AP Physics 2 · Topic 9.4

Topic 9.4: The First Law of Thermodynamics

Q: Is the first law of thermodynamics delta-U = Q + W or delta-U = Q - W?

On the AP Physics 2 equation sheet it is written as delta-U = Q + W, paired with W = -P delta-V, and in that pairing W is the work done ON the system. The CED attaches both equations to Topic 9.4, essential knowledge 9.4.B.1.ii and 9.4.B.1.iii. Many textbooks instead write delta-U = Q - W with W meaning the work done BY the gas. Both are internally consistent and give identical answers, because the work on the gas is the negative of the work by the gas. Use the sheet's version on the exam, and translate any other source before substituting numbers.

Q: On the AP Physics 2 exam, is W the work done on the gas or by the gas?

On the gas. The AP Physics 2 equation sheet prints W = -P delta-V, and the minus sign only makes sense if W is the work done on the gas: compressing a gas gives a negative change in volume, and the minus sign turns that into a positive W, which correctly says energy went into the gas. Essential knowledge 9.4.B.1.iii states it directly: W is the work done on a system by a constant or average external pressure that changes the volume of that system.

Q: What is internal energy in AP Physics 2?

Internal energy is the sum of the kinetic energy of the objects that make up a system and the potential energy of the configuration of those objects. That is essential knowledge 9.4.A.1. For an ideal gas the potential term vanishes, because 9.4.A.1.i says the atoms do not interact through conservative forces and their internal structure is not considered, so an ideal gas has no internal potential energy. For an ideal monatomic gas the internal energy is therefore just the total kinetic energy of the atoms, given on the sheet as U = (3/2) n R T = (3/2) N k_B T. Temperature is the only variable there, so two states of the same gas at the same temperature have the same internal energy however they were reached.

Q: Does U = (3/2) n R T work for any gas?

The CED states it for an ideal monatomic gas. Essential knowledge 9.4.A.1.ii says the internal energy of an ideal MONATOMIC gas is the sum of the kinetic energies of the constituent atoms, and prints U = (3/2) n R T = (3/2) N k_B T beneath that sentence. The sheet prints the same expression without the qualifier, so the CED text is the stricter of the two. Diatomic and polyatomic molecules also store energy in rotation and vibration, so their internal energy is larger at the same temperature and the factor of 3/2 does not apply.

Q: What is the difference between an isolated system and a closed system?

AP Physics 2 essential knowledge 9.4.B.1.i says that for an isolated system the total energy is constant: nothing crosses the boundary, so there is nothing to account for. 9.4.B.1.ii covers the closed system, where energy may cross the boundary but matter may not, and says the change in internal energy is the sum of energy transferred to or from the system by heating, or work done on the system, which is delta-U = Q + W. A sealed cylinder heated over a flame is closed but not isolated. Wrap the same cylinder in perfect insulation with rigid walls and it becomes isolated, and delta-U is zero.

Q: What does isovolumetric mean, and is it the same as isochoric?

They are the same process: one at constant volume. Isovolumetric is the word the AP Physics 2 required course content uses, in essential knowledge 9.4.B.3, where the special cases are listed as constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), plus adiabatic. Isochoric is the more common textbook word and appears only once in the whole course description, inside an optional sample instructional activity. Either way the change in volume is zero, so W = -P delta-V is zero and the first law collapses to delta-U = Q.

Q: Why does the CED say the absolute value of the work is the area under a PV curve?

Because an area is always positive, and the sign of the work is not carried by the area. Essential knowledge 9.4.B.2.ii says the ABSOLUTE VALUE of the work done on a gas when the gas expands or compresses equals the area underneath the curve of a plot of pressure versus volume. The area gives the magnitude; the direction of travel along the curve gives the sign. Moving right, toward larger volume, is an expansion, so the work done on the gas is negative. Moving left is a compression, so it is positive.

Unit 9: Thermodynamics15-18% of the multiple-choice section

The first law of thermodynamics is conservation of energy for a system that can be heated, cooled, or worked on. The AP Physics 2 sheet prints it as the change in U equals Q plus W, with W equal to minus P times the change in V. W is the work done on the gas, so compression makes W positive.

