Dashboard/Learning Hub/Physics HL/Chapter 10/10.2 Continued: Standard Processes and Process Equations

Physics HL · Chapter 10: Thermodynamics

10.2 Continued: Standard Processes and Process Equations

Compare isovolumetric, isobaric, isothermal, and adiabatic paths with equation-level and diagram-level reasoning.

Estimated time: 32 minutes

The Four Process Labels and Their Constraints

Isovolumetric means V is constant. Isobaric means P is constant. Isothermal means T is constant. Adiabatic means Q is zero. These labels are not decorative; each one tells you immediately which variable relation can be applied and which first-law term simplifies.

A frequent error is to treat these labels as interchangeable shortcuts for one equation family. Instead, each process is a different physical boundary condition and can produce very different temperature and energy-transfer outcomes even for similar-looking pressure-volume shifts.

Equation Set You Should Use Deliberately

isothermal: P1V1=P2V2,W=nRTln(V2V1),ΔU=0\text{isothermal: }P_1V_1=P_2V_2,\quad W=nRT\ln\left(\frac{V_2}{V_1}\right),\quad \Delta U=0

For ideal gas at constant temperature, all internal-energy change vanishes and heat equals work.

adiabatic (monatomic ideal gas): PVγ=constant,TVγ1=constant,γ=53\text{adiabatic (monatomic ideal gas): }PV^{\gamma}=\text{constant},\quad TV^{\gamma-1}=\text{constant},\quad \gamma=\frac{5}{3}

No heat transfer: Q = 0, so first law couples internal-energy change directly to work.

For isobaric paths, work is the rectangular area P Delta V and temperature changes with volume. For isovolumetric paths, work is zero and heat goes fully into internal-energy change. Writing these consequences in words before numbers keeps you from applying the wrong formula set to the wrong constraint.

Why Adiabatic Curves Are Steeper Than Isothermal Curves

Take an expansion from the same initial point. In an isothermal path, temperature is held constant by heat flow from surroundings into the gas while it expands. That heat support resists pressure drop. In an adiabatic expansion, no heat enters, so the gas must fund work output by lowering internal energy and therefore temperature, making pressure fall more rapidly with volume.

That is why adiabatic expansion curves sit below isothermal expansion curves through the same volume range. During compression from a shared start point, the opposite ordering appears: adiabatic compression rises to higher pressure at the same reduced volume because temperature increases more strongly when compression work is not offset by heat loss.

Process-Comparison Problem Routine

  • Identify the imposed constraint first (constant P, V, T, or Q = 0).
  • Write the process-specific equation and first-law simplification.
  • Use PV-geometry intuition to determine sign of work before arithmetic.
  • Check whether final temperature trend (up/down/unchanged) matches physical expectation.

Simulation: Process-Constraint Comparison

Switch process modes to compare constant-T, constant-P, and constant-V behavior, then connect those constraints back to first-law terms.

Ideal Gas Law Lab

Active law interpretation

Boyle mode keeps n and T fixed, so pressure responds inversely to volume (P*V = constant).

Container micro-view (animated gas particles + piston)

Volume: 8.00 LPressure: 343.0 kPaTemperature: 330.0 K

P-V map with isotherms

V (L)P (kPa)

State diagnostics

P

343.0 kPa

3.38 atm

T

330.0 K

56.9 deg C

V

8.00 L

8.0000e-3 m^3

Pressure index23%
Thermal index29%
Volume fraction16%

Microscopic metrics

RMS speed c_rms: 542.2 m s^-1

Mean molecular kinetic energy: 6.834e-21 J

Density estimate: 3.500 kg m^-3

Monatomic internal energy estimate: 4115.4 J

Model validity note

Ideal-model range: this state is in a typical low-density, moderate-temperature regime where PV = nRT is usually reliable.

Try this workflow: hold n fixed, then switch between Boyle/Charles/Gay-Lussac modes and verify each ratio form before returning to full-state PV = nRT checks.

Test Yourself

An isobaric expansion occurs at P = 200 kPa from 4.0 L to 9.0 L. Enter the work done by the gas in joules.

Hint: For isobaric change, W = P Delta V and 1 kPa x 1 L = 1 J.