Isobaric vs Isochoric: The Difference

An isobaric process holds the pressure constant, so the gas changes volume and work is done: the work on the gas is minus the pressure times the volume change. An isochoric process holds the volume constant, so no work is done at all and every joule transferred by heating goes into internal energy.

AP Physics: Unit 9 (topics 9.4 The First Law of Thermodynamics). Both processes are named in AP Physics 2 essential knowledge 9.4.B.3, which reads 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). The CED's required course content uses isovolumetric rather than isochoric; the word isochoric appears in the AP Physics 2 CED only in an optional sample instructional activity for Topic 9.4, which gives isochoric heating, isothermal expansion, isobaric cooling as an example cycle. A neighboring activity refers to the eight thermodynamic processes as isobaric, isovolumetric, isothermal, and adiabatic, each in both possible directions. The supporting content is 9.4.B.1.ii, which gives the first law as delta U equals Q plus W and describes it as the change in internal energy of a closed system being the sum of energy transferred to or from the system by heating, or work done on the system; 9.4.B.1.iii, which defines the work done on a system by a constant or average external pressure that changes the volume of that system as W equals minus P delta V, with a piston compressing a gas as its example; 9.4.B.2, on PV diagrams as representations of thermodynamic processes, with 9.4.B.2.i naming isotherms and 9.4.B.2.ii giving the absolute value of the work as the area underneath the curve of a plot of pressure vs. volume; and 9.4.A.1.ii, which gives the internal energy of an ideal monatomic gas as U equals three halves n R T equals three halves N k sub B T. The exam conventions box on the AP Physics 2 equation sheet states that ideal gases are monatomic, which is what fixes the three fifths and five thirds ratios on this page. No molar heat capacity is printed on that sheet, so both are derived here from the printed first law, work, gas law and internal energy equations. Topic 9.4 carries no boundary statement, and neither do Topics 9.2 and 9.3. Unit 9 prints three boundary statements in all: under Topic 9.1, that AP Physics 2 only expects students to perform qualitative and quantitative analysis of collisions in one and two dimensions and that students are not expected to know the functional form of the Maxwell-Boltzmann distribution but are expected to be familiar with how features of the distribution are related to the temperature of the gas; under Topic 9.5, that AP Physics 2 will model specific heat as independent of temperature; and under Topic 9.6, that only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2. Unit 9 is weighted at 15 to 18 percent of the multiple-choice section across a suggested 10 to 16 class periods, and the suggested skills listed for Topic 9.4 are 1.C, 2.A, 2.C and 3.C.

The distinction, stated once

One of these processes does work and the other cannot, and that single fact decides every question about the pair.

Isobaric means constant pressure. The volume is free to change, so the boundary moves, so work crosses it. Essential knowledge 9.4.B.1.iii defines 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, as

W = -P\Delta V

In an isobaric process this is not an approximation. The pressure genuinely is constant, so it comes straight out of the product with no averaging required.

Isochoric means constant volume. The boundary does not move. Substituting $\Delta V = 0$ into the same printed equation gives $W = 0$ , exactly and always, and the first law collapses:

\Delta U = Q

Every joule that arrives by heating becomes internal energy, one for one, and every joule of internal energy the gas loses left by cooling. There is nowhere else for it to go.

So the difference is not really about which variable is held fixed. It is about whether the work term exists. Isochoric is the process that switches one of the two terms in $\Delta U = Q + W$ off, and isobaric is the process that makes the surviving term as easy to compute as it ever gets.

A note on names before anything else. The CED's required course content calls the constant-volume case isovolumetric, and that is the word printed in essential knowledge 9.4.B.3. The word isochoric appears in the AP Physics 2 CED only inside an optional sample instructional activity, never in a learning objective or a piece of essential knowledge. The two words mean the same thing and both are standard, but if you are quoting the framework, isovolumetric is the CED's term. The PV diagram guide runs the procedures for both processes; this page is about what separates them.