AP Physics: Unit 9 (topics 9.4 The First Law of Thermodynamics). AP Physics 2 Unit 9, Topic 9.4. Two learning objectives and twelve essential knowledge statements. 9.4.A asks students to describe the internal energy of a system, supported by 9.4.A.1 (internal energy is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects), 9.4.A.1.i (the atoms in an ideal gas do not interact via conservative forces and internal structure is not considered, so an ideal gas has no internal potential energy), 9.4.A.1.ii (the internal energy of an ideal monatomic gas is the sum of the kinetic energies of its atoms, with U = 3/2 nRT = 3/2 N k_B T), and 9.4.A.2 (changes to internal energy can change the internal structure and behavior of a system without changing the motion of its center of mass). 9.4.B asks students to describe the behavior of a system using thermodynamic processes, supported by 9.4.B.1 (the first law is a restatement of conservation of energy accounting for energy transferred by work, heating, or cooling), 9.4.B.1.i (for an isolated system the total energy is constant), 9.4.B.1.ii (for a closed system the change in internal energy is the sum of energy transferred by heating, or work done on the system, with deltaU = Q + W), 9.4.B.1.iii (the work done on a system by a constant or average external pressure that changes its volume is W = -P deltaV), 9.4.B.2 (pressure-volume graphs represent thermodynamic processes), 9.4.B.2.i (lines of constant temperature on a PV diagram are isotherms), 9.4.B.2.ii (the absolute value of the work done on a gas as it expands or compresses equals the area under the curve of pressure vs. volume), and 9.4.B.3 (special cases include constant volume (isovolumetric), constant temperature (isothermal), constant pressure (isobaric), and adiabatic processes where no energy is transferred through thermal processes). The topic prints no boundary statement. The CED's suggested skills here are 1.C, 2.A, 2.C, and 3.C. Unit 9 carries 15 to 18 percent of the multiple-choice section and a suggested 10 to 16 class periods. Sign convention throughout: the AP Physics 2 equation sheet's deltaU = Q + W with W = -P deltaV, so W is the work done ON the gas and compression makes W positive.

What Topic 9.4 requires

Topic 9.4 carries two learning objectives and twelve essential knowledge statements between them. Counting the boxed statements topic by topic across Unit 9, that is more than any other topic in the unit: 9.3 and 9.6 have eight each, 9.1 has seven, and 9.2 and 9.5 have four each. The unit is 15 to 18 percent of the multiple-choice section and a suggested 10 to 16 class periods.

9.4.A, describe the internal energy of a system.

9.4.A.1 states that the internal energy of a system is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects.
9.4.A.1.i states that the atoms in an ideal gas do not interact with each other via conservative forces, and the internal structure is not considered, and that therefore an ideal gas does not have internal potential energy.
9.4.A.1.ii states that the internal energy of an ideal monatomic gas is the sum of the kinetic energies of the constituent atoms in the gas, with the relevant equation $U = \frac{3}{2} n R T = \frac{3}{2} N k_B T$ .
9.4.A.2 states that changes to a system's internal energy can result in changes to the internal structure and internal behavior of that system without changing the motion of the system's center of mass.

9.4.B, describe the behavior of a system using thermodynamic processes.

9.4.B.1 states that the first law of thermodynamics is a restatement of conservation of energy that accounts for energy transferred into or out of a system by work, heating, or cooling.
9.4.B.1.i states that for an isolated system, the total energy is constant.
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, with the relevant equation $\Delta U = Q + W$ .
9.4.B.1.iii states that 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 $W = -P\Delta V$ .
9.4.B.2 states that pressure-volume graphs, also known as PV diagrams, are representations used to represent thermodynamic processes.
9.4.B.2.i states that lines of constant temperature on a PV diagram are called isotherms.
9.4.B.2.ii states 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.
9.4.B.3 states that special cases of thermal processes depend on the relationship between the configuration of the system, the nature of the work done on the system, and the system's surroundings, and that these include constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), as well as processes where no energy is transferred to or from the system through thermal processes (adiabatic).

