Heat vs Work: The Difference in Thermodynamics

Heat and work are two ways of moving energy across the boundary of a system. Heat moves it through a temperature difference, by conduction, convection or radiation. Work moves it through a force acting over a distance. On the AP Physics 2 sheet both are positive when energy goes in.

AP Physics: Unit 9 (topics 9.3 Thermal Energy Transfer and Equilibrium, 9.4 The First Law of Thermodynamics). The sign convention on this page is the AP Physics 2 equation sheet's own: its Thermal Physics panel prints the first law as the change in internal energy equalling Q plus W, prints the pressure-volume work as W equals minus P times delta V, and defines Q in its symbol list as the energy transferred to a system by heating and W as the work done on a system. Essential knowledge 9.4.B.1.ii matches that wording, and 9.4.B.1.iii defines the work as done on a system by a constant or average external pressure that changes the volume of that system, with a piston compressing a gas as its example. So both terms are positive when energy enters, and this is the opposite of the common textbook form in which W is the work done by the gas. Heat's mechanism comes from Topic 9.3: 9.3.A.2 lists conduction, convection and radiation, and 9.3.A.3 fixes the spontaneous direction from higher to lower temperature. 9.4.B.1 frames the first law as a restatement of conservation of energy accounting for energy transferred into or out of a system by work, heating, or cooling; 9.4.B.2.ii gives the absolute value of the work as the area under a pressure against volume curve; and 9.4.B.3 names the special cases constant volume (isovolumetric), constant temperature (isothermal), constant pressure (isobaric), and adiabatic. The CED's appendix, in its Heat vs. Heating vs. Cooling section, supplies the parallel: heat is a thermodynamic analog to work and an object cannot have heat any more than an object may have work. Neither Topic 9.3 nor Topic 9.4 prints a boundary statement. Unit 9 carries 15 to 18 percent of the multiple-choice section across a suggested 10 to 16 class periods.

The sign convention, stated before anything else

This page is about a pair of quantities whose signs are the entire content, and half the textbooks in print use the opposite convention from the AP exam. So the convention comes first, and everything on this page obeys it.

The AP Physics 2 equation sheet's Thermal Physics panel prints the first law as

\Delta U = Q + W

and prints the pressure-volume work as

W = -P\Delta V

and its symbol list on the same panel defines the two letters explicitly: $Q$ is the energy transferred to a system by heating, and $W$ is the work done on a system. Essential knowledge 9.4.B.1.ii is worded to match, saying 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. 9.4.B.1.iii defines the work as done on a system by a constant or average external pressure that changes the volume of that system, and gives the piston compressing a gas as its example.

So both terms are positive when energy enters the system, and the plus sign between them is not a typographical accident. It is what makes the two quantities parallel.

The consequences you will use:

Compression means $\Delta V < 0$ , so $W = -P\Delta V$ is positive. Work is done on the gas and its internal energy rises.
Expansion means $\Delta V > 0$ , so $W$ is negative. The gas does work on its surroundings and its internal energy falls.
Energy in by heating is $Q > 0$ ; energy out by cooling is $Q < 0$ .

Many textbooks write the first law as $\Delta U = Q - W$ , where their $W$ is the work done by the gas. Both versions describe the same physics and give the same answer, because the two definitions of $W$ differ by exactly a minus sign. What produces wrong answers is picking one equation and the other definition. Write down which $W$ you are computing before you compute it, and use the AP sheet's version on the exam.

The distinction, stated once

Both are transfers, not properties. The CED's appendix on vocabulary makes them explicitly parallel: it says that in physics the term heat has a very specific definition that is a thermodynamic analog to work, and that similar to the amount of work done, heat is the amount of energy transferred into or out of a thermodynamic system through thermal processes such as conduction, convection, or radiation. It then adds that an object cannot have heat any more than an object may have work.

So the difference is not in what they are. It is in how the energy crosses the boundary.

Heat crosses through a thermal process. 9.3.A.2 lists all three: conduction, convection, and radiation. What drives it is a temperature difference, and 9.3.A.3 fixes the direction, spontaneously from the higher-temperature system to the lower-temperature one.
Work crosses through a force acting over a distance. In Unit 9 that force is a pressure on a moving boundary, which is why 9.4.B.1.iii's example is a piston. What drives it is whatever is pushing the piston, and there is no temperature difference in the story at all.

