will be able to:

• Recall and apply the expression W = p_{\text{ext}}\,\Delta V for the work done when a gas changes volume at constant external pressure.
• Write the work term in the first‑law equation with the correct sign convention (work by the gas vs. work on the gas).
• Extend the idea to processes where the external pressure varies (integral form).
• Distinguish reversible (quasi‑static) from irreversible processes.

Key Concepts

Internal Energy (U)

• U is the total microscopic kinetic and potential energy of all particles in a closed system.
• For an ideal gas, U depends only on temperature:

  \[

  \Delta U = nC_{V}\,\Delta T\qquad\text{(so }U\propto T\text{).}

  \]

• Through the ideal‑gas equation \(pV=nRT\), a change in temperature also implies a change in pressure or volume, but the energy change is given by the above relation.

First Law of Thermodynamics

For a closed system:

\[

\Delta U = q + W_{\text{on}}

\]

• \(q\) – heat transferred to the system (positive when added, negative when removed).
• \(W_{\text{on}}\) – work done on the system (positive when the surroundings do work on the gas, negative when the gas does work on the surroundings).
• Because many textbooks use the opposite sign convention, it is useful to define the work done by the gas:

  \[

  W{\text{by}} = -\,W{\text{on}} .

  \]

Work at Constant External Pressure

\[

W = p_{\text{ext}}\,\Delta V

\]

• \(p_{\text{ext}}\) – the pressure exerted by the surroundings (the “resisting” pressure). It equals the internal pressure only for a reversible (quasi‑static) expansion or compression.
• \(\Delta V = V{\text{f}}-V{\text{i}}\).
• This formula applies only when \(p_{\text{ext}}\) remains constant during the whole process.

Work at Variable External Pressure (Integral Form)

If the external pressure changes during the volume change, the work is obtained by integrating:

\[

W = \int{Vi}^{Vf} p{\text{ext}}(V)\,dV .

\]

• For a linear change, e.g. \(p{\text{ext}} = p0 + kV\), the integral can be evaluated analytically.
• In a free‑expansion into a vacuum, \(p_{\text{ext}} = 0\) everywhere, so \(W = 0\) (an irreversible process).

Reversible vs. Irreversible Processes

• Reversible (quasi‑static): The gas pressure is always infinitesimally different from the external pressure; the path can be retraced by an infinitesimal change of conditions. In this case \(p{\text{ext}} \approx p{\text{int}}\) and the simple formula \(W = p\Delta V\) is valid.
• Irreversible: The gas pressure differs markedly from the external pressure (e.g. rapid expansion, free expansion, sudden compression). Work must be calculated with the actual \(p_{\text{ext}}(V)\) profile or recognised as zero for free expansion.

Sign Conventions

The table summarises the sign of work for the two viewpoints.

Derivation of \(W = p_{\text{ext}}\Delta V\) (Constant External Pressure)

1. Consider a piston of cross‑sectional area \(A\) that moves a distance \(d\) under the action of the external pressure.
2. The change in gas volume is \(\displaystyle \Delta V = A\,d\).
3. The external pressure exerts a force \(F = p_{\text{ext}}A\) on the piston.
4. Work done by the gas on the surroundings:

   \[

   W{\text{by}} = F\,d = p{\text{ext}}A\,d = p_{\text{ext}}\Delta V .

   \]

5. Work done on the gas is the negative of this:

   \[

   W{\text{on}} = -\,p{\text{ext}}\Delta V .

   \]

Worked Examples

Example 1 – Constant External Pressure

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 and the work done on the gas.

1. Convert volumes to cubic metres:

   \(Vi = 2.0\times10^{-3}\ \text{m}^3,\; Vf = 5.0\times10^{-3}\ \text{m}^3\).

2. \(\Delta V = Vf - Vi = 3.0\times10^{-3}\ \text{m}^3\).
3. Work by the gas:

   \(W{\text{by}} = p{\text{ext}}\Delta V = (1.0\times10^{5})(3.0\times10^{-3}) = 300\ \text{J}\).

4. Work on the gas:

   \(W_{\text{on}} = -300\ \text{J}\).

Example 2 – Linearly Varying External Pressure

Problem: A gas expands from \(1.0\ \text{L}\) to \(3.0\ \text{L}\). The external pressure falls linearly from \(2.0\times10^{5}\ \text{Pa}\) to \(1.0\times10^{5}\ \text{Pa}\). Find the work done by the gas.

1. Express the external pressure as a function of volume:

   \[

   p{\text{ext}}(V) = pi + \frac{pf-pi}{Vf-Vi}(V-V_i)

   = 2.0\times10^{5} - \frac{1.0\times10^{5}}{2.0\ \text{L}}(V-1.0\ \text{L}).

   \]

2. Integrate:

   \[

   W{\text{by}} = \int{Vi}^{Vf} p_{\text{ext}}(V)\,dV

   = \frac{(pi+pf)}{2}\,(Vf-Vi)

   = \frac{(2.0\times10^{5}+1.0\times10^{5})}{2}\,(2.0\times10^{-3})

   = 3.0\times10^{5}\times10^{-3}=300\ \text{J}.

   \]

3. Thus \(W{\text{by}} = 300\ \text{J}\) and \(W{\text{on}} = -300\ \text{J}\).

Common Misconceptions & How to Avoid Them

• Sign of work: Remember the viewpoint. Expansion → positive \(W{\text{by}}\), negative \(W{\text{on}}\). Compression → opposite.
• Which pressure to use: Use the external pressure that resists the change. For a reversible process this equals the internal pressure, but the syllabus emphasises the external value.
• Variable pressure: When \(p{\text{ext}}\) is not constant, replace \(p{\text{ext}}\Delta V\) with \(\displaystyle \int p_{\text{ext}}\,dV\). A common error is to treat a non‑constant pressure as if it were constant.
• Free expansion: Because \(p_{\text{ext}}=0\) throughout, the work is zero even though the volume changes.
• Units: Pressure in pascals (Pa), volume in cubic metres (m³); their product gives joules (J).
• Heat symbol: The Cambridge syllabus uses \(q\) (lower‑case) for heat transferred to the system. Positive \(q\) adds energy, negative \(q\) removes energy.

Summary

• The first law links internal energy, heat and work: \(\displaystyle \Delta U = q + W_{\text{on}}\).
• For a constant external pressure, mechanical work simplifies to \(\displaystyle W = p_{\text{ext}}\Delta V\).
• If the external pressure varies, use the integral form \(\displaystyle W = \int p_{\text{ext}}\,dV\).
• Sign conventions distinguish energy leaving the system (work by the gas) from energy entering the system (work on the gas).
• Remember that the simple \(p\Delta V\) expression is strictly valid for reversible (quasi‑static) processes; otherwise the actual external pressure profile must be considered.