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The First Law of Thermodynamics

Source lecture(s): SC133 Lec 27

Intuition

Energy conservation, finally with the books complete. Mechanics kept losing energy to "heat" — the first law brings it home: a system's internal energy (the microscopic KE + PE of its molecules) changes only through two currencies, heat in and work out. Track both and nothing is ever lost. Every engine, refrigerator, and living cell is an entry in this ledger.

The law

\[\boxed{\,\Delta E_\text{int} = Q - W\,}\]
  • \(Q\): heat added to the system (positive in)
  • \(W\): work done by the system (positive out) — watch conventions; some texts flip the sign of \(W\)
  • \(E_\text{int}\): internal energy — a state function (depends only on the current state), while \(Q\) and \(W\) are path functions (depend on how you got there)

For an ideal gas, \(E_\text{int}\) depends on temperature alone: \(E_\text{int} = \tfrac{f}{2}nRT\) (with \(f\) degrees of freedom; \(f = 3\) monatomic).

Work by a gas

A gas expanding against a piston does work

\[W = \int_{V_i}^{V_f} P\,dV \quad = \text{area under the curve on a } P\text{–}V \text{ diagram}\]

Different paths between the same endpoints enclose different areas — hence work (and heat) are path-dependent while \(\Delta E_\text{int}\) is not. This is why engines run in cycles: loop area = net work per cycle.

The standard processes (ideal gas)

Process Constant First law becomes Work
Isochoric \(V\) \(\Delta E = Q\) \(W = 0\)
Isobaric \(P\) \(\Delta E = Q - P\Delta V\) \(W = P\Delta V\)
Isothermal \(T\) \(Q = W\) (\(\Delta E = 0\)) \(W = nRT\ln(V_f/V_i)\)
Adiabatic \(Q = 0\) \(\Delta E = -W\) gas cools as it expands

Adiabatic (fast or insulated) processes obey \(PV^\gamma = \text{const}\) — the physics of diesel ignition (compress → hot), cloud formation (rising air expands → cools), and the adiabatic atmosphere in PC316.

Worked example: three ways between two states

One mole of monatomic gas goes from (P₀, V₀) to (P₀, 2V₀) isobarically.

Work: \(W = P_0V_0\). Temperature doubles (\(PV = nRT\)), so \(\Delta E = \tfrac32 nR\,\Delta T = \tfrac32 P_0V_0\). Heat: \(Q = \Delta E + W = \tfrac52 P_0V_0\).

The same \(\Delta E\) via a different path (isochoric heat then isothermal expansion) would need different \(Q\) and \(W\) — but always the same difference. State vs path, in numbers.

Common mistakes

  • Sign chaos. Decide the convention (\(W\) = work by the system) and audit every term. An expanding gas does positive work; compressing it, negative.
  • "Adiabatic" ≠ "isothermal". Adiabatic means no heat flow — the temperature usually changes precisely because \(Q = 0\).
  • Treating \(Q\) or \(W\) as state functions — "the heat in a gas" is as meaningless as "the work in a gas".
  • Forgetting \(\Delta E_\text{int} = 0\) around any complete cycle — so per cycle, net heat in = net work out: the engine's whole business model.

Knowledge graph position

Prerequisites: Temperature & heat, Conservation of energy. Leads to: Ideal gas, Second law, engines & refrigerators.

Quiz

Q1 (computational). A gas absorbs 800 J of heat while doing 300 J of work. Change in internal energy?

Answer

\(\Delta E = Q - W = 800 - 300 = 500\,\text{J}\).

Q2 (conceptual). Why does a bicycle pump grow hot at the bottom even before friction matters?

Answer

Rapid compression is nearly adiabatic: \(Q \approx 0\), work is done on the gas (\(W < 0\)), so \(\Delta E > 0\) — the air heats. The reverse (adiabatic cooling on expansion) frosts CO₂ fire extinguishers and spray-can nozzles.

Q3 (multiple choice). In an isothermal expansion of an ideal gas: (a) \(Q = 0\) (b) \(W = 0\) (c) \(Q = W\)

Answer

(c). Constant \(T\) ⇒ constant \(E_\text{int}\) ⇒ every joule of work done by the gas is imported as heat.