Topic 9.4 prints no boundary statement. Within Unit 9, only 9.1, 9.5, and 9.6 do.

The CED's suggested skills here are 1.C (create qualitative sketches of graphs that represent features of a model or the behavior of a physical system), 2.A (derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway), 2.C (compare physical quantities between two or more scenarios or at different times and locations in a single scenario), and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

Read that list carefully. 1.C is sketching, not plotting, and Topic 9.2 is the one that gets 1.B, create quantitative graphs with scales and units. And 2.B, calculate an unknown quantity, is absent while 2.A, derive a symbolic expression, is present. The CED wants an expression and a correctly shaped graph, not primarily a number.

The sign convention, declared once and kept

This is the one place on this topic where a page can quietly poison every answer that follows, so the convention goes here, before any physics.

The AP Physics 2 equation sheet prints the first law as $\Delta U = Q + W$ , and prints $W = -P\Delta V$ . In that pairing, $W$ is the work done ON the system. Both appear in the sheet's Thermal Physics group, which holds eight equations in total, and the CED attaches them to essential knowledge 9.4.B.1.ii and 9.4.B.1.iii.

Everything else falls out of that one decision:

Situation	Sign of $\Delta V$	Resulting sign	Reading
The gas is compressed	negative	$W$ positive	energy enters the gas as work
The gas expands	positive	$W$ negative	the gas gives up energy as work
The volume is fixed	zero	$W$ zero	no work is done on or by the gas
Energy is added by heating	not fixed	$Q$ positive	energy enters the gas
Energy leaves by cooling	not fixed	$Q$ negative	energy leaves the gas

The sentence that keeps it straight: squeeze a gas and you put energy into it, so $W$ is positive. The minus sign in $W = -P\Delta V$ exists precisely so that a negative $\Delta V$ produces a positive $W$ .

The other convention is real and you will meet it. Many textbooks write the first law as $\Delta U = Q - W$ , where $W$ means the work done BY the gas on its surroundings. That version is internally consistent too. It is not wrong, just a different bookkeeping choice, and the two are related by one line:

W_{\text{on}} = -W_{\text{by}}

Substitute that into $\Delta U = Q + W_{\text{on}}$ and you get $\Delta U = Q - W_{\text{by}}$ . Same physics, same numerical answers, opposite sign on one symbol.

What is genuinely wrong is mixing them inside one problem: reading $W$ as positive for an expansion because a source said so, then adding it as though it were work on the gas. That gives an internal energy change with the wrong sign and it still looks plausible. On the exam, use the sheet's version, because the sheet is what you have in front of you. The last worked example below translates a problem written the other way.

What internal energy is, and the ideal-gas simplification (9.4.A.1)

9.4.A.1 gives a general definition: the internal energy of a system is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects. Two terms, kinetic and potential, and both are internal to the system.

9.4.A.1.i then knocks out the second term for the one system AP Physics 2 cares most about. The atoms in an ideal gas do not interact with each other via conservative forces, and the internal structure of those atoms is not considered, so an ideal gas has no internal potential energy at all. What is left is purely kinetic.

9.4.A.1.ii states the result and prints the equation:

U = \frac{3}{2} n R T = \frac{3}{2} N k_B T

Read the qualifier in the CED's own sentence: it is about an ideal monatomic gas. The sheet prints the equation with no qualifier attached, so the CED text is the stricter of the two, and a question about a diatomic gas can catch an answer that reached for $\frac{3}{2}nRT$ out of habit.

Three consequences that show up constantly:

Internal energy depends only on temperature for an ideal gas. Not on volume, not on pressure separately, not on the path taken. Two states at the same temperature have the same $U$ .
Therefore $\Delta T = 0$ implies $\Delta U = 0$ . That one line does most of the work in isothermal problems.
$U$ can be written in terms of $P$ and $V$ . Combine $U = \frac{3}{2}nRT$ with the ideal gas law $PV = nRT$ from Topic 9.2 and you get $U = \frac{3}{2}PV$ , which reads the internal energy straight off a point on a PV diagram.

