Isobaric process

Picture a bicycle pump left standing upright with a weight resting on the piston, or a gas trapped under a heavy, freely-sliding piston in a cylinder over a flame. As the flame heats the gas, the piston does not stay put — it climbs, slowly, steadily, and the gas beneath it swells in volume. Yet through all of this the pressure inside never changes by a single pascal, because the only thing balancing it — the weight of the piston plus whatever sits on it, together with atmospheric pressure — never changes either. This is an isobaric process: a change of state at constant pressure, and it is one of the four "named" gas processes every CBSE Class 11 student must be able to recognise, sketch, and calculate through. What makes the isobaric process worth its own spotlight is that it is the one process where both heat and work are doing real, non-zero jobs at the same time, and where a specific heat you have not used before — $C_P$ , the molar specific heat at constant pressure — finally earns its keep. Unlike an isochoric (constant-volume) change, where all the supplied heat goes straight into internal energy, here the gas insists on doing mechanical work as it pushes the piston outward, so some of the heat you pour in "leaks away" into work done on the surroundings before it can raise the internal energy. Getting the bookkeeping right — how much heat, how much work, how much internal-energy change — is exactly what the first law of thermodynamics was built for, and the isobaric process is the cleanest place to practise it, because the work integral collapses from an area under a curve into a single multiplication, $P\Delta V$ . This section builds the isobaric process from the first law and the ideal gas equation, keeps every worked number from the syllabus-anchored examples you may already have seen, and adds a derivation, two fully worked problems (board-level and competitive-level), and the traps that reliably cost marks in exams.

Key takeaways

An isobaric process is a change of a gas's volume and temperature at constant pressure ( $\Delta P = 0$ ) — the classic picture is a gas heated in a cylinder closed by a freely-movable, weighted piston.
Work done by the gas is $\Delta W = P\Delta V$ exactly, because $P$ never changes during the process; on a $P$ – $V$ diagram it is a horizontal line, and the work is the rectangular area beneath it.
Constant $P$ in $PV = nRT$ forces $V/T = \text{constant}$ — this is Charles's law, so volume and absolute temperature rise and fall together.
Heat supplied is found with the constant-pressure molar specific heat, $\Delta Q = nC_P\Delta T$ , which is always larger than $nC_V\Delta T$ for the same $\Delta T$ because some heat must also pay for the expansion work.
The first law $\Delta Q = \Delta U + \Delta W$ still governs everything, and combined with Mayer's relation $C_P - C_V = R$ it shows that $\Delta U = nC_V\Delta T$ — exactly the same internal-energy formula as for any other process, because $U$ depends only on temperature for an ideal gas.
The two named constant-pressure/constant-volume processes are mirror images of each other: isochoric has $\Delta W = 0$ and all heat becomes internal energy; isobaric has $\Delta W = P\Delta V \neq 0$ and the heat splits between work and internal energy.
Key idea: at constant pressure, work done equals $P\Delta V$ , heat supplied equals $nC_P\Delta T$ , and the first law then forces the internal-energy change to still be $nC_V\Delta T$ , exactly as in every other ideal-gas process.

The core insight

In an isobaric process the pressure stays fixed so work is simply $\Delta W = P\Delta V$ and heat is $\Delta Q = nC_P\Delta T$ , and applying the first law with Mayer's relation $C_P - C_V = R$ shows the internal-energy change is still $\Delta U = nC_V\Delta T$ .

Where it fits in the syllabus

The isobaric process sits inside the CBSE Class 11 Physics Thermodynamics chapter's treatment of "change of state of a gas," alongside the isothermal, isochoric, adiabatic, and cyclic processes. It leans directly on three earlier results in the same chapter: the work formula $\Delta W = P\Delta V$ , the first law $\Delta Q = \Delta U + \Delta W$ , and the ideal gas equation $PV = nRT$ from Kinetic Theory. The board syllabus expects you to identify $\Delta P = 0$ as the defining condition, compute work using $\Delta W = P\Delta V$ , use $\Delta Q = nC_P\Delta T$ for heat, and combine these through the first law — nothing about heat engines, efficiency, or entropy is needed or in scope here; those are kept for later study.

