The First Law and Gas Processes

Take Q as energy transferred into the system by heating, ΔU as the increase in internal energy, and W as work done by the system. Expansion normally gives positive W; compression gives negative W. For an ideal gas, internal energy depends only on temperature.

If a gas receives 100 J by heating and does 40 J of work, its internal energy rises by 60 J. If it is compressed without heat transfer, the negative W gives a positive ΔU. A sign convention explained once is more reliable than memorising a separate rule for every process.

Scroll the diagram to see the full view.

At constant volume, the gas does no boundary work, so Q=ΔU. At constant pressure, work is $p\Delta V$ and heating generally changes both volume and temperature. An isothermal ideal-gas process keeps temperature and internal energy constant, so Q=W; it requires heat exchange during expansion or compression.

An adiabatic process transfers no energy by heating, so ΔU=−W. For a fixed amount of ideal gas the adiabatic relation is $pV^\gamma=$ constant. During expansion its temperature falls and its p–V curve drops more steeply than an isotherm through the same starting point. “Fast” can help make a real process approximately adiabatic, but the defining condition is negligible heat transfer.

The area under a p–V curve is the work done by the gas during expansion. For a varying pressure estimate the area with rectangles or trapezia if numerical data are provided. On compression the signed work is negative.

The area enclosed by a clockwise cycle is net work output; an anticlockwise cycle needs net work input. Since the system returns to its initial state after a full cycle, total ΔU=0 and net heat input equals net work output. The route matters for Q and W even though ΔU depends only on the endpoints.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.