Isobaric against isochoric, row by row

Property	Isobaric	Isochoric
What is held constant	pressure	volume
The CED's name for it, from 9.4.B.3	constant pressure (isobaric)	constant volume (isovolumetric)
Work done on the gas	$W = -P\Delta V$ , exactly	$W = 0$ , exactly
First law reduces to	nothing, both terms survive	$\Delta U = Q$
Shape on a PV diagram	a horizontal line	a vertical line
Area under that line	a rectangle, $P\lvert \Delta V \rvert$	none, the line has no width
Gas law relation	$V/T$ constant	$P/T$ constant
Expansion means	the gas warms, and $W < 0$	the volume cannot change, so there is no expansion
Fraction of heat that becomes internal energy	$3/5$ for an ideal monatomic gas	all of it
Heat needed for a given temperature rise	$\frac{5}{2}nR\Delta T$	$\frac{3}{2}nR\Delta T$
Can the temperature change	yes	yes

Row six is the one that turns up on diagrams. An isochoric leg is drawn as a vertical segment, and the area under a vertical segment is zero because it spans no volume. That is the graphical statement of $W = 0$ , and 9.4.B.2.ii is what licenses it: 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 vs. volume. No volume span, no area, no work.

Row eleven is the one students get wrong in reverse. Constant volume does not mean constant temperature. Heat a sealed rigid tank and the temperature climbs, the pressure climbs with it, and only the volume stays put. The process that fixes the temperature is the isothermal one, covered in isothermal vs adiabatic.

Rows nine and ten are the quantitative heart of the comparison, and the next two sections are about where they come from.

On a PV diagram: a rectangle and a line with no width

Both processes are straight lines on a pressure against volume plot, which is part of why they get run together. They are straight in perpendicular directions, and that is the whole difference.

The isobaric line is horizontal. Pressure is fixed, so the point moves left or right at constant height. The area under it is a rectangle of height $P$ and width $\lvert \Delta V \rvert$ , so the magnitude of the work is $P \lvert \Delta V \rvert$ . The sign comes from the direction of travel, not from the area: rightward is expansion, so the work done on the gas is negative, and leftward is compression, so it is positive.

The isochoric line is vertical. Volume is fixed, so the point moves straight up or straight down. There is no area under a vertical line, because area needs a width along the volume axis and this segment has none. So $W = 0$ , and it is zero whether the pressure triples or collapses.

That asymmetry has a consequence worth stating plainly. On an isochoric leg, the pressure can change enormously and still do no work. Work is not paid for by pressure, it is paid for by a moving boundary, and 9.4.B.1.iii's $W = -P\Delta V$ has the volume change as a factor. A gas can go from $1.0 \times 10^5\ \text{Pa}$ to $5.0 \times 10^5\ \text{Pa}$ in a sealed steel tank and the work is still exactly zero.

The reverse asymmetry holds for the isobaric leg. The pressure never changes and yet the work is nonzero, because the volume did the moving.

This is also why the two processes are the standard building blocks for cycles. A rectangular loop on a PV diagram is two isobaric legs and two isochoric legs, and all of the net work comes from the two horizontal ones. The CED's own sample activities lean on exactly this: one asks students to describe a three-step cyclical thermodynamic process, giving isochoric heating, isothermal expansion, isobaric cooling as its example, and to draw the PV, PT and VT diagrams of it. Another asks groups to demonstrate each of the eight thermodynamic processes, which it lists as isobaric, isovolumetric, isothermal, and adiabatic, each in both possible directions.

The case that separates them: the same heat, two different temperature rises

Give two identical gases the same amount of energy by heating and they end at different temperatures. That is the sharpest way to see that the work term is real.

Take $0.30\ \text{mol}$ of a monatomic ideal gas at $300\ \text{K}$ , twice, and transfer $500\ \text{J}$ into each by heating.

Sample A, in a rigid container. Isochoric, so $W = 0$ and $\Delta U = Q = 500\ \text{J}$ . Using $\Delta U = \frac{3}{2}nR\Delta T$ with $\frac{3}{2}nR = 3.7395\ \text{J/K}$ :

\Delta T = \frac{500}{3.7395} = 133.7\ \text{K}

The gas reaches $433.7\ \text{K}$ .

