In short

An isobaric process is one that proceeds at constant pressure. For n moles of an ideal gas changing volume from V_1 to V_2 at pressure P, the work done by the gas is simply W = P\,\Delta V; the heat absorbed is Q = n C_P\, \Delta T, where C_P is the molar specific heat at constant pressure. An isochoric (also called isovolumetric) process holds the volume constant. Because no volume change means no pressure-volume work, W = 0, so the first law collapses to \Delta U = Q = n C_V\, \Delta T, where C_V is the molar specific heat at constant volume. Combining these two with the ideal gas law gives the Mayer relation

\boxed{\;C_P - C_V \;=\; R,\;}

whose physical content is simple: to raise the temperature of a gas by 1 K at constant pressure, you must provide not only C_V to raise the internal energy but an extra R to pay for the work the gas does while expanding. The ratio \gamma = C_P/C_V — the adiabatic index — takes the values \gamma = 5/3 for a monatomic gas (helium, argon), \gamma = 7/5 = 1.4 for a diatomic gas at room temperature (air, nitrogen, oxygen), and \gamma = 4/3 for many polyatomic gases (CO₂, methane). These two processes are the simplest bricks from which every more interesting thermodynamic cycle is built.

An Indian home kitchen makes idli on Sunday morning. You load the steamer with three plates, pour two cups of water into the base, clamp the lid down loosely so steam can escape through the gap around its rim, and switch on the flame. For the next fifteen minutes, the water below boils, steam rises through the plates, cooks the batter, and exits — and the pressure inside the steamer stays at atmospheric pressure the whole time, because the lid is loose and any extra steam simply leaks out. What happens inside is an approximately isobaric process: heat pours in, water vaporises, the total gas volume swells, and the rising steam does mechanical work pushing air and steam upward out of the cooker. Pressure stays the same because the lid cannot hold any excess.

Now swap the scene. You take a sealed aluminium can of coke out of the fridge in Delhi, put it on the dashboard of a car parked in the sun, and walk away. The can's walls are rigid (you can't compress them with your hand); the seal does not leak. As the car heats up, the gas and liquid inside heat up too — but their volume is fixed by the rigid can. No piston moves. No work crosses the boundary. The heat from the Delhi sun goes entirely into raising the internal energy of the fluid, which shows up as rising temperature and — by Gay-Lussac's law — rising pressure. That is an isochoric process. If the sun is strong enough, the can eventually splits.

These two scenarios — constant pressure and constant volume — are the two simplest ways a thermodynamic process can proceed, and they are the workhorses of every gas-law problem on a JEE paper. The first law of thermodynamics, \Delta U = Q - W, becomes particularly clean in each case: for the idli cooker, W = P\, \Delta V is just pressure times volume change; for the coke can, W = 0 and all the heat deposits as internal energy. From these two building blocks emerge the molar specific heats C_V and C_P, the Mayer relation C_P - C_V = R that links them, and the adiabatic index \gamma = C_P/C_V that shows up everywhere from sound propagation to diesel engines.

This article derives all of that from scratch. You will see why the heat required to raise the temperature of a gas depends on which path you take, how the R in Mayer's relation is exactly the work you would have to do to expand one mole of ideal gas by one kelvin's worth of thermal swelling at atmospheric pressure, and why \gamma takes the specific numerical values it does for different gas types.

The four canonical processes — where we are in the landscape

A thermodynamic process is any path the system takes between two equilibrium states. You can in principle follow any curve you like in the pressure-volume plane. In practice, four named processes cover almost every problem in a physics course:

PV diagram showing the four canonical processes from a common initial stateOn a pressure versus volume plane with pressure increasing upward and volume increasing rightward, four curves leave a shared initial point in the middle of the diagram. A horizontal line labelled isobaric extends to the right at the same pressure. A vertical line labelled isochoric extends upward at the same volume. A hyperbola-shaped curve labelled isothermal falls smoothly as volume grows. A steeper hyperbola labelled adiabatic falls faster than the isotherm. The initial state is marked with a red dot and labelled 'start'.V (volume)P (pressure)startisobaric (P const)isochoric (V const)isothermaladiabatic
The four canonical processes, all starting from the same state. An isobaric process is a horizontal line (pressure never changes). An isochoric process is a vertical line (volume never changes). An isothermal process is a hyperbola in $P$ and $V$. An adiabatic process is a steeper hyperbola — for reasons this article and the next one derive.

