Two ways in, one quantity out

Fix the convention once and then obey it:

$$\Delta U = Q + W, \qquad W = -P\,\Delta V \ \text{(constant pressure)}.$$

Compress the gas and $\Delta V$ is negative, so $W$ comes out positive and the gas gains internal energy. That matches the picture: you pushed on it, so you did work on it. Let it expand and $\Delta V$ is positive, $W$ is negative, and the gas spends energy pushing its surroundings back.

Two habits produce the wrong sign. One is importing a reflex from mechanics, where $W$ usually means the work a system does. The other is reading a work-done-by-the-gas value off a diagram and adding it unchanged. Both flip the predicted temperature change, and the physical check catches both: an insulated gas gets hotter when compressed and colder when it expands. A bicycle pump warms up; a spray can chills.

Read the name literally and the algebra follows.

1. Isothermal. $T$ fixed, so for an ideal gas $\Delta U = 0$ and therefore $Q = -W$. An isothermal expansion needs steady heating just to keep the temperature from dropping.
2. Adiabatic. $Q = 0$, so $\Delta U = W$. Here the temperature must change: the process is insulated, not isothermal.
3. Isovolumetric. $\Delta V = 0$, so $W = 0$ and $\Delta U = Q$. A vertical line on a $PV$ diagram encloses no area.
4. Isobaric. $P$ fixed, so $W = -P\,\Delta V$ outright.

The costly confusion is between the first two. "The temperature is standing still" gets heard as "nothing is being transferred", which is exactly backward: the temperature stands still because the heating is matching the work. Before touching a number, write down which of $Q$, $W$ and $\Delta U$ the named process sets to zero.

For an ideal gas the internal energy is fixed by the temperature alone:

$$U = \tfrac{3}{2}nRT \ \text{(monatomic)}, \qquad \text{and} \ T \ \text{follows} \ PV = nRT.$$

So $\Delta U$ between two points on a $PV$ diagram is the same for every path connecting them, and you can read it off the endpoints without knowing the route at all. $Q$ and $W$ are the path-dependent pair, and the first law tells you their split: pick the path, get $W$ from the area, and $Q$ is whatever it takes to reach the $\Delta U$ the endpoints already fixed.

Around a closed cycle the gas returns to its starting state, so $\Delta U = 0$. That does not make $Q$ and $W$ zero; it forces $Q_{\text{net}} = -W_{\text{net}}$, and the area enclosed by the cycle says how large both are. An engine is exactly this: a loop that comes back to where it started and does net work every time around.

The magnitude of the work is the area under the path on a $PV$ diagram, between the two volumes. $P\,\Delta V$ equals that area only when $P$ is constant, meaning a horizontal path.

1. Horizontal. $|W| = P\,\Delta V$ directly.
2. Straight and sloped. Use the average of the two pressures. A path from $300$ kPa to $100$ kPa gives $|W| = 200\ \text{kPa} \times \Delta V$, not $300$ and not $100$.
3. Vertical. No width, so no area, so $W = 0$ however far the pressure moved.
4. Curved, such as an isotherm. The area has to come from the graph or from a given value.

Grabbing the starting pressure and multiplying by $\Delta V$ overestimates the work on a falling path and underestimates it on a rising one. Reading the pressure change on a vertical path as work is the same error mirrored, and it is worth naming: a quantity moving a lot is not a quantity doing work.