Sample B, under a free piston at constant pressure. Isobaric, so the gas expands as it warms and pushes the piston out, and some of the $500\ \text{J}$ leaves again as work. Only $300\ \text{J}$ stays as internal energy, with $200\ \text{J}$ leaving through the piston, so

\Delta T = \frac{300}{3.7395} = 80.2\ \text{K}

The gas reaches $380.2\ \text{K}$ .

Same gas, same energy in, and a temperature rise larger by a factor of $133.7/80.2 = 5/3$ in the rigid container. The isobaric sample was leaking energy out of the other side of the first law the entire time it was being warmed.

The ratio is exact and it is worth knowing where it comes from, because the AP Physics 2 sheet prints no molar heat capacities and you have to build it from what is there. For an isobaric process on an ideal gas, $PV = nRT$ with $P$ fixed gives $P\Delta V = nR\Delta T$ , so

W = -P\Delta V = -nR\Delta T

and the first law then gives

Q = \Delta U - W = \tfrac{3}{2}nR\Delta T + nR\Delta T = \tfrac{5}{2}nR\Delta T

against the isochoric $Q = \Delta U = \frac{3}{2}nR\Delta T$ . The ratio of the two is $5/3$ , and the fraction of the isobaric heat that survives as internal energy is $\frac{3/2}{5/2} = 3/5$ , that is $60\ \%$ , with the other $40\ \%$ going out as work. Both numbers are properties of a monatomic ideal gas, which the AP Physics 2 exam conventions box says is the only kind this course uses.

When it costs a mark

Computing work on an isochoric leg from the pressure change. Work needs a moving boundary. $W = -P\Delta V$ with $\Delta V = 0$ is zero however violently the pressure moves, and there is no $-V\Delta P$ term anywhere on the sheet.
Assuming constant volume means constant temperature. It does not, and it usually does not. Heat a rigid tank and $\Delta U = Q$ puts every joule into the temperature. The process that fixes the temperature is isothermal.
Assuming constant pressure means constant internal energy. It does not either. An isobaric expansion warms an ideal gas, because $P\Delta V = nR\Delta T$ makes $\Delta T$ positive whenever $\Delta V$ is.
Using $Q = \frac{3}{2}nR\Delta T$ for an isobaric process. That is the isochoric result. Isobaric needs $\frac{5}{2}nR\Delta T$ , and the extra $nR\Delta T$ is precisely the work the gas exported.
Getting the isobaric sign backwards. Expansion means $\Delta V > 0$ , so $W = -P\Delta V$ is negative, meaning the gas did work on its surroundings. Compute $\Delta V = V_f - V_i$ as a signed number before touching the equation.
Reading the area under a PV curve as a signed work. 9.4.B.2.ii gives the absolute value only. For an isobaric leg the area is $P\lvert \Delta V \rvert$ and the direction of travel supplies the sign.
Calling a vertical line on a PV diagram isobaric. Horizontal is constant pressure, because pressure is the vertical axis and a horizontal line holds it fixed. Vertical is constant volume. Half the errors on this pair are this axis mix-up rather than any physics.
Writing isochoric in a free-response answer and expecting the framework's wording. Both terms are standard and either will be understood, but the CED's required content says constant volume (isovolumetric), so that is the safer phrase to mirror.

Reading the setup to name the process

Almost every question on this pair identifies the process in the scenario rather than in the question stem, so the translation is worth having written out.