The connection to Topic 9.1 is worth making explicit. $K_{\text{avg}} = \frac{3}{2}k_B T$ is the energy of one average atom; $U = \frac{3}{2}N k_B T$ is that times the number of atoms. Temperature is the per-atom quantity, internal energy is the total.

Internal energy changes without the system going anywhere (9.4.A.2)

9.4.A.2 is short and easy to skim past: changes to a system's internal energy can result in changes to the internal structure and internal behavior of that system without changing the motion of the system's center of mass.

That sentence is doing a job. AP Physics 1 spent a year treating objects as point particles at their center of mass, tracking the kinetic energy of that point plus the potential energy of its position. Heat a sealed cylinder of gas on a bench and none of those change: it does not move, its center of mass stays put, its kinetic energy stays zero. Yet something clearly happened.

Internal energy is the accounting line for that something. It is the energy of the motion and configuration inside the system boundary, invisible to a center-of-mass description. This is why the first law needs its own statement rather than being a corollary of the work-energy theorem.

If the system model from Topic 2.1 of AP Physics 1 is still fresh, this is the same system-boundary discipline applied to a new energy account, and the work chapter is where the sign habits for $W$ were first built.

Isolated versus closed, the distinction 9.4.B.1 makes and most sources skip

9.4.B.1 says the first law of thermodynamics is a restatement of conservation of energy that accounts for energy transferred into or out of a system by work, heating, or cooling. Three transfer routes, named explicitly.

Then the CED splits into two cases, and the split is the useful part.

9.4.B.1.i, for an isolated system, the total energy is constant. Nothing crosses the boundary at all, so there is nothing to account for. This is the statement that lets you write "total energy before equals total energy after" for a rigid insulated container.

9.4.B.1.ii, 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. Energy may cross the boundary; matter may not. That is the case $\Delta U = Q + W$ describes.

Getting these the right way round matters when a stem says "insulated" or "rigid" and expects you to zero out a term. Insulated kills $Q$ . Rigid kills $W$ , because $\Delta V = 0$ . Isolated kills both, so $\Delta U = 0$ .

Notice also that "closed" in thermodynamics is about matter, not about energy. A closed system is not a sealed-off one. A sealed cylinder of gas being heated over a flame is a closed system all the way through, and its internal energy is changing the whole time.

Work as area under a PV curve, and why the CED says absolute value (9.4.B.2)

9.4.B.2 introduces the representation: pressure-volume graphs, also known as PV diagrams, are used to represent thermodynamic processes. Pressure on the vertical axis, volume on the horizontal, one point per state.

9.4.B.2.i names one family of curves: lines of constant temperature on a PV diagram are called isotherms. For an ideal gas, $PV = nRT$ means an isotherm is a curve of constant $PV$ , so it falls away from the origin as a hyperbola, and isotherms further from the origin are hotter.

9.4.B.2.ii is the one to read word by word: 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.

The CED writes "absolute value" on purpose. An area on a graph is a positive number. The sign of the work is not in the area, it is in the direction you travel along the curve:

Move right, to larger volume: the gas expanded, so $W$ is negative and the area gives its magnitude.
Move left, to smaller volume: the gas was compressed, so $W$ is positive and the area gives its magnitude.
A vertical line: no area, no volume change, no work.

Two consequences follow. Work is path-dependent: two routes between the same pair of states enclose different areas and so involve different work, even though $\Delta U$ matches for both because $U$ depends only on temperature. And a closed cycle returns to its starting state, so $\Delta U = 0$ for the cycle, leaving the net work as the enclosed area with a sign set by the direction of travel.

Because 9.4.B.1.iii defines $W = -P\Delta V$ for a constant or average external pressure, the rectangle shortcut is exact for an isobaric step and usable with an average pressure elsewhere. The PV diagrams guide carries the full step-by-step routine for reading work off a diagram, cycles included, and is the right page to practice on.

The four processes the CED names, and the word it uses (9.4.B.3)

9.4.B.3 lists the special cases: constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), plus processes where no energy is transferred to or from the system through thermal processes (adiabatic). Four names.

