Mathematical Statement of the First Law

Starting from the general first-law statement $\Delta U=q+w$ (7.7), four special cases -- each corresponding to one of the process types from Section 7.2.3 -- simplify the equation considerably.

Case 1: Cyclic process (isothermal expansion of an ideal gas around a full cycle). Since the system returns exactly to its starting state, $\Delta U=0$; substituting into (7.7) gives $q=-w$ -- during a cyclic process, the heat absorbed by the system exactly equals the work done BY the system.

Case 2: Isochoric process ($\Delta V=0$, no volume change, hence no expansion work possible). Starting from $\Delta U=q+w=q-P\Delta V$ and setting $\Delta V=0$ gives simply $\Delta U=q_V$ -- at constant volume, all the heat supplied to the system goes directly into raising its internal energy, none of it diverted into P-V work.

Case 3: Adiabatic process ($q=0$, no heat exchange at all). Substituting into (7.7): $\Delta U=w$ -- the fall in a system's internal energy during an adiabatic process is exactly equal to the work it does on its surroundings (there being no heat available to replace the lost energy).

Case 4: Isobaric process (constant pressure). Here $\Delta U=q+w=q-P\Delta V$ -- part of the heat absorbed by the system is spent doing P-V expansion work against the constant external pressure, and the rest is added to the system's internal energy.

Worked example (Problem 7.1). A gas in a frictionless piston-cylinder expands against a constant external pressure of 1 atm, from an initial volume of 5 L to a final volume of 10 L, absorbing 400 J of heat from its surroundings in the process. Since heat is given to the system ($+q=400$ J) and work is done by the system as it expands (using the isobaric-case formula): …