Work Done On vs Work Done By a Gas: AP Signs

The AP sheet settles this, and it settles it one way

Two quantities are in play and they are numerically opposite:

On the AP Physics 2 exam the bare letter $W$ always means the first one. The Table of Information's Thermal Physics symbol key defines it in words, with no ambiguity: $W$ is the work done on a system. Alongside it the key defines $Q$ as energy transferred to a system by heating and $U$ as internal energy. The two equations in that box are

The CED says the same thing in prose. Essential knowledge 9.4.B.1.ii states that for a closed system, the change in internal energy is the sum of energy transferred to or from the system by heating, or work done on the system. EK 9.4.B.1.iii then defines the work term: the work done on a system by a constant or average external pressure that changes the volume of that system, for example a piston compressing a gas in a container, is defined as $W = -P\Delta V$.

Read the minus sign as physics rather than notation. When the gas expands, $\Delta V = V_f - V_i$ is positive, so $W$ is negative: the gas pushed the piston out, spent energy doing it, and lost that energy. When the gas is compressed, $\Delta V$ is negative, so $W$ is positive: something outside pushed energy in. The sign of $W$ tells you which way energy crossed the boundary, and it always reports that from the gas's point of view.

That is the entire page. Everything below is the consequences and the ways students lose the sign anyway.

Work done on the gas vs work done by the gas, side by side

The row worth reading twice is the first-law row, because it is where the two conventions look like a contradiction and are not. Both equations describe the same physics. They differ only in which of the two work quantities the letter $W$ names. Substitute $W_{\text{by}} = -W$ into the second and you recover the first exactly.

What cannot survive is mixing them, and that mixture is the defect this page exists to prevent: computing $W_{\text{by}}$ as a positive number for an expansion and then substituting it into $\Delta U = Q + W$. That produces an internal-energy change wrong by twice the work, and it usually produces a temperature change in the wrong direction as well.

The sign map, three cases and nothing else

Every constant-pressure step falls into one of three cases, and there is no fourth.

The constant-volume row is the one students skip and then need. EK 9.4.B.3 lists the special thermal processes: constant volume, which the CED calls isovolumetric, constant temperature (isothermal), constant pressure (isobaric), and processes with no energy transferred through thermal processes (adiabatic). A constant-volume step does exactly zero work no matter how violently the pressure changes, because $\Delta V = 0$ kills the product. On a PV diagram it is a vertical line, and a vertical line encloses no area.

A mnemonic that survives exam pressure: the gas is the accountant. Ask what happened to the gas's energy. Squeezed, it gained some; expanded, it spent some. Then set the sign of $W$ to match. This beats memorising "expansion is negative" as a bare rule, because the rule is what you forget at minute 68 and the picture is not.

One warning on $W = -P\Delta V$ itself. It applies when the pressure is constant, or when you are handed an average pressure to use. If the pressure varies along the path, that equation is the wrong tool and the area under the curve is the right one, which is the next section.

The area under a PV curve gives the size, never the sign

The CED is precise about this in a way that most sources are not. EK 9.4.B.2.ii says that the absolute value of the work done on a gas when the gas expands or compresses is equal to the area underneath the curve of a plot of pressure vs. volume for the gas.

That wording is doing real work. The area under a curve is a positive number: it is a pressure times a volume, and both axes are positive throughout. So the area answers "how much" and answers nothing else. You supply the sign yourself, from the direction the process runs along the volume axis.

The procedure, in the order that avoids errors:

1. Find the area between the curve and the volume axis, using rectangles, triangles and trapezoids. Its units come out in joules when pressure is in pascals and volume in cubic meters.
2. Look at which way the process travels. Rightward on the diagram means the volume grew, so the gas expanded.
3. Attach the sign: rightward gives $W$ negative, leftward gives $W$ positive.
4. Only now write $\Delta U = Q + W$.

This matters because a sloped or curved path has no single $P$ to feed into $-P\Delta V$. The third worked example below runs a straight sloped path where the pressure falls as the volume grows: the area is a trapezoid, the arithmetic is easy, and the only difficulty in the whole problem is the sign.

For the full procedure on PV diagrams, including multi-step paths and cycles, the PV diagrams and the first law guide owns it. This page owns the question of which of the two work quantities you just calculated.

Why the other convention exists, and why it is not wrong

Look this up outside an AP resource and you will often meet

and a claim that $W$ is the work done by the gas. That version is not an error. It is the older engineering convention, and it exists because thermodynamics grew up around heat engines, where the useful output is the work the gas delivers to the world. Making that output the positive quantity is convenient if your job is to sell engines.

The two conventions agree on every physical prediction. They disagree on one thing only: what the letter $W$ names.

Notice the last row. The convention for $Q$ is the same in both, which is exactly why the mixed-up version is so easy to write: only one of the two terms flips, so a half-remembered rule leaves you with a plausible-looking equation that is wrong.

In the exam room, the AP sheet is the one that counts. It is printed in front of you, it defines $W$ as the work done on a system in its own symbol key, and a reader of your free-response answer marks against that definition. If a homework problem or a video uses the other one, translate it before you use it: read off the work done by the gas, flip its sign, and call the result $W$.