This article focuses on the first two. Isothermal and adiabatic are covered separately in thermodynamic processes — isothermal and adiabatic because their work integrals require a little more algebra.

The isobaric process

An isobaric process is one in which the pressure of the system is held constant throughout. The canonical physical setup is a gas in a cylinder sealed by a piston that has a fixed weight on top of it. The atmosphere plus the weight exert a constant downward pressure on the piston; if the gas is free to push the piston up or down, it reaches equilibrium at that same pressure. Any heat added or removed changes V and T while P stays pinned.

Isobaric setup — piston with fixed weight above a gas cylinderA vertical cylinder contains gas. A piston floats on top of the gas; on top of the piston rests a labelled weight W. An arrow labelled Q enters from a flame at the bottom. Another arrow labelled W equals P times delta V points upward from the piston, representing work done on the surroundings. The pressure inside is labelled P equals atmospheric plus weight over area, a constant.weight Mggas (P constant)QW = P ΔVP = P₀ + Mg/A = const
The textbook isobaric setup: a frictionless piston of area $A$ carries a fixed weight $Mg$. Whatever the gas does inside, the pressure it supports at the top is $P = P_{\text{atm}} + Mg/A$, a constant. As heat is added, the gas expands and the piston rises — but the pressure cannot change.

Work done by the gas

Start from the general expression for work, which the first law of thermodynamics article derived from "force times distance":

W \;=\; \int_{V_1}^{V_2} P\, dV.

Because P is constant in an isobaric process, it comes outside the integral:

W \;=\; P \int_{V_1}^{V_2} dV \;=\; P\,(V_2 - V_1) \;=\; \boxed{\;P\, \Delta V.\;}

Why: a constant under an integral factors out; the remaining integral is just the difference in volume. Geometrically, the area under a horizontal line from V_1 to V_2 on a PV diagram is the rectangle P \times (V_2 - V_1).

Heat absorbed — introducing C_P

The heat required to raise the temperature of n moles of a gas by \Delta T at constant pressure is defined by

Q \;=\; n\, C_P\, \Delta T,

Why: this is the definition of the molar specific heat at constant pressure. It says "give me an amount, multiply by the temperature change, and I'll tell you the heat needed when the process is isobaric".

The subscript P is essential — C_P is not the same as the specific heat at constant volume, precisely because the pressure-constancy constraint is different and forces the gas to do work on its surroundings.

Change in internal energy

For an ideal gas, the internal energy depends only on T:

\Delta U \;=\; n\, C_V\, \Delta T.

Why: this works for any process in an ideal gas, not just isochoric ones, because U = U(T) — the C_V formula is how you turn a temperature change into an internal-energy change regardless of path. The subscript V in C_V simply reminds you how to measure C_V experimentally (hold V fixed so \Delta U = Q); the formula \Delta U = n C_V \Delta T itself is path-independent for ideal gases.

Applying the first law to an isobaric process

Substitute everything into \Delta U = Q - W:

n\, C_V\, \Delta T \;=\; n\, C_P\, \Delta T \;-\; P\,\Delta V.

The ideal gas law says that at constant pressure P\, V = n R T, so P\, \Delta V = n R\, \Delta T. Substitute:

n\, C_V\, \Delta T \;=\; n\, C_P\, \Delta T \;-\; n R\, \Delta T.

Cancel n\, \Delta T (nonzero for a nontrivial process):

C_V \;=\; C_P \;-\; R \quad\Longrightarrow\quad \boxed{\;C_P \;-\; C_V \;=\; R.\;}

Why: this is Mayer's relation, derived as a direct consequence of the first law applied to an isobaric process of an ideal gas. The physics reading: to raise one mole of ideal gas by one kelvin, you must supply C_V joules to raise its internal energy plus R \approx 8.314 joules to pay for the work it does pushing the piston. That pushing work is always R joules per mole per kelvin at constant pressure — no matter what gas you are using — because the ideal gas law P\, dV = n R\, dT fixes the push exactly.

