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Thermodynamics

Specific heats and thermodynamic processes

Calculate heat and work for isothermal, adiabatic, isobaric and isochoric processes.

A gas has no single specific heat: the heat needed per degree depends on what else happens as it warms. Two values are used, per mole:

  • C_v at constant volume: the gas does no work, so all the heat raises U.
  • C_p at constant pressure: the gas also expands and does work, so it needs more heat. For an ideal gas, C_p – C_v = R (Mayer's relation).
  • Monatomic gas: C_v = (3/2)R; diatomic: C_v = (5/2)R. Their ratio is γ = C_p/C_v (5/3 and 7/5).

On a P–V graph (indicator diagram), the work done by the gas is the area under the curve: positive when it expands, negative when it is compressed. Round a closed cycle, the net work is the area enclosed, positive if the cycle runs clockwise.

The four standard processes:

ProcessWhat is fixedKey facts
IsothermalTPV = constant; ΔU = 0, so Q = W; W = μRT ln(V₂/V₁)
Adiabaticno heat flows (Q = 0)PV^γ = constant; W = –ΔU = μR(T₁ – T₂)/(γ – 1)
IsobaricPW = PΔV; Q = μC_pΔT
IsochoricVW = 0; Q = ΔU = μC_vΔT

An isothermal change must be slow, through walls that conduct, so heat has time to flow in or out. An adiabatic change is sudden, or behind insulating walls, so no heat can flow: a bursting tyre's air cools, and air pumped hard into a bicycle tyre warms up.

Example: a gas at 50 °C (323 K) and 75 cm of mercury is squeezed to a quarter of its volume (γ = 1.5).

  • Slowly (isothermal): P₁V₁ = P₂V₂, so P₂ = 4 × 75 = 300 cm of Hg, still at 50 °C.
  • Suddenly (adiabatic): P₂ = 75 × 4^1.5 = 600 cm of Hg, and from TV^(γ–1) = constant, T₂ = 323 × 4^0.5 = 646 K (373 °C). Always use kelvin in these formulas.

Example of isothermal work: 3 mol of gas at 300 K compressed from 4 L to 1 L. W = μRT ln(V₂/V₁) = 3 × 8.31 × 300 × ln(1/4) ≈ –1.04 × 10⁴ J. The minus sign means the work is done on the gas, and the same amount of heat flows out.

Work = area under P-V; adiabatic means PV^γ = constant.

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