That is why an insulated cylinder can still change a gas's temperature. Wrap the walls so no thermal process is available, push the piston in, and the internal energy rises with $Q = 0$ throughout. The reverse also works: hold the volume fixed so no work is possible and put the container on a hot plate, and the internal energy rises with $W = 0$ . Same destination, two different doors.

And that is the reason the first law needs two terms rather than one. 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 words, two terms, one equation.

Heat against work, row by row

Property	Heat, $Q$	Work, $W$
Definition on the AP sheet	energy transferred to a system by heating	work done on a system
Mechanism	conduction, convection or radiation (9.3.A.2)	a force acting over a distance, in Unit 9 a pressure on a moving boundary
What drives it	a temperature difference (9.3.A.3)	anything that moves the boundary
Printed equations	$Q = mc\Delta T$ , $Q/\Delta t = kA\Delta T/L$	$W = -P\Delta V$ , and in the mechanics panel $W = F_{\parallel}d = Fd\cos\theta$
Sign when energy enters	positive	positive
Zero when	the system is insulated, or at the same temperature as its surroundings	the volume does not change
Name for the process where it is zero	adiabatic (9.4.B.3)	isovolumetric, also called isochoric (9.4.B.3)
On a PV diagram	not shown	the area under the curve gives its magnitude (9.4.B.2.ii)
Can a system possess it	no	no
Depends on the path taken	yes	yes

Rows five and nine are the two that matter most and they pull in opposite directions from a student's intuition. The signs are the same, both positive inward, which is easy to get wrong because the mechanics convention for work and the common textbook thermodynamics convention disagree. And neither is a stored quantity, which means neither has a value until you name a process.

Row eight is worth a caution. 9.4.B.2.ii says 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. It gives you the magnitude only. The sign comes from the direction of travel along the curve, not from the area, and $W = -P\Delta V$ is the shortcut that only works when the pressure is constant. The PV diagram guide runs the procedure for the general case and for cycles.

Row ten is the deepest one. Internal energy is a property of a state: 9.4.A.1.ii gives $U = \frac{3}{2}nRT$ for an ideal monatomic gas, so knowing $n$ and $T$ fixes $U$ completely, with no reference to how the gas got there. Neither $Q$ nor $W$ works like that, which is exactly what the next section demonstrates.

The case that separates them: one final state, two different transfers

Take $0.50\ \text{mol}$ of a monatomic ideal gas at $300\ \text{K}$ and raise its internal energy by $300\ \text{J}$ , twice, by two different routes.

Route A, a rigid container on a hot plate. The volume cannot change, so $\Delta V = 0$ and $W = -P\Delta V = 0$ . The first law becomes $\Delta U = Q$ , so $Q = +300\ \text{J}$ . All of it arrived by heating.

Route B, an insulated cylinder with a piston pushed in. No thermal process is available, so $Q = 0$ . The first law becomes $\Delta U = W$ , so $W = +300\ \text{J}$ . All of it arrived as work done on the gas, and the plus sign is right because compression makes $\Delta V$ negative and $W = -P\Delta V$ positive.

Now compare the end states. Both gases have the same $n$ and the same $\Delta U$ , so by 9.4.A.1.ii both have the same temperature: $\Delta T = \Delta U / (\frac{3}{2}nR) = 300/6.2325 = 48.1\ \text{K}$ , giving $348\ \text{K}$ in each case. Measure the temperature afterwards and the two gases are indistinguishable.

Measure the transfers and they have nothing in common. Route A had $Q = 300\ \text{J}$ and $W = 0$ ; route B had $Q = 0$ and $W = 300\ \text{J}$ .

That is the separation, and it says two things at once.

First, heat and work are interchangeable as far as the system is concerned. The gas keeps no record of which door the energy came through. This is why they appear symmetrically in $\Delta U = Q + W$ with the same sign.

Second, neither one is a property of the gas. $\Delta U$ is determined by the initial and final states alone. $Q$ and $W$ are not: they depend on the process, and asking for "the heat of the gas at 348 K" is asking for a number that does not exist. The CED's appendix makes the same point in words when it says an object cannot have heat any more than an object may have work.

One caveat on the comparison. The two routes do not end at the same volume or the same pressure, only at the same temperature and internal energy. Route A held the volume fixed and let the pressure rise; route B reduced the volume. The AP Physics 2 sheet prints no relation for the shape of an adiabatic curve, so route B's final pressure and volume cannot be pinned down from the given information, and a question would have to supply them.