Note the vocabulary. The CED's required course content calls the constant-volume case isovolumetric. The word "isochoric", which most textbooks use for the same thing, appears exactly once in the whole AP Physics 2 course description, inside an optional sample instructional activity. They mean the same process; isovolumetric is the word the required content uses, so it is the one to recognize on a stem.

Process	Held constant	What the sheet gives you
Isovolumetric	volume	$\Delta V = 0$ , so $W = 0$ and $\Delta U = Q$
Isobaric	pressure	$W = -P\Delta V$ with $P$ outside the change
Isothermal	temperature	$\Delta T = 0$ , so $\Delta U = 0$ and $Q = -W$
Adiabatic	no thermal transfer	$Q = 0$ , so $\Delta U = W$

Two of those rows deserve a second look.

The isothermal row relies on $U$ depending only on $T$ for an ideal gas, which is 9.4.A.1.ii, not a general truth about all systems. An isothermal expansion has $\Delta U = 0$ , negative $W$ , and therefore positive $Q$ : the gas must absorb energy to hold its temperature up while it pushes outward.

The adiabatic row is the one confused with isothermal, because both feel like "nothing is being added". Adiabatic means no energy crosses by a thermal process, which is $Q = 0$ . It does not mean constant temperature. An adiabatic compression has $W > 0$ , so $\Delta U > 0$ , so the temperature rises even though nothing was heated. That is why the tip of a bicycle pump gets hot.

The CED introduces these four as special cases that depend on the configuration of the system, the nature of the work done on it, and its surroundings, so they are not the only possibilities. A process may be none of the four.

Skill 2.A: derive a symbolic expression, do not reach for a number

Of this topic's four suggested skills, 2.A is the one that changes how you should practice. It asks you to derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway, and the Unit 9 opener flags deriving expressions from fundamental principles as an emphasis for the unit.

In practice that means an answer built from equations, not arithmetic. The pathway usually runs through three moves:

Write the first law for the process, $\Delta U = Q + W$ , and cross out whichever term the process kills.
Replace $\Delta U$ with $\frac{3}{2}nR\Delta T$ or $\frac{3}{2}\Delta(PV)$ , and $W$ with $-P\Delta V$ .
Use $PV = nRT$ to trade any variable you were not given for one you were.

Skill 1.C pairs with it: sketch the process on a PV diagram before writing anything. The sketch fixes the direction of travel, which fixes the sign of $W$ , which is where most derivations go wrong. Skill 3.C then asks you to justify the result, and a derived expression is its own evidence: pointing out that $\Delta V$ is negative in your expression, so $W$ came out positive, beats restating the conclusion. The first worked example below is a pure 2.A derivation, with numbers used only as a check at the end.

How Topic 9.4 is tested, and where it leads

The task types recur.

Given $Q$ and $W$ , or a process and one of them, find $\Delta U$ and say whether the temperature rose or fell (skill 2.C).
Sketch a described process, or a cycle, on a PV diagram, and mark the direction of travel (skill 1.C).
Read the magnitude of work off an area and attach the sign from the direction of travel.
Derive a symbolic relationship for a named process, such as $Q$ in terms of $P$ and $\Delta V$ for an isobaric step (skill 2.A).
Compare two processes between the same pair of states and explain why $\Delta U$ matches while $Q$ and $W$ do not.
Explain in words why an adiabatic compression raises the temperature, or why an isothermal expansion requires heating (skill 3.C).

Two counted details from the CED itself. AP Classroom's Progress Check for Unit 9 lists about 18 multiple-choice questions and 4 free-response questions. And of the five optional sample instructional activities the CED prints for this unit, three are attached to Topic 9.4 and the other two to Topic 9.2, with none for the remaining four topics.

Where it leads: Topic 9.5 supplies the material properties that turn $Q$ into a temperature change, through $Q = mc\Delta T$ . Topic 9.6 asks the question the first law cannot answer, which is why some energy-conserving processes never happen in reverse, and answers it with entropy and the second law. Backwards, Topic 9.3 defines the heating and cooling that $Q$ measures, and Topic 9.2 supplies the ideal gas law that every substitution above depends on.

