Isothermal process

Constant $T$. On a pressure-volume diagram the path is an isotherm, which essential knowledge 9.4.B.2.i defines as a line of constant temperature; for an ideal gas $PV = nRT$ makes it a hyperbola, and isotherms further from the origin are hotter.

The physics comes from internal energy depending only on temperature for an ideal gas, which is 9.4.A.1.ii. So:

using the sheet's $\Delta U = Q + W$ with $W = -P\Delta V$, where $W$ is the work done on the gas.

Run that through an expansion. The gas pushes outward, so $\Delta V > 0$ and $W < 0$. Then $Q$ has to be positive by the same amount: the gas must be heated continuously just to hold its temperature steady while it gives energy up as work. Compress it isothermally instead and $W > 0$, so $Q < 0$ and energy has to be carried away.

Two things this does not mean. It does not mean nothing crosses the boundary; energy flows through the whole time, and the net is zero only because $Q$ and $W$ cancel. And it is not adiabatic, which is the case with $Q = 0$ and a temperature that moves.

The shortcut also depends on the ideal gas model. Zero internal energy change follows from $U = \frac{3}{2}nRT$, not from a general truth about every system.

Reading work off the area under the curve is the PV diagrams guide's job.