When it costs a mark

Using $\Delta U = Q - W$ with the AP sheet's $W$ . The sheet's $W$ is work done on the system, so it belongs in $\Delta U = Q + W$ . Combining the textbook equation with the AP definition double-counts the minus sign and flips the answer. If a question gives you "the work done by the gas", negate it before substituting.
Getting the sign of a compression wrong. Compression means the volume decreases, so $\Delta V = V_f - V_i$ is negative, so $W = -P\Delta V$ is positive. Work done on the gas is positive and its internal energy rises. Compute $\Delta V$ as a signed number before touching the work equation.
Reading the area under a PV curve as signed. 9.4.B.2.ii gives the absolute value only. An expansion and a compression between the same two states have the same area and opposite signs for $W$ .
Using $W = -P\Delta V$ when the pressure changes. 9.4.B.1.iii specifies a constant or average external pressure. On an isotherm the pressure varies continuously, so the work comes from the area instead.
Calling any energy transfer heat. 9.3.A.2's list of thermal processes is conduction, convection and radiation. Energy delivered by a moving piston is not on that list and is not $Q$ , which is why an insulated compression has $Q = 0$ and a rising temperature at the same time.
Saying an insulated system cannot change temperature. It cannot exchange heat. It can still be worked on, and $\Delta U = W$ then puts every joule into the internal energy, which for an ideal gas means straight into the temperature.
Treating $Q$ and $W$ as properties of a state. They are properties of a process. A question asking for the heat transferred has to specify the path, and one asking for the change in internal energy does not, because 9.4.A.1.ii makes $U$ a function of $n$ and $T$ alone.

Reading the wording of a question into signs

Most lost marks on this pair are translation errors rather than physics errors, so it is worth having the phrasebook written out.

The question says	Then
the gas absorbs $400\ \text{J}$ of energy	$Q = +400\ \text{J}$
the gas releases $400\ \text{J}$	$Q = -400\ \text{J}$
the gas is compressed by an external pressure	$\Delta V < 0$ , so $W > 0$
the gas expands against the atmosphere	$\Delta V > 0$ , so $W < 0$
$500\ \text{J}$ of work is done on the gas	$W = +500\ \text{J}$
the gas does $500\ \text{J}$ of work	$W = -500\ \text{J}$
the container is rigid or the volume is fixed	$W = 0$ , so $\Delta U = Q$
the container is insulated, or the process is adiabatic	$Q = 0$ , so $\Delta U = W$
the temperature is unchanged and the gas is ideal	$\Delta U = 0$ , so $Q = -W$
the process is a complete cycle	$\Delta U = 0$ , so $Q = -W$ over the whole cycle

The last two rows land on the same relationship for different reasons, and both come from 9.4.A.1.ii. An isothermal process has $\Delta U = 0$ because the temperature is unchanged and the internal energy of an ideal gas depends only on temperature. A cycle has $\Delta U = 0$ because it returns to its starting state. Isothermal vs adiabatic is built on the first of those.

One more phrase to watch, because it is the one that does not translate. "The gas is heated" tells you $Q > 0$ . "The gas gets hotter" tells you $\Delta T > 0$ and therefore $\Delta U > 0$ , which is a statement about the internal energy and says nothing about $Q$ on its own. Those two sentences look interchangeable in English and are different givens in physics.

When they coincide, and why that lulls you

In the two commonest classroom setups, one of the two terms is zero, so a student can work correctly for weeks without ever needing to keep them apart.

A sealed rigid container has $W = 0$ and everything is $Q$ . An insulated cylinder has $Q = 0$ and everything is $W$ . In both cases $\Delta U$ equals whichever term survives, the sign question never arises, and the two quantities feel like the same quantity wearing different labels.

There is a second, subtler coincidence. Because both terms carry the same sign convention on the AP sheet, adding them is always the right operation, so a student who never thinks about the mechanism still gets $\Delta U$ right. The distinction only produces a different number when a question asks for one of the terms rather than for their sum, which is exactly when it appears on an exam: give $Q$ and $\Delta U$ and ask for $W$ , or give a PV diagram and ask how much energy was transferred by heating.

The third coincidence is the dangerous one, and it is not really a coincidence at all. Both quantities are measured in joules, both can be positive or negative, both are path dependent, and both appear on the right of the same equation with a plus sign. Everything about the mathematics says treat them alike, and the mathematics is right. The one place they part company is the mechanism, and the mechanism is what a question tests when it says "insulated", "rigid", "in thermal contact with a reservoir", or "in a vacuum flask". Those four words are the question. Heat vs temperature covers the other half of the confusion, which is the one that treats heat as a stored quantity in the first place.