The problem says	The process is	Because
a rigid container, or a sealed steel tank	isochoric	the walls cannot move, so $\Delta V = 0$
a fixed, clamped or locked piston	isochoric	same reason, the boundary is held
the volume is unchanged	isochoric	stated directly
a freely sliding piston with the atmosphere above	isobaric	the piston finds the pressure that balances it, and holds it
a piston loaded with a fixed weight	isobaric	the load and the atmosphere together fix the pressure
a gas in a balloon at ambient pressure	isobaric	the surroundings set the pressure
the pressure is unchanged	isobaric	stated directly
a vertical line on a PV diagram	isochoric	no volume span
a horizontal line on a PV diagram	isobaric	constant height on the pressure axis

The fourth and fifth rows are the ones people miss, because nothing in the wording says pressure. A piston free to slide will move until the gas pressure matches whatever is pressing on the other side, and if that combination does not change, neither does the gas pressure. The CED's own sample activity for Topic 9.2 builds exactly this apparatus: set a capped syringe to $10\ \text{mL}$ , stand it vertically, stack books on top, and calculate the absolute pressure of the books plus atmosphere before making a pressure versus volume graph. Note that it says absolute pressure, which is a reminder that a gas law wants the total and not the excess over the atmosphere. Gauge vs absolute pressure is the page for that.

One more reading tip. If a question gives you a pressure and a volume change, it is handing you an isobaric process and expects $W = -P\Delta V$ . If it gives you a pressure change and no volume change, it is handing you an isochoric process and expects you to notice that the pressure change is a distraction as far as work is concerned. It is not a distraction for the temperature, which is what the question will actually ask for.

When they overlap, and why that lulls you

Strictly, these two processes never coincide. Holding the pressure and the volume both constant means holding $PV$ constant, and by $PV = nRT$ that means holding the temperature constant too, so nothing happens at all. Any question in which both are fixed is a question about a system that is not doing anything.

What does happen is that they get confused for four reasons that have nothing to do with the physics agreeing.

They look alike on a diagram. Both are straight lines, both are trivially easy to draw, and both are described by one word beginning with iso. Which axis each one is parallel to is the entire content, and it is the detail most easily transposed under time pressure.

They appear together. A rectangular cycle is two of each, alternating, so most diagrams a student sees contain both and the eye stops distinguishing them. That is also where the difference pays: the whole enclosed area of a rectangular cycle comes from the two isobaric legs, and the two isochoric legs contribute nothing to the net work at all.

Both have an easy work formula, which makes the work step feel like a formality. It is a formality in different ways. Isochoric has $W = 0$ because the equation vanishes. Isobaric has $W = -P\Delta V$ because the pressure factors out of an integral you never have to do. Neither is the general case, and on an isothermal or a curved path you need the area instead.

Both let the temperature change, so neither one can be identified from a temperature reading. Only the volume or the pressure staying fixed identifies them, which means you have to read the apparatus and not the outcome.

There is one honest partial overlap. In the limit of a very small change, an isobaric step and an isochoric step starting from the same state both take the gas somewhere with almost the same internal energy per joule of heating, because the work term is proportional to $\Delta V$ and shrinks with it. That is not a coincidence you can use, but it does explain why the difference feels invisible in a demonstration where nothing moves very far and becomes obvious in one where the piston travels.

What the CED asks of Topic 9.4

Both processes are named in Topic 9.4, The First Law of Thermodynamics, inside Unit 9, Thermodynamics, which carries $15$ to $18\ \%$ of the AP Physics 2 multiple-choice section across a suggested $10$ to $16$ class periods.

The naming happens in a single piece of essential knowledge, 9.4.B.3, and it is worth quoting whole because it lists all four special cases at once: 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; 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).

Four cases, and note how the CED groups them. The first three are named by the state variable held constant. The fourth is named by what does not cross the boundary. That is a real distinction and not just phrasing: isobaric and isochoric are constraints on the system's state, while adiabatic is a constraint on its surroundings.

The supporting content is:

9.4.B.1.ii, the first law itself, $\Delta U = Q + W$ , described as the change in internal energy of a closed system being the sum of energy transferred to or from the system by heating, or work done on the system.
9.4.B.1.iii, the definition of the work as done by a constant or average external pressure that changes the volume of that system, $W = -P\Delta V$ , with a piston compressing a gas as its example.
9.4.B.2, PV diagrams as representations of thermodynamic processes, with 9.4.B.2.i naming isotherms and 9.4.B.2.ii giving the absolute value of the work as the area under the pressure against volume curve.
9.4.A.1.ii, the internal energy of an ideal monatomic gas, $U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$ , which is what turns a temperature change into an energy change.

