B.4.9 (HL)—Gas processes

Process	Fixed quantity or condition	First-law consequence
Isovolumetric	$V$	$W=0$ , so $Q=\Delta U$
Isobaric	$P$	$W=P\Delta V$
Isothermal	$T$	ideal gas has $\Delta U=0$ , so $Q=W$
Adiabatic	$Q=0$	$\Delta U=-W$

Connect to the first law

Use $Q=\Delta U+W$ after identifying the fixed quantity. For an isothermal ideal-gas process, $\Delta U=0$ , so Q=W. For an adiabatic process, Q=0, so $\Delta U=-W$ .

Read the PV path

An isovolumetric process is vertical on a PV diagram; an isobaric process is horizontal. Isothermal and adiabatic curves both change P and V, but the adiabatic curve is steeper than the isothermal curve for an ideal gas.

Common trap

“Adiabatic” means no thermal energy transfer, not constant temperature. Temperature usually changes when an ideal gas expands adiabatically.

How it is assessed

The evidence tests recognizing a Carnot-cycle PV diagram and explaining why a theoretical cycle is impractical.

Command terms

Outline / Determine

What earns marks

Classify each stage by its fixed variable or heat-transfer condition. On a P–V diagram identify isovolumetric, isothermal and adiabatic paths, and use direction/area to determine work. Explain practical limits such as very slow isothermal and very rapid insulated adiabatic operation.

Watch for

Confusing adiabatic with isothermal or ignoring the process direction on the PV diagram.

Representative question

[Maximum number: 1]

A cyclic process for an ideal gas is shown. The cycle has three stages: isovolumetric, adiabatic, and isothermal. Work is done on the gas during the isothermal stage.

Which stage is isothermal and what is the direction of the cyclic process?

Stage

Direction

Anti-clockwise

Clockwise

B.4.9 (HL)—Gas processes

Classify Thermodynamic Gas Processes

B.4.9 (HL) Exam Analysis

Assessment in practice

Question 1