Mayer's relation is one of the most remarkable small results in thermodynamics. It connects three quantities — a heat capacity measured at constant pressure, a heat capacity measured at constant volume, and the universal gas constant — and tells you that the first two differ by the third, for every ideal gas, regardless of whether it's monatomic, diatomic, or polyatomic.

Concrete C_P and C_V values for Indian-classroom gases

From degrees of freedom and equipartition, for an ideal gas the internal energy per mole is U_m = \frac{f}{2} R T, where f is the number of active quadratic degrees of freedom. The specific heat at constant volume is then

C_V \;=\; \frac{1}{n}\, \frac{\partial U}{\partial T} \;=\; \frac{f}{2}\, R.

Combine with Mayer's relation C_P = C_V + R to get

C_P \;=\; \frac{f + 2}{2}\, R.
Gas type f C_V C_P \gamma = C_P/C_V
Monatomic (He, Ne, Ar) 3 \tfrac{3}{2} R = 12.5 J K⁻¹ mol⁻¹ \tfrac{5}{2} R = 20.8 5/3 \approx 1.667
Diatomic at room temp (N₂, O₂, H₂) 5 \tfrac{5}{2} R = 20.8 \tfrac{7}{2} R = 29.1 7/5 = 1.400
Linear triatomic / rigid polyatomic (CO₂ approx.) 6 3 R = 24.9 4 R = 33.3 4/3 \approx 1.333

The table reads right off the equipartition formula for each gas type; the quoted numerical specific heats are at ordinary atmospheric temperatures, where vibrational modes of diatomic gases have not yet become active.

The adiabatic index \gamma is the single most important specific-heat quantity in thermodynamics. It appears in the adiabatic equation PV^\gamma = \text{const}, in the formula for the speed of sound in a gas (v = \sqrt{\gamma P/\rho}), in the efficiency of Otto and Diesel cycles, and in countless other places. The values 5/3 and 7/5 are worth memorising — they come up in every JEE thermodynamics problem.

The isochoric process

An isochoric (equivalently, isovolumetric or isometric) process is one in which the volume of the system is held constant. The canonical physical setup is a gas in a rigid sealed container. Heat may enter or leave through the walls, but the walls do not flex, so the gas cannot do any pressure-volume work on the surroundings.

Isochoric setup — rigid sealed container with heat in, no work outA thick-walled sealed rectangular container holds a gas. An arrow labelled Q enters from below, representing heat in from a flame. An arrow labelled W equals zero with a strikethrough points upward from the top, indicating no work crosses the boundary. Inside the box, the label reads V constant. Below the figure, the formulas W equals zero and delta U equals Q equals n Cv delta T are written.rigid containerV = constantQ inW = 0 (walls rigid)
The textbook isochoric setup: a rigid container whose walls cannot move. The piston is gone, replaced by steel. $W = \int P\, dV = 0$ identically because $dV = 0$. All heat added or removed shows up as a change in internal energy.

Work, heat, and internal-energy change

Because V is constant, dV = 0 throughout, so

W \;=\; \int_{V_1}^{V_2} P\, dV \;=\; 0.

Heat absorbed in an isochoric process of an ideal gas is, by definition of C_V:

Q \;=\; n\, C_V\, \Delta T.

The first law \Delta U = Q - W with W = 0 gives

\Delta U \;=\; Q \;=\; n\, C_V\, \Delta T.

Why: no volume change means no work crosses the boundary; every joule of heat goes into internal energy. This is actually how C_V is measured in a lab — put the gas in a rigid container, add a known amount of heat, measure the temperature rise.

Pressure change at constant volume — Gay-Lussac's law

For an ideal gas with fixed n and V, the ideal gas law PV = nRT becomes P = (nR/V)\,T, i.e. P \propto T:

\frac{P_2}{P_1} \;=\; \frac{T_2}{T_1}.