One final state, two routes, and the signs that prove it

A cylinder holds $0.50\ \text{mol}$ of a monatomic ideal gas at $300\ \text{K}$ . Take $R = 8.31\ \text{J/(mol} \cdot \text{K)}$ . (a) Find its initial internal energy. (b) Route A: the cylinder is rigid and placed on a hot plate until the internal energy rises by $300\ \text{J}$ . Find $Q$ and $W$ . (c) Route B: the cylinder is insulated and the piston is pushed in until the internal energy rises by the same $300\ \text{J}$ . Find $Q$ and $W$ . (d) Find the final temperature in each case.

(a) 9.4.A.1.ii gives the internal energy of an ideal monatomic gas as $U = \frac{3}{2}nRT$ , and the sheet's exam conventions state that ideal gases are monatomic. So $U_i = \frac{3}{2}(0.50)(8.31)(300)$ .
Compute the coefficient once, because it recurs: $\frac{3}{2}(0.50)(8.31) = (0.75)(8.31) = 6.2325\ \text{J/K}$ . Then $U_i = 6.2325 \times 300 = 1869.75\ \text{J}$ , or $1.87 \times 10^3\ \text{J}$ .
(b) Route A has a rigid cylinder, so $\Delta V = 0$ and the printed $W = -P\Delta V$ gives $W = 0$ . The first law $\Delta U = Q + W$ becomes $\Delta U = Q$ , so $Q = +300\ \text{J}$ . The plus sign says energy entered by heating, which matches a hot plate.
(c) Route B is insulated, so no thermal process is available and $Q = 0$ . That is the adiabatic case named in 9.4.B.3. The first law becomes $\Delta U = W$ , so $W = +300\ \text{J}$ .
Check that sign against $W = -P\Delta V$ independently. Pushing the piston in decreases the volume, so $\Delta V = V_f - V_i < 0$ , so $-P\Delta V > 0$ . Positive $W$ , work done on the gas, internal energy up. The two routes to the sign agree.
(d) Both routes have $\Delta U = +300\ \text{J}$ , and $\Delta U = \frac{3}{2}nR\Delta T$ with the same coefficient as before. So $\Delta T = \dfrac{300}{6.2325} = 48.135\ \text{K}$ , which is $48.1\ \text{K}$ .
The final temperature is $300 + 48.135 = 348.135\ \text{K}$ , or $348\ \text{K}$ to three significant figures, in both cases.
Check the final internal energy directly: $6.2325 \times 348.135 = 2169.75\ \text{J}$ , and $1869.75 + 300 = 2169.75\ \text{J}$ . They agree exactly.
The point of the pair. Two gases now sit at the same temperature with the same internal energy, one of which received $300\ \text{J}$ entirely by heating and the other $300\ \text{J}$ entirely as work. Nothing you can measure about the final state distinguishes them, which is why $Q$ and $W$ are not properties of a state.

(a) $U_i = 1.87 \times 10^3\ \text{J}$ . (b) Route A: $W = 0$ and $Q = +300\ \text{J}$ . (c) Route B: $Q = 0$ and $W = +300\ \text{J}$ , positive because compression makes $\Delta V$ negative in $W = -P\Delta V$ . (d) $348\ \text{K}$ in both cases.

An isobaric expansion, where both terms are nonzero and they disagree in sign

A cylinder holds $0.50\ \text{mol}$ of a monatomic ideal gas. It expands at a constant pressure of $1.0 \times 10^5\ \text{Pa}$ from $1.0 \times 10^{-2}\ \text{m}^3$ to $1.5 \times 10^{-2}\ \text{m}^3$ . Take $R = 8.31\ \text{J/(mol} \cdot \text{K)}$ . Find the work done on the gas, the initial and final temperatures, the change in internal energy, and the energy transferred by heating.

