10.2 First Law and Work on P-V Diagrams

Consider a gas in a cylinder with a movable piston. During a small displacement dV, the gas exerts pressure P on the piston. If the displacement is outward, the gas does positive work on surroundings; if inward, surroundings do work on the gas. This boundary-work idea is the mechanical part of thermodynamic energy transfer.

In SI units, pressure in pascal and volume in cubic meter give joules directly. On many classroom graphs pressure is in kPa and volume in liters; that still yields joules because 1 kPa x 1 L = 1 J. This unit shortcut is useful for fast checks in exam conditions.

This single equation is the conservation-of-energy statement for thermal systems. It says internal-energy increase can happen either by heat entering the system or by surroundings doing work on the system (which appears as negative W in our sign convention). Internal-energy decrease can occur when the system exports heat, does work, or both.

Do not memorize disconnected sentence rules for each process. Instead, build the sign from geometry and words. Expansion means V2 > V1 so area under P-V path is positive and W > 0. Compression means V2 < V1 so W < 0. Then evaluate Q from Q = Delta U + W if needed.

Between the same initial and final states, different process paths can produce different work values because the pressure profile versus volume changes. If one path runs at higher pressure over most of the same expansion interval, area under the curve is larger and the gas does more work. Heat transfer then adjusts so that first law still lands on the same Delta U between those shared states.

This is a high-yield exam concept: same endpoints do not imply same Q or W. They imply the same Delta U for ideal-gas monatomic treatment, because Delta U is state-based through temperature. Whenever a problem shows multiple routes between two states, first identify which quantity is path-dependent before doing algebra.

If a gas expands while cooling, Delta U is negative and W is positive. Therefore Q must be less than W and can even be negative. If a gas is compressed adiabatically, Q = 0 and W is negative, so Delta U must be positive. These one-line sign checks catch arithmetic mistakes before they propagate through multi-part questions.