B.4.10 (HL)—Adiabatic ideal gas

Adiabatic monatomic gas

For an adiabatic process of a monatomic ideal gas,

$$PV^{5/3}=\text{constant}$$

so $$P_1V_1^{5/3}=P_2V_2^{5/3}$$.

Solve a state change

Write the two-state equation first, keep volumes in the same units, and rearrange for the unknown pressure or volume. The exponent 5/3 belongs to a monatomic ideal gas in this syllabus.

Interpret the expansion

During adiabatic expansion, the gas does work without receiving thermal energy, so its internal energy and temperature fall. The pressure drops more steeply with volume than along an isothermal path.

Worked example from the mapped local textbook

A monatomic ideal gas is compressed adiabatically to one eighth of its initial volume, so $V_1/V_2=8$.

$$\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^{5/3}=8^{5/3}=32$$

The pressure increases by a factor of 32. An isothermal compression by the same volume factor would increase pressure only by a factor of 8.

Common trap

Do not use $$PV=\text{constant}$$ for an adiabatic change; that is the isothermal relation. Do not use 5/3 for a non-monatomic gas unless the model specifies it.