Topic 9.4 prints no boundary statement, and neither do Topics 9.2 and 9.3. Unit 9 carries three boundary statements in total, under Topics 9.1, 9.5 and 9.6, and none of them touches the two processes on this page. For the record they are: under 9.1, that AP Physics 2 only expects students to perform qualitative and quantitative analysis of collisions in one and two dimensions, and that students are not expected to know the functional form of the Maxwell-Boltzmann distribution but are expected to be familiar with how features of the distribution are related to the temperature of the gas; under 9.5, that AP Physics 2 will model specific heat as independent of temperature; and under 9.6, that only qualitative treatment of the second law of thermodynamics is within the scope of AP Physics 2.

The suggested skills listed for Topic 9.4 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; 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. Skill 2.C is this page: two scenarios, one energy input, two answers. Topic 9.4 has the framing in full, and heat vs internal energy covers why only one side of $\Delta U = Q + W$ carries a delta.

The same 500 J into two identical gases

Two identical samples of $0.30\ \text{mol}$ of monatomic ideal gas both start at $300\ \text{K}$ . Sample A is sealed in a rigid container. Sample B sits under a freely sliding piston at constant pressure. Each absorbs $500\ \text{J}$ by heating. Take $R = 8.31\ \text{J/(mol} \cdot \text{K)}$ . (a) Find the final temperature of sample A. (b) Find the work done on sample B and its change in internal energy. (c) Find the final temperature of sample B. (d) State the ratio of the two temperature rises and where it comes from.

Compute the two coefficients once, since both recur: $nR = (0.30)(8.31) = 2.493\ \text{J/K}$ , and $\frac{3}{2}nR = 3.7395\ \text{J/K}$ .
(a) Sample A is at constant volume, so $\Delta V = 0$ and the printed $W = -P\Delta V$ gives $W = 0$ . The first law becomes $\Delta U = Q = +500\ \text{J}$ .
From 9.4.A.1.ii, $\Delta U = \frac{3}{2}nR\Delta T$ , so $\Delta T = 500/3.7395 = 133.71\ \text{K}$ , and the final temperature is $300 + 133.71 = 433.7\ \text{K}$ .
(b) Sample B is at constant pressure. Both $Q$ and $W$ are nonzero, so solve for $\Delta T$ first using the isobaric relation. With $P$ constant, $PV = nRT$ gives $P\Delta V = nR\Delta T$ , so $W = -P\Delta V = -nR\Delta T$ .
Substituting into $\Delta U = Q + W$ with $\Delta U = \frac{3}{2}nR\Delta T$ : $\frac{3}{2}nR\Delta T = Q - nR\Delta T$ , so $\frac{5}{2}nR\Delta T = Q$ .
$\frac{5}{2}nR = (2.5)(2.493) = 6.2325\ \text{J/K}$ , so $\Delta T = 500/6.2325 = 80.225\ \text{K}$ .
Now the two energies separately. $\Delta U = \frac{3}{2}nR\Delta T = (3.7395)(80.225) = 300.0\ \text{J}$ , and $W = -nR\Delta T = -(2.493)(80.225) = -200.0\ \text{J}$ .
Check them against the first law as given: $Q = \Delta U - W = 300.0 - (-200.0) = +500\ \text{J}$ , which is the $500\ \text{J}$ the problem supplied. The split is $300\ \text{J}$ retained and $200\ \text{J}$ exported.
(c) Sample B's final temperature is $300 + 80.225 = 380.2\ \text{K}$ .
(d) The rises are $133.71\ \text{K}$ and $80.225\ \text{K}$ , a ratio of $133.71/80.225 = 1.667$ , which is exactly $5/3$ . It comes from the two heat requirements derived above, $\frac{3}{2}nR\Delta T$ at constant volume against $\frac{5}{2}nR\Delta T$ at constant pressure.
The physical reading: sample B spent $200\ \text{J}$ of its $500\ \text{J}$ pushing the piston out, so only $300\ \text{J}$ , that is $60\ \%$ , was left to raise its temperature. Sample A had no piston to push and kept all of it.

