B4-3. Thermodynamic Processes and Adiabatic Modelling HL Only

1. Mathematical Summary of the Four Fundamental Processes

By restricting specific state variables, ideal gases undergo four classic thermodynamic paths. Each uniquely alters the variables within the First Law of Thermodynamics ($Q = \Delta U + W$):

Process Name Constant Variable First Law Impact Physical Mechanism
Isobaric Pressure ($\Delta p = 0$) $$Q = \Delta U + p\Delta V$$ Gas expands/contracts freely against a constant external boundary load.
Isovolumetric Volume ($\Delta V = 0$) $$W = 0 \Rightarrow Q = \Delta U$$ Rigid container. No boundary work is done; all heat directly changes internal kinetic energy.
Isothermal Temperature ($\Delta T = 0$) $$\Delta U = 0 \Rightarrow Q = W$$ Extremely slow expansion/compression maintained in perfect contact with a thermal reservoir.
Adiabatic Thermal Energy ($Q = 0$) $$W = -\Delta U$$ Perfectly insulated system or an ultra-rapid change leaving no time for heat exchange.
Isotherm TH Isotherm TC p V V1 V2,ad V2,it State 1 (p1, V1, TH) Isobaric (Δp = 0) Isovolumetric (ΔV = 0) W = 0 Isothermal (ΔT = 0) Follows TH Adiabatic (Q = 0) Reference Isotherms Work Projection Bounds

2. Advanced Modelling of Adiabatic Processes

During an adiabatic process, no thermal energy passes across the boundaries ($Q = 0$). As a gas expands adiabatically, it performs work at the direct expense of its internal molecular kinetic energy, causing the temperature to plunge. On a $p\text{-}V$ plot, the adiabatic curve is steeper than an isothermal curve. For a monatomic ideal gas, the path is governed by a specific ratio of specific heats ($\gamma = \dfrac{5}{3}$):

Adiabatic State Equations (Monatomic Gas):

$$pV^{\frac{5}{3}} = \text{constant} \quad \Rightarrow \quad p_1 V_1^{\frac{5}{3}} = p_2 V_2^{\frac{5}{3}}$$

By substituting the ideal gas law ($p = \dfrac{nRT}{V}$), we derive the corresponding temperature-volume relationship:

$$TV^{\frac{2}{3}} = \text{constant} \quad \Rightarrow \quad T_1 V_1^{\frac{2}{3}} = T_2 V_2^{\frac{2}{3}}$$

Example 1: Mathematical Treatment of Adiabatic Changes

Problem: A sample of monatomic ideal gas inside a diesel engine cylinder begins at an initial pressure of $1.20 \times 10^5$ Pa and occupies a volume of $6.0 \times 10^{-4}$ m³. The gas is compressed rapidly (adiabatically) to a tiny final volume of $1.0 \times 10^{-4}$ m³. Calculate the final pressure ($p_2$) of the compressed gas.

Solution:

  • Step 1: Set up the governing equation.
    $$p_1 V_1^{\frac{5}{3}} = p_2 V_2^{\frac{5}{3}} \implies p_2 = p_1 \left(\dfrac{V_1}{V_2}\right)^{\frac{5}{3}}$$
  • Step 2: Determine the volume compression ratio.
    $$\dfrac{V_1}{V_2} = \dfrac{6.0 \times 10^{-4}}{1.0 \times 10^{-4}} = 6$$
  • Step 3: Evaluate the exponent and compute $p_2$.
    $$p_2 = (1.20 \times 10^5 \text{ Pa}) \times (6)^{\frac{5}{3}}$$
    Using an algebraic calculator: $6^{\frac{5}{3}} = 6^{1.6667} \approx 19.812$
    $$p_2 = (1.20 \times 10^5) \times 19.812 = \mathbf{2.38 \times 10^6 \text{ Pa}}$$ (or $2.38$ MPa)