Why: constant V and constant n make the ratio P/T a fixed quantity, so P and T rise together in strict proportion. Double the temperature on the kelvin scale and the pressure doubles.

This is why the coke can on your dashboard can explode on a hot day: a can-interior temperature rise from, say, 295 K (a cold can fresh from the fridge) to 345 K (a very hot dashboard, about 72 °C in the sun) multiplies the internal pressure by 345/295 \approx 1.17. The can was already pressurised by carbon dioxide at about 3 atm; it now carries about 3.5 atm. For a thin aluminium can that can rupture at around 6 atm, this is well within the safety margin — but a partially shaken can, a poorly sealed can, or a slightly weaker spot can fail. The physics is simple; the engineering margin is all that stands between you and a wet dashboard.

PV diagrams at a glance — work as area

On a pressure-volume diagram, the work done by the gas is the area under the process curve. This is directly visible for our two special processes and makes their contrast immediate:

Work as area on PV diagrams — isobaric vs isochoricTwo PV diagrams side by side. Left: an isobaric process drawn as a horizontal line from V1 to V2 at height P; the rectangle under it, of width V2 minus V1 and height P, is shaded red, labelled W equals P delta V, and represents work done by the gas. Right: an isochoric process drawn as a vertical line from P1 to P2 at fixed V; no area is enclosed between the curve and the V-axis because the curve is vertical; the figure is labelled W equals 0.IsobaricVPV₁V₂PW = P·ΔVIsochoricVPVP₁P₂W = 0(no area)
Left: in an isobaric process the work is the shaded rectangle, $W = P\,\Delta V$. Right: in an isochoric process the process curve is vertical, so there is no area under it and $W = 0$. The full geometry of PV work is treated in [PV diagrams and work](/wiki/pv-diagrams-and-work).

Worked examples

Example 1: Heating an idli steamer at constant pressure

An idli steamer in a Chennai kitchen contains air initially occupying V_1 = 2.0\text{ L} = 2.0 \times 10^{-3}\text{ m}^3 at atmospheric pressure P = 1.01 \times 10^5\text{ Pa} and temperature T_1 = 300\text{ K}. The steamer's loosely fitting lid keeps the internal pressure at 1 atm throughout. Heat is added until the gas temperature rises to T_2 = 360\text{ K} (ignoring the water and steam as an idealisation). Treat the gas as an ideal diatomic gas (C_V = \frac{5}{2} R, C_P = \frac{7}{2} R). Find Q, W, and \Delta U for the process, and the new volume.

Isobaric heating of a gas — before and after statesTwo cylinder diagrams side by side show the same gas before and after heating. Before: piston at low height, volume labelled V1 = 2 L, temperature T1 = 300 K, pressure P = 1 atm. After: piston higher, volume labelled V2, temperature T2 = 360 K, pressure P still equal to 1 atm. An arrow labelled Q from a flame below enters the 'before' cylinder. An arrow labelled W equals P times delta V points up from the top of the 'after' cylinder.BeforeV₁ = 2 LT₁ = 300 KP = 1 atmAfterV₂ = ?T₂ = 360 KP = 1 atmQW = PΔV
The gas heats up at constant pressure. The piston rises, expanding the gas, and some of the added heat goes into the work of pushing the piston up; the rest raises the internal energy.

Step 1. Count moles of gas from the initial state.

n \;=\; \frac{P V_1}{R T_1} \;=\; \frac{(1.01 \times 10^5)(2.0 \times 10^{-3})}{(8.314)(300)} \;=\; \frac{202}{2494.2} \;\approx\; 0.0810\text{ mol}.

Why: we need n to use the molar specific heats later. Read it off from the initial conditions by inverting PV = nRT.

Step 2. Find the final volume from the isobaric condition.

At constant pressure, V/T = nR/P is constant, so V \propto T:

V_2 \;=\; V_1 \cdot \frac{T_2}{T_1} \;=\; 2.0 \times 10^{-3} \times \frac{360}{300} \;=\; 2.4 \times 10^{-3}\text{ m}^3 \;=\; 2.4\text{ L}.