Find the signed volume change first, before any equation: $\Delta V = 1.5 \times 10^{-2} - 1.0 \times 10^{-2} = +5.0 \times 10^{-3}\ \text{m}^3$ . It is positive because the gas expanded.
The pressure is constant, so 9.4.B.1.iii's $W = -P\Delta V$ applies directly: $W = -(1.0 \times 10^5)(5.0 \times 10^{-3}) = -500\ \text{J}$ .
Read that sign. $W$ is negative, meaning negative work is done on the gas, meaning the gas did $500\ \text{J}$ of work on its surroundings as it pushed the piston out. A textbook using the other convention would report its $W$ as $+500\ \text{J}$ and pair it with a minus sign in the first law, arriving at the same physics.
Now the temperatures, from the printed $PV = nRT$ . Initially, $T_i = \dfrac{PV_i}{nR} = \dfrac{(1.0 \times 10^5)(1.0 \times 10^{-2})}{(0.50)(8.31)} = \dfrac{1000}{4.155} = 240.67\ \text{K}$ .
Finally, $T_f = \dfrac{PV_f}{nR} = \dfrac{(1.0 \times 10^5)(1.5 \times 10^{-2})}{4.155} = \dfrac{1500}{4.155} = 361.01\ \text{K}$ . So $\Delta T = +120.34\ \text{K}$ and the gas got hotter while expanding.
The change in internal energy, from 9.4.A.1.ii: $\Delta U = \frac{3}{2}nR\Delta T = 6.2325 \times 120.34 = 750\ \text{J}$ .
Check that against a shortcut, since the pressure was constant: $\Delta U = \frac{3}{2}\Delta(PV) = \frac{3}{2}P\Delta V = 1.5(1.0 \times 10^5)(5.0 \times 10^{-3}) = 750\ \text{J}$ exactly. The two routes agree, so the $120.34\ \text{K}$ is right.
Finally the heating, by rearranging $\Delta U = Q + W$ to $Q = \Delta U - W$ : $Q = 750 - (-500) = +1250\ \text{J}$ .
Sanity-check the energy bookkeeping. $1250\ \text{J}$ entered by heating, $500\ \text{J}$ left as work done on the surroundings, and $750\ \text{J}$ stayed behind as internal energy. $1250 - 500 = 750$ . Nothing is missing.
Notice that the two transfers have opposite signs and different sizes, and that neither one equals $\Delta U$ . This is the ordinary case, and the earlier routes where one term vanished were the special ones.

$W = -500\ \text{J}$ , so the gas did $500\ \text{J}$ of work on its surroundings. $T_i = 241\ \text{K}$ , $T_f = 361\ \text{K}$ , $\Delta U = +750\ \text{J}$ , and $Q = +1250\ \text{J}$ transferred in by heating. The transfers have opposite signs and the internal energy change is the sum.

Translating a textbook-worded question into the AP convention

A question states: "During a process, a gas does $500\ \text{J}$ of work on its surroundings while absorbing $200\ \text{J}$ of energy by heating." (a) Find the change in internal energy using the AP Physics 2 sheet's convention. (b) Show that the textbook convention gives the same answer. (c) Show what answer a student gets by mixing the two, and by how much it is wrong.

(a) Translate the wording first. The gas does work on its surroundings, so the work done on the gas is the negative of that: $W = -500\ \text{J}$ . Absorbing energy by heating gives $Q = +200\ \text{J}$ .
Substitute into the sheet's $\Delta U = Q + W$ : $\Delta U = 200 + (-500) = -300\ \text{J}$ . The internal energy fell, so for an ideal gas the temperature fell.
Check that this is physically sensible. The gas gave away $500\ \text{J}$ and took in only $200\ \text{J}$ , so it is $300\ \text{J}$ poorer. It is.
(b) The textbook form is $\Delta U = Q - W_{\text{by}}$ , where $W_{\text{by}}$ is the work done by the gas, here $+500\ \text{J}$ . So $\Delta U = 200 - 500 = -300\ \text{J}$ .
The two agree exactly, as they must: the two definitions of $W$ differ by a minus sign and the two equations differ by a minus sign, and the two cancel. Neither convention is more correct; they are the same statement written twice.
(c) Now the mixture. A student who takes the AP equation $\Delta U = Q + W$ but substitutes the work done by the gas gets $\Delta U = 200 + 500 = +700\ \text{J}$ .
That is wrong by $700 - (-300) = 1000\ \text{J}$ , and it is wrong in sign as well as size: it says the gas got hotter when in fact it got colder. The gap is exactly $2W_{\text{by}} = 2(500) = 1000\ \text{J}$ , because the work term was entered with the wrong sign rather than merely mis-sized.
The habit that prevents it: write the words "work done ON the gas" next to your $W$ before you substitute, and check that its sign matches the physical direction. Compression gives positive; expansion gives negative.
If the same process had instead been described as " $500\ \text{J}$ of work is done on the gas while it releases $200\ \text{J}$ ", the substitution would be $W = +500\ \text{J}$ and $Q = -200\ \text{J}$ , giving $\Delta U = +300\ \text{J}$ . Same magnitudes, every sign reversed, and only the wording told you which.