(a) Sample A reaches $433.7\ \text{K}$ , a rise of $133.7\ \text{K}$ . (b) $W = -200\ \text{J}$ and $\Delta U = +300\ \text{J}$ . (c) Sample B reaches $380.2\ \text{K}$ , a rise of $80.2\ \text{K}$ . (d) The ratio is $5/3$ , from $\frac{5}{2}nR\Delta T$ of heat needed at constant pressure against $\frac{3}{2}nR\Delta T$ at constant volume.

One isobaric expansion, read off the diagram

A monatomic ideal gas expands at a constant pressure of $8.0 \times 10^4\ \text{Pa}$ from $2.5 \times 10^{-3}\ \text{m}^3$ to $7.5 \times 10^{-3}\ \text{m}^3$ . (a) Sketch the path on a PV diagram and find the work done on the gas from the area. (b) Find the change in internal energy. (c) Find the energy transferred by heating, and the fraction of it that stayed as internal energy. (d) Find the ratio of the final temperature to the initial temperature.

(a) Constant pressure is a horizontal line on a PV diagram, running rightward from $2.5 \times 10^{-3}$ to $7.5 \times 10^{-3}\ \text{m}^3$ at a height of $8.0 \times 10^4\ \text{Pa}$ . The area beneath it is a rectangle.
$\Delta V = 7.5 \times 10^{-3} - 2.5 \times 10^{-3} = +5.0 \times 10^{-3}\ \text{m}^3$ , positive because the gas expanded. The rectangle's area is $(8.0 \times 10^4)(5.0 \times 10^{-3}) = 400\ \text{J}$ , which by 9.4.B.2.ii is the magnitude of the work.
The travel is rightward, an expansion, so the work done on the gas is negative: $W = -P\Delta V = -400\ \text{J}$ . The gas did $400\ \text{J}$ of work on its surroundings.
(b) For an ideal monatomic gas, combining $U = \frac{3}{2}nRT$ with $PV = nRT$ gives $U = \frac{3}{2}PV$ , so no value of $n$ is needed. $PV$ starts at $(8.0 \times 10^4)(2.5 \times 10^{-3}) = 200\ \text{J}$ and finishes at $(8.0 \times 10^4)(7.5 \times 10^{-3}) = 600\ \text{J}$ .
$\Delta U = \frac{3}{2}(600 - 200) = \frac{3}{2}(400) = +600\ \text{J}$ . The gas got hotter while expanding, which is what an isobaric expansion always does.
(c) Rearranging the first law: $Q = \Delta U - W = 600 - (-400) = +1000\ \text{J}$ .
The fraction that stayed as internal energy is $600/1000 = 0.60$ , and $400/1000 = 0.40$ left as work. Those are the $3/5$ and $2/5$ that any isobaric process on a monatomic ideal gas gives, independent of the numbers.
Check the split symbolically: $\Delta U = \frac{3}{2}P\Delta V$ and $\lvert W \rvert = P\Delta V$ , so $Q = \frac{5}{2}P\Delta V$ and $\Delta U / Q = 3/5$ exactly. With $P\Delta V = 400\ \text{J}$ , that predicts $\Delta U = 600\ \text{J}$ and $Q = 1000\ \text{J}$ , matching.
(d) At constant pressure, $PV = nRT$ makes $T$ proportional to $V$ , so $T_f/T_i = V_f/V_i = 7.5 \times 10^{-3} / 2.5 \times 10^{-3} = 3.0$ . The gas ends at three times its starting absolute temperature.
Now the counterfactual that makes the pair concrete. Had the same gas been held at constant volume instead, the work would have been zero, and the same $1000\ \text{J}$ of heating would have produced $\Delta U = 1000\ \text{J}$ rather than $600\ \text{J}$ , a temperature rise five thirds as large.

