Work done on a system is energy transferred to a gas by mechanical means, like a piston compressing it. For a constant or average external pressure, W = -PΔV, so compression (negative ΔV) means positive work on the gas, which raises its internal energy in the first law, ΔU = Q + W.
Work done on a system is one of the two ways energy crosses a system's boundary in thermodynamics (the other is heat, Q). When an external force pushes on a gas and changes its volume, think of a piston squeezing gas in a cylinder, energy flows into the gas as work. For a constant or average external pressure, the CED gives you W = -PΔV.
That negative sign is doing real physics, not decoration. If the gas is compressed, ΔV is negative, so W comes out positive. Positive W means energy went INTO the gas, and its internal energy goes up (the gas particles speed up, temperature rises). If the gas expands, ΔV is positive, W is negative, and the gas lost energy by pushing its surroundings outward. AP Physics 2 uses the "work done ON the system" convention everywhere, so the first law is written ΔU = Q + W with a plus sign.
This term lives in Topic 9.4, The First Law of Thermodynamics, in Unit 9. It directly supports learning objective AP Physics 2 Revised 9.4.B, describing the behavior of a system using thermodynamic processes, and it feeds into 9.4.A because work changes a system's internal energy. The first law is just conservation of energy bookkeeping, and W is one of the two entries in the ledger. If you can't get the sign of W right, every ΔU = Q + W calculation falls apart. Almost every P-V diagram problem, thermodynamic cycle question, and isothermal/adiabatic/isobaric process comparison in Unit 9 quietly tests whether you know what W means and which way the energy flowed.
Keep studying AP® Physics 2 Unit 9
Work done by a gas (Unit 9)
Same physical process, opposite bookkeeping. Work done BY the gas is just the negative of work done ON the gas. When a gas expands, it does positive work on the surroundings, which means W (on the system) is negative. AP Physics 2's equation sheet uses the "on the system" convention, so translate any "work done by" question into this language before plugging in.
ΔU = Q + W, the first law (Unit 9)
Work done on a system is the W in the first law. The first law is conservation of energy restated for thermodynamics, and W is the mechanical channel for energy transfer while Q is the thermal one. A practice question like "how does work done on a system affect its internal energy?" has a one-line answer because of this equation. Positive W raises U directly.
P-V diagram (Unit 9)
On a pressure-volume graph, the magnitude of the work is the area under the curve. W = -PΔV is the constant-pressure case, which shows up as a horizontal line on the diagram. Direction matters too. Moving left (compression) means positive work on the gas, moving right (expansion) means negative work on the gas.
Thermodynamic cycle (Unit 9)
In a full cycle the gas returns to its starting state, so ΔU = 0 over the loop. That forces Q = -W for the whole cycle, and the net work corresponds to the area enclosed by the loop on the P-V diagram. Cycle problems are really just work-done-on-a-system problems applied leg by leg.
Expect multiple-choice stems that hand you a P-V diagram or describe a piston and ask for the work done, the sign of the work, or the resulting change in internal energy. Fiveable practice questions hit exactly these angles, asking how work done on a system affects internal energy and what its mathematical expression is (W = -PΔV for constant or average external pressure). On free-response questions, this concept shows up inside first-law analyses. You might justify why a compressed gas heats up, calculate W from the area under a P-V curve, or argue qualitatively whether U increases or decreases during a process. The most common point lost is a sign error, so always state your convention. In AP Physics 2, W is the work done ON the gas, and ΔU = Q + W.
These are the same energy transfer viewed from opposite sides, and they differ only by a sign. Work done ON the system is positive during compression (energy enters the gas). Work done BY the gas is positive during expansion (the gas spends energy pushing outward). The AP Physics 2 equation sheet uses the "on the system" convention, ΔU = Q + W with W = -PΔV. Some textbooks write ΔU = Q - W using the "by the gas" convention, which is the classic source of sign-error chaos. On the exam, stick with W = work done on the gas.
Work done on a system is energy transferred to the system by mechanical means, like a piston compressing a gas.
For a constant or average external pressure, the work done on the system is W = -PΔV, so compression gives positive W and expansion gives negative W.
Positive work done on a gas increases its internal energy through the first law, ΔU = Q + W.
On a P-V diagram, the magnitude of the work equals the area under the process curve, and the direction of the process tells you the sign.
AP Physics 2 always defines W as work done ON the system, which is why the first law appears as ΔU = Q + W rather than ΔU = Q - W.
In a constant volume process, ΔV = 0, so no work is done on the system and any change in internal energy comes entirely from heating or cooling.
It's energy transferred to a system through mechanical means, such as an external pressure compressing a gas. For constant or average external pressure, it's calculated as W = -PΔV, and it appears as the W in the first law of thermodynamics, ΔU = Q + W.
Positive. Compression means ΔV is negative, and since W = -PΔV, the work done on the gas comes out positive. Energy flows into the gas, so its internal energy and temperature increase if no heat escapes.
They're opposite signs of the same energy transfer. Work done on the gas is positive when the gas is compressed; work done by the gas is positive when the gas expands. AP Physics 2 uses the "on the gas" convention, which is why its first law is ΔU = Q + W.
Because the AP equation sheet defines W as work done ON the system, not by it. With that convention, positive W adds energy to the gas, so it's added to Q. The ΔU = Q - W version you might see in other books defines W as work done BY the gas instead.
No. With ΔV = 0 in a constant volume process, W = -PΔV gives zero work. Any change in internal energy during a constant volume process comes entirely from heating or cooling, so ΔU = Q.
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