Mistake Master
Student view — seeing the site as a student does
CED objectives

The First Law of Thermodynamics

▶︎  Watch it animatedinteractive step-through · ~3 min · optional ⚙︎  Open the appletPV Path Lab · choose the path between two states and read the internal energy change off the endpoints, the work off the area, and what heating did off the first law

The first law is $\Delta U = Q + W$ with $W$ the work done on the gas, so compression carries a positive $W$ and an insulated gas warms when it is squeezed. Each named process pins one quantity: isothermal fixes $T$ and so $\Delta U = 0$ and $Q = -W$; adiabatic fixes $Q = 0$ and so $\Delta U = W$; isovolumetric kills the work; isobaric makes $W = -P\Delta V$ exact. Internal energy is a state quantity, fixed by $U = \tfrac{3}{2}nRT$ for a monatomic ideal gas, so $\Delta U$ depends only on the endpoints while $Q$ and $W$ depend on the path. The magnitude of the work is the area under that path.

Five errors dominate. Signing the work as though compression drained energy from the gas. Setting $Q = 0$ for an isothermal process, or expecting no temperature change in an adiabatic one. Letting $\Delta U$ depend on the route, or concluding that a closed cycle with $\Delta U = 0$ must have zero net work and zero net heat. Using $P\Delta V$ with a single pressure on a sloped path, or crediting a vertical path with work because the pressure moved. And assuming energy transferred in by heating has to show up as a temperature rise, when an isothermal expansion spends every joule of it on work.

same endpoints, same ΔU, different W and different Q P V A B path 1: drop P first, then expand path 2: expand first, then drop P ΔU is read off A and B alone: U = (3/2)nRT, and T comes from PV the two shaded areas differ, so W differs, so Q differs by exactly as much
Only the shaded areas change between the two routes. Whatever the path does to W, the required Q moves the opposite way and the total lands on the same number.
read the name, write down what it zeroes P V isobaric isovolumetric isothermal adiabatic (steeper) isobaric: W = −PΔV exactly isovolumetric: W = 0, ΔU = Q isothermal: ΔU = 0, Q = −W adiabatic: Q = 0, ΔU = W the two blue-violet curves are the pair that get swapped an isothermal expansion needs steady heating; an adiabatic expansion cools the gas
The isothermal and adiabatic curves both fall to the right and mean opposite things about Q. The steeper one is the insulated case.

The work

3 ways in · any order
Lesson
The First Law of Thermodynamics

Fixes the first-law sign convention against the physical check that compression warms an insulated gas, sorts the four named processes by what each one zeroes, and treats work as an area rather than a product.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: flipping the work sign on a compression, swapping isothermal for adiabatic, letting the internal energy change depend on the path, using one pressure on a sloped path, and expecting every joule transferred in to raise the temperature. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not started · 10 items · ~15 min
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Take the diagnostic to identify your misconceptions