(a) $W = -400\ \text{J}$ , the area of a rectangle $8.0 \times 10^4\ \text{Pa}$ high and $5.0 \times 10^{-3}\ \text{m}^3$ wide, negative because the gas expanded. (b) $\Delta U = +600\ \text{J}$ . (c) $Q = +1000\ \text{J}$ , of which $60\ \%$ stayed as internal energy. (d) $T_f/T_i = 3.0$ .

The same temperature rise, two different heat bills

A sample of $0.30\ \text{mol}$ of monatomic ideal gas is to be warmed by exactly $100\ \text{K}$ . Take $R = 8.31\ \text{J/(mol} \cdot \text{K)}$ . (a) Find the heat required if the volume is held constant. (b) Find the heat required if the pressure is held constant. (c) Find the difference, and account for it exactly. (d) State which process is cheaper and why the other one costs more.

As before, $nR = (0.30)(8.31) = 2.493\ \text{J/K}$ , $\frac{3}{2}nR = 3.7395\ \text{J/K}$ and $\frac{5}{2}nR = 6.2325\ \text{J/K}$ .
(a) At constant volume, $W = 0$ and the first law gives $Q = \Delta U = \frac{3}{2}nR\Delta T = (3.7395)(100) = 373.95\ \text{J}$ .
(b) At constant pressure, the derivation from the previous example applies: $Q = \frac{5}{2}nR\Delta T = (6.2325)(100) = 623.25\ \text{J}$ .
(c) The difference is $623.25 - 373.95 = 249.30\ \text{J}$ .
Account for it. In the isobaric case the gas expanded, and the work done on it was $W = -nR\Delta T = -(2.493)(100) = -249.30\ \text{J}$ . So the gas exported $249.30\ \text{J}$ through the moving boundary, which is exactly the extra heat that had to be supplied. Nothing is unexplained.
Check the internal energy is the same in both, since the temperature change is the same: $\Delta U = (3.7395)(100) = 373.95\ \text{J}$ in each case. For the isobaric run, $\Delta U = Q + W = 623.25 + (-249.30) = 373.95\ \text{J}$ . It agrees.
The ratio of the two heat bills is $623.25/373.95 = 1.667$ , that is $5/3$ again, the same constant that appeared when the heat was fixed instead of the temperature rise. It has to be the same, because the two questions are the same relation read in opposite directions.
(d) Constant volume is cheaper. It costs $373.95\ \text{J}$ against $623.25\ \text{J}$ for the same $100\ \text{K}$ , because none of the energy is diverted into pushing a boundary.
The general statement, worth carrying: warming a gas at constant pressure always costs more than warming it at constant volume by exactly $nR\Delta T$ , the work the expansion exports. There is no equivalent penalty in the other direction, because a constant-volume process has no boundary to push.
One caution on scope. These two heat requirements are what other courses call the molar heat capacities at constant volume and at constant pressure. Neither appears on the AP Physics 2 equation sheet, which is why both were built here from $\Delta U = Q + W$ , $W = -P\Delta V$ , $PV = nRT$ and $U = \frac{3}{2}nRT$ , all four of which are printed.

(a) $Q = 373.95\ \text{J}$ at constant volume. (b) $Q = 623.25\ \text{J}$ at constant pressure. (c) The difference of $249.30\ \text{J}$ is exactly the work the expanding gas did on its surroundings, $nR\Delta T$ . (d) Constant volume is cheaper by a factor of $5/3$ , because no energy leaves through a moving boundary.

Frequently asked questions

What is the difference between an isobaric and an isochoric process?