Outside the unit, the first law is conservation of energy from AP Physics 1 with one extra account added. The PV diagrams guide is the calculation drill for this topic, and /formulas/ap-physics-2 prints the whole sheet.

Derive the heat transferred in an isobaric process (skill 2.A)

A fixed quantity of ideal monatomic gas is taken through a process at constant pressure $P$ , during which its volume changes by $\Delta V$ . Derive an expression for the energy $Q$ transferred by heating, in terms of $P$ and $\Delta V$ only. Then evaluate it for $P = 4.0 \times 10^4\ \mathrm{Pa}$ and an expansion of $\Delta V = +2.5 \times 10^{-3}\ \mathrm{m^3}$ .

Declare the convention first: $\Delta U = Q + W$ with $W = -P\Delta V$ , both from the sheet, so $W$ is the work done on the gas. Every sign below follows from that.
Start from the internal energy of an ideal monatomic gas, $U = \frac{3}{2} n R T$ , and eliminate $nRT$ using the ideal gas law $PV = nRT$ . That gives $U = \frac{3}{2} P V$ .
Take the change. The pressure is constant, so it comes outside: $\Delta U = \frac{3}{2}\Delta(PV) = \frac{3}{2} P \Delta V$ .
Write the work for the same step: $W = -P\Delta V$ .
Rearrange the first law for $Q$ and substitute both results: $Q = \Delta U - W = \frac{3}{2} P \Delta V - (-P \Delta V) = \frac{3}{2} P \Delta V + P \Delta V$ .
$Q = \frac{5}{2} P \Delta V$
Read the expression before evaluating it. For an expansion, $\Delta V > 0$ , so $Q > 0$ : an isobaric expansion always requires heating, which makes sense because the gas both warms up and pushes outward. The factor $\frac{5}{2}$ splits as $\frac{3}{2}$ into internal energy and $1$ into work against the surroundings.
Now the numbers. $P\Delta V = (4.0 \times 10^4)(2.5 \times 10^{-3}) = 100\ \mathrm{J}$ . So $W = -100\ \mathrm{J}$ , negative because the gas expanded; $\Delta U = \frac{3}{2}(100) = +150\ \mathrm{J}$ ; and $Q = \frac{5}{2}(100) = +250\ \mathrm{J}$ .
Check against the first law directly: $Q + W = 250 + (-100) = 150\ \mathrm{J} = \Delta U$ . Consistent.

$Q = \frac{5}{2} P \Delta V$ . For the given values, $W = -100\ \mathrm{J}$ , $\Delta U = +150\ \mathrm{J}$ , and $Q = +250\ \mathrm{J}$ . Of the 250 J supplied, 150 J went into internal energy and raised the temperature, and 100 J left as work done by the gas on its surroundings.

Two different processes, the same change in internal energy (skill 2.C)

A sample of $0.40\ \mathrm{mol}$ of ideal monatomic gas starts at temperature $T_0$ . In process X it is compressed adiabatically, and $750\ \mathrm{J}$ of work is done on it. In process Y, starting from the same state, the gas is held at constant volume while $750\ \mathrm{J}$ is transferred to it by heating. (a) Find $\Delta U$ and $\Delta T$ for each process. (b) Explain what the comparison shows. (c) What would a third process, isothermal, require? Use $R = 8.31\ \mathrm{J/(mol \cdot K)}$ from the sheet.