The physical picture

An isobaric process holds the pressure $P$ of a fixed amount of gas constant while its volume and temperature are free to change — physically, this happens whenever the only forces resisting the gas (a weighted piston, or the atmosphere itself) do not themselves change during the process. Because $P$ never varies, the work done by the gas as it expands or is compressed from $V_1$ to $V_2$ collapses from the general area-under-the-curve integral into ordinary multiplication:

\Delta W = P\,\Delta V = P(V_2 - V_1)

where: $P$ is the constant pressure (Pa), $V_1$ and $V_2$ are the initial and final volumes (m³), and $\Delta W$ is the work done by the gas (J) — positive when the gas expands ( $V_2 > V_1$ ) and negative when it is compressed. Physically, this is the same $P\Delta V$ formula used to introduce work earlier in the chapter, except here it applies exactly, without any need to average or integrate, because pressure is fixed throughout. Substituting the constant $P$ into the ideal gas equation $PV = nRT$ shows that $V$ and $T$ are directly proportional whenever $n$ and $P$ are fixed — this constant-pressure relation is Charles's law:

\frac{V_1}{T_1} = \frac{V_2}{T_2}

where: $T_1$ , $T_2$ are the initial and final absolute temperatures (K), always measured on the Kelvin scale. The physical reading is simple: heat a gas at constant pressure and it must expand in direct proportion to its rise in absolute temperature — double the absolute temperature and the volume doubles too. Heat exchanged during an isobaric process is computed using the molar specific heat at constant pressure, $C_P$ , not $C_V$ :

\Delta Q = nC_P\Delta T

where: $n$ is the number of moles and $\Delta T = T_2 - T_1$ . $C_P$ is always larger than $C_V$ for the same gas because, at constant pressure, part of the heat supplied must do expansion work in addition to raising the internal energy — Mayer's relation makes this precise: $C_P - C_V = R$ . Applying the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$ , and rearranging for the internal-energy change:

\Delta U = \Delta Q - \Delta W = nC_P\Delta T - P\Delta V

Since $P\Delta V = nR\Delta T$ at constant pressure (directly from $PV = nRT$ ), substitute and use Mayer's relation $C_P - C_V = R$ :

\Delta U = nC_P\Delta T - nR\Delta T = n(C_P - R)\Delta T = nC_V\Delta T

where: the striking result is that $\Delta U = nC_V\Delta T$ turns out to hold for any process on an ideal gas, isobaric or otherwise, because internal energy depends only on temperature. The constant-pressure identity of the heat you supply ( $nC_P\Delta T$ ) is not the same as the internal-energy change — the gap between them is exactly the work done, $P\Delta V$ . On a $P$ – $V$ diagram, an isobaric process traces a horizontal straight line at the fixed height $P$ , running from $V_1$ to $V_2$ . The work done is the rectangular area trapped between this line, the horizontal axis, and the two vertical lines at $V_1$ and $V_2$ — the simplest possible "area under the curve" in the whole chapter.

Figure

A pressure volume diagram showing a horizontal line at constant pressure from volume V1 to volume V2, with the rectangular area beneath it shaded to represent the work done by the gas

The main derivation

Derivation AGetting the change in internal energy from the first law at constant pressure

The goal here is to show, starting only from the ideal gas equation and the first law of thermodynamics, that even though the heat supplied at constant pressure is $nC_P\Delta T$ , the actual rise in internal energy is the smaller quantity $nC_V\Delta T$ — and to make explicit exactly where the missing piece of heat goes, so that this identity is not just memorised but understood as a direct bookkeeping consequence of work being done during expansion.

Step 1. Write down the work done by the gas at constant pressure. Since $P$ is fixed throughout the process, $\Delta W = P\Delta V$ , with no need to integrate because the pressure never varies as $V$ changes from $V_1$ to $V_2$ .

Step 2. Express $P\Delta V$ using the ideal gas equation. From $PV = nRT$ , at constant $n$ and constant $P$ , differencing the equation between the initial and final states gives $P(V_2 - V_1) = nR(T_2 - T_1)$ , so $P\Delta V = nR\Delta T$ . This converts the mechanical work into a purely thermal quantity — proportional to the same $\Delta T$ that drives the heat and internal-energy changes.

Step 3. Write the heat supplied using the constant-pressure specific heat. By definition, $\Delta Q = nC_P\Delta T$ is the heat needed to raise $n$ moles of the gas through $\Delta T$ while pressure is held fixed.

Step 4. Apply the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$ , and substitute Steps 2 and 3: $\Delta U = \Delta Q - \Delta W = nC_P\Delta T - nR\Delta T = n(C_P - R)\Delta T$ .

Step 5. Bring in Mayer's relation. Since $C_P - C_V = R$ for an ideal gas (a result you should already know from Kinetic Theory), substitute $C_P - R = C_V$ directly into the Step 4 result, giving $\Delta U = nC_V\Delta T$ .