Turning a question's wording into a sign

Exam questions rarely hand you "$W = -3000$ J". They describe what happened. Here is the translation table, with the signs already resolved into the AP convention.

The free-expansion row is a genuine trap: the volume changes, so $\Delta V$ is not zero, yet the work is zero because the external pressure is zero. That is a reminder that $W = -P\Delta V$ carries an external pressure, not the gas's own pressure, which is what EK 9.4.B.1.iii means by "a constant or average external pressure."

Two more habits pay for themselves.

• Write down which quantity you computed, in words, next to the number. "Work done on the gas, $-3000$ J" costs four seconds and makes the rest of the question self-checking.
• Sanity check with the temperature. For an ideal monatomic gas, $U = \frac{3}{2}nRT$, so $\Delta U$ and $\Delta T$ always share a sign. If your signs give a compressed, insulated gas a falling temperature, the work term is backwards.

Where the confusion costs a mark

This is the specific list. Each item is a scoring event, and the first two are the reason this page exists.

• Computing the work done by the gas and substituting it into $\Delta U = Q + W$. For an expansion where the gas does $3000$ J of work and absorbs $5000$ J of heat, the correct answer is $\Delta U = 5000 + (-3000) = +2000$ J. The mixed version gives $5000 + 3000 = +8000$ J, wrong by $6000$ J, which is twice the work. On a free-response question this normally loses the substitution point, the answer point, and every later point that depends on the value.
• Getting the direction of the temperature change wrong. Because $\Delta U$ and $\Delta T$ share a sign for an ideal gas, a flipped work term can turn a cooling gas into a heating one. A reasoning part that asks whether the temperature rises then fails on physics rather than arithmetic, and no partial credit survives it.
• Taking the area under a PV curve as a signed quantity. The area is a magnitude. EK 9.4.B.2.ii says so with the words "the absolute value". Read the direction of travel and attach the sign yourself.
• Using $W = -P\Delta V$ on a step where the pressure is not constant. There is no single $P$ to use. Take the area instead.
• Writing $\Delta V = V_i - V_f$. It is final minus initial, like every other delta on the sheet. Reversing it flips every sign on the page.
• Calling the work zero because the pressure is constant. Constant pressure is the case where $W = -P\Delta V$ is easiest, not the case where it vanishes. Constant volume is the one that vanishes.
• Reporting a bare number when the question asked "by the gas". If a question asks for the work done by the gas, give it the positive $3000$ J, not the sheet's $-3000$ J. Answer the question that was asked, and say which one you answered.
• Assuming a positive $Q$ means the internal energy rose. It only does if the work term does not overwhelm it. An expanding gas can absorb heat and still cool.

What the CED asks, and what the sheet prints

This is AP Physics 2 material and only AP Physics 2 material. A boundary statement under AP Physics 1 Topic 3.2 says that course only expects students to analyze the transfer of mechanical energy, although students should be aware that mechanical energy may be dissipated in the form of thermal energy or sound, and that in AP Physics 2 students will also study how thermal energy can be transferred between systems through heating or cooling. Nothing on this page is examinable in AP Physics 1, and neither Physics C sheet carries a thermal physics box at all.

Unit 9, Thermodynamics is weighted at 15 to 18 percent of the AP Physics 2 multiple-choice section over about 10 to 16 class periods. Topic 9.4, The First Law of Thermodynamics, has two learning objectives. 9.4.A asks you to describe the internal energy of a system and carries EK 9.4.A.1 through 9.4.A.2, including that an ideal gas has no internal potential energy and that the internal energy of an ideal monatomic gas is the sum of the kinetic energies of its atoms, $U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$. 9.4.B asks you to describe the behavior of a system using thermodynamic processes and carries EK 9.4.B.1 through 9.4.B.3, which is where the sign convention, the PV diagram reading and the named processes live. Suggested skills for Topic 9.4 are 1.C, 2.A, 2.C and 3.C. Skill 3.C is justifying a claim with evidence, and a justification built on a flipped sign is worth nothing.

On the AP Physics 2 equation sheet, the Thermal Physics box prints $P = \frac{F_{\perp}}{A}$, $K_{\text{avg}} = \frac{3}{2}k_BT = \frac{1}{2}mv_{\text{rms}}^2$, $\frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$, $PV = nRT = Nk_BT$, $U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$, $W = -P\Delta V$, $\Delta U = Q + W$ and $Q = mc\Delta T$. Its symbol key spells out that $W$ is the work done on a system and that $Q$ is energy transferred to a system by heating. There is no printed equation for the work done by a gas, and no printed statement of $\Delta U = Q - W_{\text{by}}$. If you want that quantity you flip the sign yourself and label it.

Going further: PV diagrams and the first law for the full procedure including cycles, heat vs work for the other half of the first law, and isothermal vs adiabatic for the two processes where one of the two terms is forced to a specific value. The CED framing is on Topic 9.4.