(a) Process X is adiabatic, so no energy crosses by a thermal process and $Q = 0$ . The first law gives $\Delta U = Q + W = 0 + 750 = +750\ \mathrm{J}$ . The work is positive because the gas was compressed, which is $W = -P\Delta V$ with $\Delta V < 0$ .
Process Y is isovolumetric, the CED's word for constant volume. Then $\Delta V = 0$ , so $W = -P\Delta V = 0$ , and $\Delta U = Q + W = 750 + 0 = +750\ \mathrm{J}$ .
Convert either result to a temperature change with $U = \frac{3}{2} n R T$ , so $\Delta U = \frac{3}{2} n R \Delta T$ and $\Delta T = \dfrac{2 \Delta U}{3 n R}$ . With $3nR = 3(0.40)(8.31) = 9.972$ , that is $\Delta T = \dfrac{2(750)}{9.972} = \dfrac{1500}{9.972} = 150.4\ \mathrm{K}$ , or $150\ \mathrm{K}$ to three significant figures. Checking back the other way, $\frac{3}{2}(0.40)(8.31)(150.42) = 750.0\ \mathrm{J}$ .
(b) The two processes have nothing in common physically. One does 750 J of work with no heating; the other transfers 750 J by heating with no work. Both give $\Delta U = +750\ \mathrm{J}$ and both raise the temperature by 150 K.
That is the point of 9.4.A.1.ii. Internal energy is fixed by temperature alone for an ideal gas, so it is a property of the state, not of the route. $Q$ and $W$ are properties of the route. The first law is the bookkeeping tying two path-dependent quantities to one path-independent one.
(c) An isothermal process has $\Delta T = 0$ , so $\Delta U = 0$ , so $Q = -W$ . If the gas expands isothermally, $W$ is negative, so $Q$ must be positive by the same amount: the gas has to be heated continuously just to hold its temperature steady while it pushes outward. No net change in internal energy, and energy flowing through the whole time.

(a) Both processes give $\Delta U = +750\ \mathrm{J}$ and $\Delta T = +150\ \mathrm{K}$ (150.4 K unrounded). (b) Internal energy depends only on the state, through temperature, so two entirely different routes with the same energy input produce the same change; $Q$ and $W$ depend on the route and here they are completely different. (c) An isothermal process needs $Q = -W$ , so an isothermal expansion must be heated by exactly the energy it gives up as work.

Translating a problem written in the other sign convention

A textbook problem reads: "A gas absorbs 900 J of heat and does 350 J of work on its surroundings. Find the change in internal energy." (a) Solve it in the textbook's convention. (b) Solve it again in the AP Physics 2 sheet's convention and confirm the answers agree. (c) Now solve the reverse case: the gas is compressed with 350 J of work done on it while it releases 900 J by cooling.

(a) The phrase "does work on its surroundings" is the tell. That textbook is using $W$ for the work done BY the gas, and its first law reads $\Delta U = Q - W_{\text{by}}$ . So $\Delta U = 900 - 350 = +550\ \mathrm{J}$ .
(b) Translate before calculating. The sheet's $W$ is the work done ON the gas, and $W_{\text{on}} = -W_{\text{by}}$ , so $W = -350\ \mathrm{J}$ . The gas expanded, so a negative $W$ is exactly what $W = -P\Delta V$ would give. $Q$ needs no translation: "absorbs 900 J of heat" is heating, positive in both conventions, so $Q = +900\ \mathrm{J}$ .
Apply the sheet: $\Delta U = Q + W = 900 + (-350) = +550\ \mathrm{J}$ . The same answer, as it has to be. The physics never depended on the bookkeeping.
Name the trap: reading $W = 350\ \mathrm{J}$ off the stem and dropping it into $\Delta U = Q + W$ without translating gives $+1250\ \mathrm{J}$ . That is wrong by 700 J, it is positive and plausible-looking, and nothing in the arithmetic flags it.
(c) Reverse case, straight into the sheet's convention. Work done on the gas, so $W = +350\ \mathrm{J}$ . Energy released, which is cooling, so $Q = -900\ \mathrm{J}$ . Then $\Delta U = -900 + 350 = -550\ \mathrm{J}$ .
The internal energy fell, so by $U = \frac{3}{2}nRT$ the temperature fell too, even though work was being done on the gas the whole time. Compressing a gas does not guarantee warming it; that follows only when the compression is adiabatic.