Step 6. Read the physical meaning. This is exactly the internal-energy formula used for an isochoric process, where all the heat goes into $\Delta U$ because no work is done. Here, at constant pressure, a larger amount of heat is supplied ( $nC_P\Delta T$ rather than $nC_V\Delta T$ ), but the internal-energy change ends up identical for the same $\Delta T$ — the extra heat is precisely accounted for by the work $nR\Delta T = P\Delta V$ done pushing the piston outward. Internal energy, for an ideal gas, is a function of temperature alone, so it cannot "know" whether the process that produced $\Delta T$ was isobaric, isochoric, or anything else.

Worked examples

Board level - heating a gas isobarically

2 moles of an ideal gas at pressure $1.8\times10^5$ Pa expand isobarically from volume $3\times10^{-3}\ \text{m}^3$ to $5\times10^{-3}\ \text{m}^3$ . Given $C_P = 29.1\ \text{J mol}^{-1}\text{K}^{-1}$ and initial temperature $T_1 = 320$ K, find the work done, the final temperature, the heat absorbed, and the change in internal energy.

Step 1. Find the work done using $\Delta W = P\Delta V$ . $\Delta W = 1.8\times10^5 \times (5\times10^{-3} - 3\times10^{-3}) = 1.8\times10^5 \times 2\times10^{-3} = 360\ \text{J}$ .

Step 2. Find the final temperature using Charles's law. $T_2 = T_1 \times \dfrac{V_2}{V_1} = 320 \times \dfrac{5\times10^{-3}}{3\times10^{-3}} = 320 \times 1.6667 = 533.3\ \text{K}$ , so $\Delta T = 533.3 - 320 = 213.3\ \text{K}$ .

Step 3. Find the heat absorbed using $\Delta Q = nC_P\Delta T$ . $\Delta Q = 2 \times 29.1 \times 213.3 = 12{,}413.9\ \text{J}$ .

Step 4. Find the change in internal energy from the first law. $\Delta U = \Delta Q - \Delta W = 12{,}413.9 - 360 = 12{,}053.9\ \text{J}$ .

The gas absorbs about 12,414 J of heat, does 360 J of work expanding, and its internal energy rises by about 12,054 J.

Competitive level - compression with a monatomic gas and a Cp check

3 moles of a monatomic ideal gas ( $C_V = \tfrac32 R$ , $C_P = \tfrac52 R$ , with $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ ) is cooled at a constant pressure of $2.5\times10^5$ Pa so its volume shrinks from $6\times10^{-3}\ \text{m}^3$ to $4\times10^{-3}\ \text{m}^3$ . Find the work done by the gas, the drop in temperature, the heat released, and the change in internal energy, and verify the first law balances.

Step 1. Work done by the gas. $\Delta W = P\Delta V = 2.5\times10^5 \times (4\times10^{-3} - 6\times10^{-3}) = 2.5\times10^5\times(-2\times10^{-3}) = -500\ \text{J}$ . The negative sign means the gas does negative work — equivalently, 500 J of work is done on the gas by the surroundings as it is compressed.

Step 2. Temperature drop from $P\Delta V = nR\Delta T$ . Rearranging, $\Delta T = \dfrac{P\Delta V}{nR} = \dfrac{-500}{3\times8.314} = \dfrac{-500}{24.94} = -20.0\ \text{K}$ (approximately), so the gas cools by about $20$ K.

Step 3. Heat released using $C_P = \tfrac52 R = 2.5\times8.314 = 20.785\ \text{J mol}^{-1}\text{K}^{-1}$ . $\Delta Q = nC_P\Delta T = 3\times20.785\times(-20.0) = -1247.1\ \text{J}$ , so about 1247 J of heat leaves the gas.

Step 4. Change in internal energy directly via $nC_V\Delta T$ , using $C_V = \tfrac32 R = 12.471\ \text{J mol}^{-1}\text{K}^{-1}$ : $\Delta U = 3\times12.471\times(-20.0) = -748.3\ \text{J}$ .

Step 5. Check the first law. $\Delta U + \Delta W = -748.3 + (-500) = -1248.3\ \text{J}$ , which matches $\Delta Q = -1247.1\ \text{J}$ to within rounding — the small mismatch (about 1 J) comes purely from carrying $\Delta T$ to three significant figures; a sharper $\Delta T = -20.03\ \text{K}$ closes the gap exactly.

The gas releases about 1247 J of heat, has 500 J of work done on it, and its internal energy falls by about 748 J, consistent with the first law. Timing note: in a competitive-exam setting, do not recompute $\Delta T$ twice from scratch — get it once from $P\Delta V = nR\Delta T$ (Step 2) and reuse that same number for both the $\Delta Q$ and $\Delta U$ lines; recomputing independently, as this example shows, only invites small rounding mismatches and wastes seconds you do not have.

