Isochoric process

Bolt the lid on a steel pressure cooker and set it over a flame without ever loosening the valve, and something interesting happens: the gas trapped inside cannot expand by even a millimetre, no matter how hot it gets. The rigid steel wall pins the volume in place, so every joule of heat you pour in has nowhere to go except straight into the gas's internal energy — and straight into its pressure. This is the isochoric process: change of state at constant volume, the plainest and, in a precise sense, the most economical of the named thermodynamic processes, because none of the heat is "wasted" doing mechanical work. It sits quietly at the intersection of two ideas you already know. From the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$ , you know heat splits between raising internal energy and doing work. From the ideal gas equation, $PV = nRT$ , you know pressure, volume and temperature are locked together. An isochoric process is what happens when you freeze one variable — $V$ — and watch the other two respond. The work term collapses to zero, the first law collapses to $\Delta Q = \Delta U$ , and the gas law collapses to Gay-Lussac's law, $P/T = \text{constant}$ . Three big ideas, one clean constraint. This article rebuilds that reasoning from the ground up, with a rigorous single derivation linking $\Delta Q = \Delta U = nC_V\Delta T$ to the molar specific heat at constant volume, a fresh full worked derivation, two extended worked problems (a board-style numeric problem and a sharper competitive-exam problem), the traps examiners love to plant, and a quick-fire quiz to check you can spot an isochoric process on sight — on a $P$ – $V$ diagram, in a word problem, or inside a sealed rigid vessel like that pressure cooker.

Key takeaways

Volume is frozen. In an isochoric (constant-volume) process, $\Delta V = 0$ throughout — the classic real-world case is gas heated or cooled inside a rigid, sealed container.
No work is ever done. Since $\Delta W = P\Delta V$ and $\Delta V = 0$ , work done by or on the gas is exactly zero at every instant of the process.
Heat equals internal-energy change. The first law $\Delta Q = \Delta U + \Delta W$ collapses to $\Delta Q = \Delta U$ — all supplied heat becomes internal energy, none is diverted into pushing a boundary.
This is the operational definition of $C_V$ . $\Delta Q = nC_V\Delta T = \Delta U$ is exactly how the molar specific heat at constant volume is measured and defined.
Pressure and temperature track each other. From $PV = nRT$ with $V$ fixed, $P/T = \text{constant}$ — this is Gay-Lussac's law, the constant-volume twin of Boyle's and Charles's laws.
On a $P$ – $V$ diagram it is a vertical line. Because $V$ never changes, the process traces straight up (heating) or straight down (cooling) at fixed $V$ , enclosing zero area — visual proof that $\Delta W = 0$ .
Key idea: because volume never changes in an isochoric process, ΔW = PΔV is identically zero, so the first law reduces to ΔQ = ΔU = nCᵥΔT while pressure and temperature move together through Gay-Lussac's law P/T = constant.

The core insight

Holding volume fixed forces $\Delta W = 0$ , so the first law collapses to $\Delta Q = \Delta U = nC_V\Delta T$ , while pressure and temperature stay locked together by Gay-Lussac's law $P/T = \text{constant}$ .

Where it fits in the syllabus

The CBSE Class 11 Physics Thermodynamics unit explicitly names isothermal ( $PV=\text{const}$ ), adiabatic ( $PV^\gamma=\text{const}$ ), reversible/irreversible, and cyclic processes as the prescribed "change of state of a gas" list — the isochoric (constant-volume) process is not called out by that label in the syllabus text itself. Even so, it is fully inside the prescribed scope by direct construction: it is nothing more than the first law $\Delta Q = \Delta U + \Delta W$ and the equation of state $PV=nRT$ , both explicitly prescribed, applied to the special case $\Delta V = 0$ . It is also the process that operationally defines $C_V$ (constant-volume molar specific heat), a quantity the Kinetic Theory chapter uses directly in $\gamma = C_P/C_V$ . Boards and competitive exams both test it routinely — this article treats it honestly as a standard, syllabus-consistent application rather than an independently named "law."

The physical picture

Picture the gas as a swarm of molecules bouncing around inside a rigid steel box — the pressure cooker from the opening, or any sealed cylinder whose piston is clamped so it cannot move. Because the walls are perfectly rigid, the gas can never push a boundary outward or let one collapse inward; there is no displacement for any force to act through. Work, in the mechanical sense used throughout this chapter, is force acting over a displacement — and here the displacement of the boundary is permanently zero.