An isobaric process holds the pressure constant while the volume changes, so work crosses the boundary and the AP Physics 2 sheet's W equals minus P delta V applies exactly. An isochoric process holds the volume constant, so delta V is zero, the work is exactly zero, and the first law reduces to the change in internal energy equalling Q. Essential knowledge 9.4.B.3 names both among the special cases of thermal processes, writing them as constant pressure (isobaric) and constant volume (isovolumetric). On a pressure against volume diagram the isobaric path is a horizontal line with a rectangular area beneath it, and the isochoric path is a vertical line with no area beneath it at all.

Why is the work zero in an isochoric process?

Because work needs a boundary that moves. Essential knowledge 9.4.B.1.iii defines the work done on a system as coming from a constant or average external pressure that changes the volume of that system, and writes it as W equals minus P delta V. If the volume does not change, delta V is zero and the product is zero, whatever the pressure is doing. Graphically it is the same statement: 9.4.B.2.ii gives the magnitude of the work as the area under the curve on a pressure against volume plot, and a vertical line spans no volume, so there is no area under it. A gas sealed in a steel tank can have its pressure multiplied several times over and still do no work.

Does an isochoric process mean the temperature stays the same?

No. Only the volume stays the same. Heat a rigid sealed container and the first law reduces to the change in internal energy equalling the energy transferred by heating, so every joule that enters raises the internal energy and therefore, for an ideal gas, the temperature. The pressure rises with it, since P over T is constant at fixed volume. The process that holds the temperature constant is the isothermal one, which is a different member of the list in essential knowledge 9.4.B.3. Confusing constant volume with constant temperature turns a process where all the heat becomes internal energy into one where none of it does.

How much heat does it take to warm a gas at constant pressure compared with constant volume?

Five thirds as much, for an ideal monatomic gas. At constant volume no work is done, so the heat required is just the internal energy change, three halves n R delta T. At constant pressure the gas also expands and exports work equal to n R delta T, so the heat required is the sum, five halves n R delta T. Warming 0.30 moles by 100 kelvin costs 373.95 joules at constant volume and 623.25 joules at constant pressure, and the difference of 249.30 joules is exactly the work the expansion did on the surroundings. Neither of these heat capacities is printed on the AP Physics 2 sheet; both are built from the first law, W equals minus P delta V, PV equals nRT, and U equals three halves n R T.

What fraction of the heat added in an isobaric process becomes internal energy?

Three fifths, or 60 percent, for an ideal monatomic gas, with the remaining 40 percent leaving as work. The derivation is short: the internal energy change is three halves P delta V, the magnitude of the work is P delta V, so the heat supplied is five halves P delta V, and the ratio of the first to the last is three fifths. Expanding a gas at 8.0 times 10 to the fourth pascals from 2.5 to 7.5 times 10 to the minus three cubic meters gives a work of minus 400 joules, an internal energy rise of 600 joules and a heat input of 1000 joules, and 600 out of 1000 is the same 60 percent. In an isochoric process the corresponding fraction is 100 percent.

Is the correct word isochoric or isovolumetric?

Both are standard and both mean constant volume. The AP Physics 2 CED uses isovolumetric in its required course content: essential knowledge 9.4.B.3 lists the special cases as constant volume (isovolumetric), constant temperature (isothermal), and constant pressure (isobaric), together with adiabatic. The word isochoric appears in that CED only inside an optional sample instructional activity, one which asks students to describe a three-step cyclical process and gives isochoric heating, isothermal expansion, isobaric cooling as its example. So isochoric is the phrasing students search for and isovolumetric is the phrasing the framework prints, and mirroring the framework is the safer choice in a written answer.

Which line on a PV diagram is isobaric and which is isochoric?

Pressure is the vertical axis, so a horizontal line holds the pressure fixed and is the isobaric path, while a vertical line holds the volume fixed and is the isochoric path. The area beneath the horizontal line is a rectangle of height P and width equal to the magnitude of the volume change, and that area is the magnitude of the work. The vertical line spans no volume, so there is no area and no work. In a rectangular cycle made of two of each, all of the net work comes from the two horizontal legs. Transposing the two axes is one of the commonest errors on this pair, and it is a reading mistake rather than a physics one.