Common traps

Confusing isobaric with isothermal. Isobaric means constant pressure; isothermal means constant temperature. In an isobaric process, temperature generally does change (per Charles's law) — students who write $\Delta U = 0$ for an isobaric process are silently importing an isothermal-process fact where it does not belong.
Using $C_V$ instead of $C_P$ for heat supplied. Since work is done during an isobaric process, more heat is needed for the same $\Delta T$ than in an isochoric process — always use $C_P$ , not $C_V$ , when computing $\Delta Q$ here.
Forgetting that $\Delta U = nC_V\Delta T$ regardless of process type. Some students think $\Delta U$ must be computed differently for isobaric processes — it never is; only the heat and work formulas change with process type, and $\Delta U$ can always be found from $nC_V\Delta T$ or equivalently by subtracting $\Delta W$ from $\Delta Q$ .
Sign errors on $\Delta V$ during compression, leading to a wrong sign for the work done — always take $V_{\text{final}} - V_{\text{initial}}$ carefully, and remember a negative $\Delta W$ means work is done on the gas.
Mixing up $C_P$ and $C_V$ values for a given gas type. For a monatomic gas $C_V = \tfrac32 R$ and $C_P = \tfrac52 R$ ; for a diatomic gas $C_V = \tfrac52 R$ and $C_P = \tfrac72 R$ — plugging in the wrong pair silently corrupts every downstream number.

You should be able to…

Define an isobaric process as one with $\Delta P = 0$ , and compute the work done using $\Delta W = P\Delta V$ .
Use Charles's law ( $V/T = \text{constant}$ ) to relate initial and final volumes and temperatures at fixed pressure.
Apply $\Delta Q = nC_P\Delta T$ to find heat absorbed or released, and know why $C_P$ is used here rather than $C_V$ .
Derive and use $\Delta U = nC_V\Delta T$ for an isobaric process, and explain why the same formula works for any ideal-gas process.
Sketch an isobaric process as a horizontal line on a $P$ – $V$ diagram and interpret the enclosed rectangular area as work done.
Solve combined problems that require Charles's law, $C_P$ , $C_V$ , Mayer's relation, and the first law together, tracking signs correctly through compression as well as expansion.

Ask the tutor: Want a side-by-side comparison of isochoric, isobaric, isothermal, and adiabatic processes starting from the same state, or more mixed practice combining Charles's law with $C_P$ and $C_V$ in one problem? Ask the AI tutor for guided isobaric process problems.

Why does an isobaric process need C_P instead of C_V if internal energy only cares about temperature anyway?

Internal energy does only depend on temperature for an ideal gas, and $\Delta U = nC_V\Delta T$ is true no matter what the process is. But $\Delta Q$ is not the same thing as $\Delta U$ — it is whatever heat you must supply to get that temperature change given the constraints of the process. At constant pressure, the gas also expands and does work, so you must feed in extra heat beyond what raises $\Delta U$ alone, just to cover that work. $C_P$ is defined precisely to include that extra amount, which is why $C_P$ is always bigger than $C_V$ .

If pressure is constant, does that mean the gas cannot do any work?

It is the opposite — constant pressure is actually the easiest case for calculating work, not a case with no work at all. Work is zero only in an isochoric (constant-volume) process, because then $\Delta V = 0$ so $P\Delta V = 0$ regardless of pressure. In an isobaric process the volume is very much changing, so $\Delta W = P\Delta V$ is generally large and non-zero; it is simply easy to compute because $P$ stays fixed, letting you multiply instead of integrate.

How is an isobaric process different from an isochoric one in terms of the first law?

In an isochoric process $\Delta V = 0$ , so $\Delta W = 0$ and the first law collapses to $\Delta Q = \Delta U = nC_V\Delta T$ — all the heat becomes internal energy. In an isobaric process $\Delta W = P\Delta V \neq 0$ , so the first law reads $\Delta Q = \Delta U + \Delta W$ with both terms present; the heat supplied, $nC_P\Delta T$ , splits between raising the internal energy (still $nC_V\Delta T$ ) and doing expansion work ( $nR\Delta T$ ). The two processes bracket the same physics from opposite sides.

Do I need to know about heat engines or Carnot cycles to answer isobaric-process questions?

No — heat engines, the Carnot cycle, efficiency, and refrigerators are not part of the current CBSE Class 11 prescription and will not appear in isobaric-process questions at this level; those ideas are studied later, in higher classes. Everything you need here comes from the first law, the ideal gas equation, and the specific heats $C_P$ and $C_V$ already introduced in this chapter and in Kinetic Theory.