\Delta W = P\,\Delta V = 0 \qquad (\text{since } \Delta V = 0 \text{ at every instant})

where: $P$ is the gas pressure and $\Delta V$ is the change in volume of the gas — the small work done in any infinitesimal step is $P\,dV$ , and integrating zero over the whole process still gives zero. The physical reading is blunt: it does not matter how much the pressure changes, how violently the gas is heated, or how long the process runs — as long as $V$ genuinely never changes, $\Delta W$ is identically zero, not merely small. Substituting into the first law of thermodynamics immediately gives the defining relation of this process:

\Delta Q = \Delta U + \Delta W = \Delta U + 0 = \Delta U

where: $\Delta Q$ is heat supplied to the gas (positive if absorbed), $\Delta U$ is the change in internal energy, and $\Delta W$ is work done by the gas. Physically, every joule of heat you feed into a rigidly sealed gas goes into raising the average kinetic energy of its molecules — none of it is spent expanding against an external pressure, because there is no expansion to spend it on. This is precisely the condition used to define the molar specific heat at constant volume $C_V$ :

\Delta Q = nC_V\,\Delta T = \Delta U

where: $n$ is the number of moles and $\Delta T$ is the temperature change. This equation is why $C_V$ , not $C_P$ , is the "pure" heat capacity of a gas — it measures internal-energy change alone, uncontaminated by any work term, which is exactly what makes it directly linked to the kinetic theory result $U = \frac{f}{2}nRT$ for $f$ degrees of freedom (a Kinetic Theory chapter result, quoted here, not re-derived). Finally, since $V$ is fixed, the ideal gas equation $PV=nRT$ reduces to a direct proportionality between $P$ and $T$ :

\frac{PV}{T} = nR = \text{constant} \quad\Rightarrow\quad \frac{P}{T} = \text{constant} \quad\Rightarrow\quad \frac{P_1}{T_1} = \frac{P_2}{T_2}

where: $P_1, T_1$ are the initial pressure and temperature and $P_2, T_2$ the final ones (all temperatures in kelvin). This is Gay-Lussac's law — heat a fixed volume of gas and its pressure climbs in exact step with its absolute temperature; cool it and both fall together. It is the constant-volume counterpart of Boyle's law ( $PV=\text{const}$ at fixed $T$ ) and Charles's law ( $V/T=\text{const}$ at fixed $P$ ).

Figure

A pressure versus volume diagram showing a single vertical line segment at a fixed volume, with an upward arrow labelled heating raising pressure and temperature and a downward arrow labelled cooling lowering them, next to a small sketch of a rigid sealed cylinder with a flame under it and a padlocked piston that cannot move

Derivation AHeat supplied in an isochoric process equals internal-energy change, defining C-sub-V

The goal of this derivation is to start from the general first law of thermodynamics and the general definition of work in a gas process, and show rigorously that fixing the volume forces the heat supplied to equal exactly the change in internal energy — and that this equality is precisely how the molar specific heat at constant volume, $C_V$ , is defined and measured in the laboratory. This matters because students often treat $\Delta Q = \Delta U$ as a memorised shortcut rather than a forced consequence of $\Delta V = 0$ , and then misapply it to processes where volume is only approximately constant.

Step 1. Begin with the general expression for small work done by a gas expanding or compressing quasi-statically against an external pressure $P$ : $dW = P\,dV$ , so over a finite change, $\Delta W = \int P\,dV$ .

Step 2. Apply the isochoric condition. By definition, an isochoric process holds volume exactly fixed: $V = \text{constant}$ , so $dV = 0$ at every instant of the process, from the initial to the final state.

Step 3. Substitute $dV = 0$ into the work integral. Since the integrand $P\,dV$ vanishes identically at every point along the process — regardless of how large or how rapidly $P$ itself is changing — the total work is $\Delta W = \int P\,(0)\,dV = 0$ .

Step 4. Insert this into the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$ . With $\Delta W = 0$ , this reduces immediately to $\Delta Q = \Delta U$ : the heat supplied to (or removed from) the gas is exactly equal to the change in its internal energy, with no correction term at all.

Step 5. Connect this to the definition of molar specific heat at constant volume. By definition, $C_V$ is the heat required per mole per kelvin rise when volume is held fixed: $\Delta Q\big|_{V} = nC_V\,\Delta T$ . Combining with Step 4, $\Delta U = nC_V\,\Delta T$ for this process — so measuring the heat needed to raise a fixed-volume gas sample's temperature is the direct experimental route to $C_V$ , and conversely $C_V$ lets you predict $\Delta Q$ for any isochoric heating or cooling once $n$ and $\Delta T$ are known.

Worked examples

Board level - heating a gas at constant volume in a sealed vessel

3 moles of an ideal gas, sealed in a rigid container, are heated so that its temperature rises from 280 K to 340 K. Given $C_V = 12.5\ \text{J mol}^{-1}\text{K}^{-1}$ for this gas, find the work done, the heat supplied, and the change in internal energy. Then find the final pressure if the initial pressure was $2.0\times10^5$ Pa.