(a) $\Delta U = +550\ \mathrm{J}$ from $\Delta U = Q - W_{\text{by}}$ . (b) $\Delta U = +550\ \mathrm{J}$ from $\Delta U = Q + W$ with $W = -350\ \mathrm{J}$ . The conventions agree once $W_{\text{on}} = -W_{\text{by}}$ is applied; skipping the translation gives $+1250\ \mathrm{J}$ , which is the classic error. (c) $\Delta U = -550\ \mathrm{J}$ , so the gas cooled despite being compressed.

Frequently asked questions

Is the first law of thermodynamics delta-U = Q + W or delta-U = Q - W?

On the AP Physics 2 equation sheet it is written as delta-U = Q + W, paired with W = -P delta-V, and in that pairing W is the work done ON the system. The CED attaches both equations to Topic 9.4, essential knowledge 9.4.B.1.ii and 9.4.B.1.iii. Many textbooks instead write delta-U = Q - W with W meaning the work done BY the gas. Both are internally consistent and give identical answers, because the work on the gas is the negative of the work by the gas. Use the sheet's version on the exam, and translate any other source before substituting numbers.

On the AP Physics 2 exam, is W the work done on the gas or by the gas?

On the gas. The AP Physics 2 equation sheet prints W = -P delta-V, and the minus sign only makes sense if W is the work done on the gas: compressing a gas gives a negative change in volume, and the minus sign turns that into a positive W, which correctly says energy went into the gas. Essential knowledge 9.4.B.1.iii states it directly: W is the work done on a system by a constant or average external pressure that changes the volume of that system.

What is internal energy in AP Physics 2?

Internal energy is the sum of the kinetic energy of the objects that make up a system and the potential energy of the configuration of those objects. That is essential knowledge 9.4.A.1. For an ideal gas the potential term vanishes, because 9.4.A.1.i says the atoms do not interact through conservative forces and their internal structure is not considered, so an ideal gas has no internal potential energy. For an ideal monatomic gas the internal energy is therefore just the total kinetic energy of the atoms, given on the sheet as U = (3/2) n R T = (3/2) N k_B T. Temperature is the only variable there, so two states of the same gas at the same temperature have the same internal energy however they were reached.

Does U = (3/2) n R T work for any gas?

The CED states it for an ideal monatomic gas. Essential knowledge 9.4.A.1.ii says the internal energy of an ideal MONATOMIC gas is the sum of the kinetic energies of the constituent atoms, and prints U = (3/2) n R T = (3/2) N k_B T beneath that sentence. The sheet prints the same expression without the qualifier, so the CED text is the stricter of the two. Diatomic and polyatomic molecules also store energy in rotation and vibration, so their internal energy is larger at the same temperature and the factor of 3/2 does not apply.

What is the difference between an isolated system and a closed system?

AP Physics 2 essential knowledge 9.4.B.1.i says that for an isolated system the total energy is constant: nothing crosses the boundary, so there is nothing to account for. 9.4.B.1.ii covers the closed system, where energy may cross the boundary but matter may not, and says the change in internal energy is the sum of energy transferred to or from the system by heating, or work done on the system, which is delta-U = Q + W. A sealed cylinder heated over a flame is closed but not isolated. Wrap the same cylinder in perfect insulation with rigid walls and it becomes isolated, and delta-U is zero.

What does isovolumetric mean, and is it the same as isochoric?

They are the same process: one at constant volume. Isovolumetric is the word the AP Physics 2 required course content uses, in essential knowledge 9.4.B.3, where the special cases are listed as constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), plus adiabatic. Isochoric is the more common textbook word and appears only once in the whole course description, inside an optional sample instructional activity. Either way the change in volume is zero, so W = -P delta-V is zero and the first law collapses to delta-U = Q.

Why does the CED say the absolute value of the work is the area under a PV curve?

Because an area is always positive, and the sign of the work is not carried by the area. Essential knowledge 9.4.B.2.ii says the ABSOLUTE VALUE of the work done on a gas when the gas expands or compresses equals the area underneath the curve of a plot of pressure versus volume. The area gives the magnitude; the direction of travel along the curve gives the sign. Moving right, toward larger volume, is an expansion, so the work done on the gas is negative. Moving left is a compression, so it is positive.