Cycles, PV Loops, and Energy Accounting Quiz

Concept: state function over a cycle. Internal energy U depends only on the state, so returning to the starting state makes ΔU=0. This is true regardless of what happens in between.

Concept: first law for cycles. Over a full cycle, ΔU=0. So 0=Q_net - W_net, which means Q_net = W_net.

Concept: PV loop area equals net work. Net work over a cycle is ∫PdV, which is the signed area inside the loop. Clockwise vs counterclockwise direction sets whether the work is positive or negative.

Concept: Q_net equals W_net in a cycle. For a cycle, ΔU=0, so Q_net=W_net. If net work done by the system is +700 J, the system must have net heat input of +700 J.

Concept: direction of PV loop. A clockwise loop indicates expansion occurs at higher pressure than compression. That yields positive net area and therefore positive net work done by the gas.

Concept: cycle energy balance. Releasing heat overall means Q_net=-200 J. Since Q_net=W_net for a cycle, W_net must also be -200 J.

Concept: algebraic rearrangement of the first law. Starting from ΔU=Q-W, add W to both sides and subtract ΔU from both sides. This gives W=Q-ΔU, useful when solving for work.

Concept: using W=Q-ΔU. Internal energy decreases means ΔU=-200 J. W=Q-ΔU=1500-(-200)=1700 J, so the system did 1700 J of work.

Concept: comparing Q and W in ΔU=Q-W. If W>Q, then Q-W is negative, so ΔU<0. That means internal energy decreases because more energy left as work than entered as heat.

Concept: substituting signs correctly. ΔU=Q-W=-400-(-100)=-400+100=-300 J. Even though compression (negative W) adds energy, the heat loss is larger, so ΔU is negative.

Concept: work sign and path dependence. Expansion typically means ΔV>0, giving W>0, while compression gives ΔV<0 and W<0. Work is path-dependent, and on PV diagrams it is represented by the area under/within curves.

Concept: no PV work at constant volume. If volume never changes, dV=0 throughout the cycle. Therefore W=∫PdV=0 for every step and the net work is zero.

Concept: cycle with ΔU=0. Over a cycle, ΔU=0. If net W=0 as well, then the first law gives Q_net=0.

Concept: solving for Q. Rearrange ΔU=Q-W to Q=ΔU+W. Q=600+900=1500 J, meaning heat input must cover both the internal energy increase and the work output.

Concept: insulation implies adiabatic. An insulated boundary prevents heat transfer between system and surroundings. Therefore Q=0 by definition of an adiabatic condition.

Concept: adiabatic expansion. Insulated means Q=0, and expansion with work done by the gas means W>0. Then ΔU=0-250=-250 J, so internal energy decreases.

Concept: competing effects of Q and W. Releasing heat makes Q negative, which tends to lower U. But if the surroundings do enough work on the system (W very negative), then ΔU=Q-W can still be positive.

Concept: solving for W from ΔU and Q. Use ΔU=Q-W: 40=-80-W. Adding 80 gives 120=-W, so W=-120 J, meaning work was done on the system.

Concept: sanity checks in thermodynamics. Cycles must give ΔU=0 and rigid containers must give W=0, so violating those is a clear setup error. Unrealistic outputs and unit mismatches often indicate sign mistakes or inconsistent conversions.

Concept: comparing magnitudes in ΔU=Q-W. If Q>W, ΔU is positive; if Q=W, ΔU is zero; if Q<W, ΔU is negative. The signs alone don’t determine ΔU—relative size matters.