Step 1. Identify the process as isochoric — the container is rigid and sealed, so $\Delta V = 0$ and therefore $\Delta W = 0$ throughout.

Step 2. Compute the heat supplied using $\Delta Q = nC_V\Delta T$ , with $n = 3\ \text{mol}$ , $C_V = 12.5\ \text{J mol}^{-1}\text{K}^{-1}$ , and $\Delta T = 340 - 280 = 60\ \text{K}$ : $\Delta Q = 3 \times 12.5 \times 60 = 2250\ \text{J}$ .

Step 3. Apply the first law with $\Delta W = 0$ to find the internal-energy change: $\Delta U = \Delta Q - \Delta W = 2250 - 0 = \mathbf{2250\ J}$ . Heat supplied and internal-energy rise are numerically identical, confirming the isochoric relation $\Delta Q = \Delta U$ derived above.

Step 4. Use Gay-Lussac's law to find the final pressure, since $V$ is fixed: $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ , so $P_2 = P_1 \times \dfrac{T_2}{T_1} = 2.0\times10^5 \times \dfrac{340}{280} = 2.0\times10^5 \times 1.214 \approx \mathbf{2.43\times10^5\ Pa}$ . So the gas absorbs 2250 J of heat, does zero work, gains 2250 J of internal energy, and its pressure climbs from $2.0\times10^5$ Pa to about $2.43\times10^5$ Pa — pressure and temperature rising together, exactly as Gay-Lussac's law predicts.

Competitive level - cooling gas in a rigid tank, finding the heat released and comparing to an isobaric alternative

A rigid, sealed tank holds 2 moles of an ideal diatomic gas at pressure $1.6\times10^5$ Pa and temperature 300 K. The gas is cooled at constant volume until the pressure drops to $1.2\times10^5$ Pa. Taking $C_V = \tfrac{5}{2}R$ for a diatomic gas with $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ , find the final temperature, the heat released, and compare this heat with what would have been released cooling the same gas through the same temperature drop at constant pressure instead (using $C_P = \tfrac{7}{2}R$ ).

Step 1. Find the final temperature using Gay-Lussac's law, since volume is fixed: $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$ , so $T_2 = T_1\times\dfrac{P_2}{P_1} = 300 \times \dfrac{1.2\times10^5}{1.6\times10^5} = 300 \times 0.75 = \mathbf{225\ K}$ .

Step 2. Compute $C_V$ numerically: $C_V = \tfrac{5}{2}\times8.314 = 20.785\ \text{J mol}^{-1}\text{K}^{-1}$ . The temperature drop is $\Delta T = 225 - 300 = -75\ \text{K}$ .

Step 3. Apply $\Delta Q = nC_V\Delta T$ (valid here because $\Delta W = 0$ ): $\Delta Q = 2 \times 20.785 \times (-75) = -3117.75\ \text{J} \approx \mathbf{-3118\ J}$ . The negative sign means the gas releases about 3118 J of heat to the surroundings as it cools, consistent with $\Delta U < 0$ and $\Delta W = 0$ .

Step 4. For comparison, an isobaric cooling through the same $\Delta T = -75\ \text{K}$ would use $C_P = \tfrac{7}{2}\times8.314 = 29.099\ \text{J mol}^{-1}\text{K}^{-1}$ , giving $\Delta Q_P = 2\times29.099\times(-75) \approx -4365\ \text{J}$ — noticeably more heat would need to be removed, because at constant pressure the gas also contracts and does negative work on itself (work is done on the gas by the surroundings), so more heat must leave to produce the same temperature drop. This is the general rule: for the same $\Delta T$ , isochoric heat exchange is always smaller in magnitude than isobaric, because $C_V < C_P$ always. Timing note: in a competitive exam, do not waste time separately deriving $C_P - C_V = R$ here — it is a standard quoted relation (Mayer's relation, from the Kinetic Theory chapter); simply plug in $C_V = \tfrac52 R$ , $C_P = \tfrac72 R$ for a diatomic gas and move straight to the arithmetic. Budget under 90 seconds for the Gay-Lussac step and under 60 seconds for each heat calculation — the comparison in Step 4 is the kind of "extra" reasoning that separates a full-marks answer from a partial one, so do not skip it even under time pressure.