(a) $\Delta U = 200 + (-500) = -300\ \text{J}$ . (b) The textbook form gives $200 - 500 = -300\ \text{J}$ , the same answer, because both the definition of $W$ and the sign in the equation are reversed together. (c) Mixing them gives $+700\ \text{J}$ , wrong by $1000\ \text{J}$ and wrong in sign, which turns a cooling gas into a warming one.

Frequently asked questions

What is the difference between heat and work in thermodynamics?

They are two mechanisms for the same thing, moving energy across the boundary of a system. Heat moves it through a thermal process, which AP Physics 2 essential knowledge 9.3.A.2 lists as conduction, convection and radiation, and what drives it is a temperature difference. Work moves it through a force acting over a distance, and in Unit 9 that means an external pressure changing the volume, which is essential knowledge 9.4.B.1.iii. The CED's appendix makes the parallel explicit, calling heat a thermodynamic analog to work and noting that an object cannot have heat any more than an object may have work. Neither is a quantity a system stores; what a system stores is internal energy.

What sign convention does AP Physics 2 use for the first law of thermodynamics?

The change in internal energy equals Q plus W, with the pressure-volume work written as W equals minus P times delta V. The equation sheet's Thermal Physics panel defines Q as the energy transferred to a system by heating and W as the work done on a system, so both terms are positive when energy enters the system. Compression makes delta V negative and therefore makes W positive: work is done on the gas and its internal energy rises. Expansion makes W negative. Essential knowledge 9.4.B.1.ii matches, describing the change in internal energy as the sum of energy transferred by heating and work done on the system.

Why does my textbook write the first law with a minus sign?

Because it defines W as the work done by the gas rather than on it. The two conventions differ by a single minus sign in the definition and a single minus sign in the equation, so they always give the same answer when applied consistently. A gas that does 500 J of work while absorbing 200 J has a change in internal energy of minus 300 J either way: the AP form gives 200 plus negative 500, and the textbook form gives 200 minus 500. What produces wrong answers is taking the AP equation and substituting the work done by the gas, which for those numbers returns plus 700 J instead, wrong in sign as well as size. Write down which work you are computing before you substitute.

Can a system's temperature change without any heat transfer?

Yes, and this is the clearest way to see that heat and work are different mechanisms. Insulate a cylinder so no conduction, convection or radiation is possible, and Q is zero throughout, which is the adiabatic case in essential knowledge 9.4.B.3. Push the piston in and the first law reduces to the change in internal energy equalling the work done on the gas, so every joule of work raises the internal energy and, for an ideal gas, the temperature. The CED's Unit 9 sample activities give the demonstration: cap a syringe and push the plunger hard, an example the CED points at under the name fire syringe.

Is the area under a PV curve the work done on the gas or by the gas?

Neither, on its own. Essential knowledge 9.4.B.2.ii says 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, so the area gives you the magnitude only. The sign comes from the direction of travel: moving right on the diagram is an expansion, so the work done on the gas is negative, and moving left is a compression, so it is positive. Note also that the shortcut W equals minus P times delta V requires a constant or average pressure, per 9.4.B.1.iii, so on a curve where the pressure varies you need the area rather than the shortcut.

Are heat and work path dependent?

Yes, both of them, and that is what separates them from internal energy. Essential knowledge 9.4.A.1.ii gives the internal energy of an ideal monatomic gas as three halves of n R T, so it is fixed by the amount of gas and the temperature with no reference to how the gas arrived. Q and W have no such property. Raise a gas's internal energy by 300 J in a rigid container and Q is 300 J with W zero; do it in an insulated cylinder with a piston and W is 300 J with Q zero. The two gases end up identical and the two sets of transfers have nothing in common.

When is the work done on a gas zero, and when is the heat zero?

The work is zero when the volume does not change, because W equals minus P times delta V. Essential knowledge 9.4.B.3 names that process constant volume, or isovolumetric, and it is also commonly called isochoric; the first law then reduces to the change in internal energy equalling Q. The heat is zero when no energy is transferred to or from the system through thermal processes, which 9.4.B.3 names adiabatic, and the first law then reduces to the change in internal energy equalling W. Those two cases are the reason a student can work correctly for a long time without ever having to keep the two terms apart, since one of them keeps vanishing.