Common traps

Confusing isochoric with isobaric. Isochoric fixes volume; isobaric fixes pressure. In an isochoric process, pressure and temperature are both free to change (together, proportionally); mixing this up with the constant-pressure case is the single most common error.
Applying $\Delta W = P\Delta V$ with the wrong $\Delta V$ . Some problems disguise a small but nonzero volume change (a piston that moves slightly, a valve that opens briefly) as "constant volume." Always confirm $\Delta V$ is genuinely, exactly zero before invoking the isochoric shortcut — do not assume it from a rigid-sounding description alone.
Using $C_P$ instead of $C_V$ . Since no work is done, only $C_V$ enters $\Delta Q = nC_V\Delta T$ for an isochoric process; substituting $C_P$ silently overestimates the heat required or released, a very common numeric slip under time pressure.
Forgetting to convert temperature to kelvin before applying Gay-Lussac's law. $P/T = \text{constant}$ only holds for absolute temperature; using Celsius values directly gives a wrong ratio.
Drawing the wrong line on a $P$ – $V$ diagram. An isochoric process is a strictly vertical line at fixed $V$ ; a sloped or horizontal line wrongly implies the volume is changing, and immediately signals a conceptual error to an examiner.
Ignoring the sign of $\Delta Q$ . Cooling at constant volume releases heat ( $\Delta Q < 0$ ); a plugged-in positive $C_V\Delta T$ magnitude without tracking the sign of $\Delta T$ leads to the wrong physical conclusion about whether heat was absorbed or released.

Common trapIsochoric does not mean nothing changes — it means volume is what stays fixed

The single biggest misconception is thinking that an isochoric process is somehow a "do-nothing" process because work is zero. In reality, pressure and temperature both change substantially — often dramatically, as inside a pressure cooker — while only volume is pinned. Students who internalise "isochoric = boring, nothing happens" often then wrongly assume $\Delta U = 0$ or $\Delta T = 0$ as well, when in fact $\Delta U = \Delta Q$ can be large and nonzero. The correct mental model is: volume is the one variable held hostage; pressure, temperature, and internal energy are all free to move, and move together in the specific proportional way fixed by $PV = nRT$ and $\Delta Q = nC_V\Delta T$ .

You should be able to…

State the defining condition of an isochoric process ( $\Delta V = 0$ ) and show that it forces $\Delta W = 0$ .
Derive and apply $\Delta Q = \Delta U = nC_V\Delta T$ for an isochoric process, starting from the first law.
Use Gay-Lussac's law ( $P/T = \text{constant}$ ) to relate initial and final pressure and temperature at fixed volume.
Sketch an isochoric process as a vertical line on a $P$ – $V$ diagram and explain why it encloses zero area.
Determine whether heat is absorbed or released in an isochoric process from the sign of the temperature change.
Compare isochoric and isobaric heat exchange for the same temperature change, using $C_V < C_P$ .

Ask the tutor: Want to work through more mixed problems combining Gay-Lussac's law with the first law, or see a side-by-side comparison of isochoric versus isobaric heating on the same $P$ – $V$ axes? Ask the AI tutor for guided isochoric process practice.

Is a pressure cooker really a perfect example of an isochoric process?

It is a very good practical approximation, though not perfect. While the lid stays sealed and the metal walls barely expand, the volume of gas trapped inside is essentially fixed, so treating the heating phase as isochoric is standard and reasonable for problem-solving. In reality the safety valve eventually releases gas once pressure exceeds a threshold, which technically ends the strict constant-volume phase — but up to that point, and for exam purposes, a sealed rigid vessel heated by a flame is exactly the textbook isochoric scenario.

If no work is done in an isochoric process, does that mean it is somehow more efficient at heating a gas?

In one specific sense, yes: an isochoric process delivers all of the supplied heat into internal energy, with none diverted into mechanical work, so for the same heat input you get a larger internal-energy (and temperature) rise than you would at constant pressure, where some heat is spent expanding the gas. This is exactly why $C_V$ is smaller than $C_P$ for the same gas — less heat is needed per degree of temperature rise when no work has to be done. It is not "efficiency" in the everyday sense, just a direct bookkeeping consequence of the first law with $\Delta W = 0$ .

Can an isochoric process ever be adiabatic at the same time?

Only in the trivial, uninteresting case. An adiabatic process has $\Delta Q = 0$ , and an isochoric process has $\Delta W = 0$ ; combining both in the first law gives $\Delta U = \Delta Q - \Delta W$ ... actually with both zero, $\Delta U = 0$ too, meaning nothing changes at all — no heat in or out, no work, no change in internal energy or temperature. So a process that is genuinely both isochoric and adiabatic is simply the gas sitting still in equilibrium, not a meaningful "process" in the exam sense. Exam questions treat isochoric and adiabatic as two distinct, separately testable named processes.

How is an isochoric process different from just heating a gas in an open beaker?

Heating gas in an open beaker (or any container with a free surface or movable boundary) generally lets the gas expand as it warms, so both $V$ and $P$ can change — that is neither isochoric nor isobaric in general, and would need a more complex treatment. The isochoric process specifically requires a rigid, sealed container (or an external constraint that actively prevents any volume change) so that $\Delta V = 0$ is guaranteed by the setup itself, not just approximately true. Always check the problem statement for words like "rigid," "sealed," "fixed volume," or "container that does not expand" before assuming a process is isochoric.