Why: Charles's law — at constant pressure, volume is directly proportional to temperature on the kelvin scale. A 20 % rise in temperature produces a 20 % rise in volume.

Step 3. Compute the work done by the gas.

W \;=\; P\,\Delta V \;=\; (1.01 \times 10^5)(2.4 - 2.0) \times 10^{-3} \;=\; (1.01 \times 10^5)(4.0 \times 10^{-4}) \;=\; 40.4\text{ J}.

Why: for an isobaric process W = P\,\Delta V directly. The sign is positive because the gas expanded against atmospheric pressure. Alternatively, W = n R \,\Delta T = 0.0810 \times 8.314 \times 60 \approx 40.4 J — same answer, as it must be.

Step 4. Compute the heat absorbed.

Q \;=\; n C_P\, \Delta T \;=\; 0.0810 \times \tfrac{7}{2} \times 8.314 \times 60 \;=\; 0.0810 \times 29.10 \times 60 \;\approx\; 141.4\text{ J}.

Why: at constant pressure, use C_P. For a diatomic gas, C_P = \frac{7}{2} R. The product \frac{7}{2} R = 29.1 J mol⁻¹ K⁻¹ — a number you should have at your fingertips.

Step 5. Compute the change in internal energy, as a check.

\Delta U \;=\; n C_V \Delta T \;=\; 0.0810 \times \tfrac{5}{2} \times 8.314 \times 60 \;\approx\; 101.0\text{ J}.

And verify the first law:

Q - W \;=\; 141.4 - 40.4 \;=\; 101.0\text{ J} \;=\; \Delta U. \quad\checkmark

Why: Q - W must equal \Delta U for any process. Computing \Delta U independently and seeing the two match is a sanity check worth doing on every thermo problem.

Result. Q = 141.4\text{ J}, W = 40.4\text{ J}, \Delta U = 101.0\text{ J}, V_2 = 2.4\text{ L}.

What this shows. Of the 141 J of heat you poured in, about 101 J went into raising the internal energy (and hence the temperature from 300 K to 360 K) and about 40 J paid for the work of pushing air out of the steamer. Notice the ratio: 40/141 ≈ 28 %. For a diatomic gas at constant pressure, a fraction R/C_P = 2/7 \approx 28.6\% of the heat input always goes to work regardless of the temperature change. That is the geometric meaning of Mayer's relation.

Example 2: A sealed aluminium can on a Delhi dashboard

A 330 mL aluminium can of carbonated water (assume for simplicity that the can contains only gas) is sealed at pressure P_1 = 3.0 \times 10^5\text{ Pa} and temperature T_1 = 278\text{ K} (taken straight from a refrigerator). It is placed on the dashboard of a closed car in Delhi. The can's walls are rigid. Over a few minutes the gas inside heats to T_2 = 345\text{ K}. Treat the gas as ideal diatomic. The can's volume is V = 330\text{ cm}^3 = 3.3 \times 10^{-4}\text{ m}^3. Find the heat absorbed by the gas, the work done, the change in internal energy, and the final pressure.

Sealed aluminium can on a sunny dashboardA sealed aluminium can is drawn as a tall rectangle with rounded top. Inside, labels read V = 330 mL, T1 = 278 K, P1 = 3 atm, rising to T2 = 345 K with pressure increasing. Arrows labelled Q enter from the sides representing radiant heat from a sun symbol drawn above. An arrow labelled W equals zero with strikethrough, because walls are rigid. A note reads 'rigid walls, V constant'.V = 330 mLT₁ = 278 KP₁ = 3 atm→ T₂ = 345 KP₂ = ?QQW = 0 (rigid walls)
The car dashboard radiates heat through the can's walls, but the walls do not flex. All heat goes into raising $U$; none goes into mechanical work.

Step 1. Moles of gas.

n \;=\; \frac{P_1 V}{R T_1} \;=\; \frac{(3.0 \times 10^5)(3.3 \times 10^{-4})}{(8.314)(278)} \;=\; \frac{99}{2311.3} \;\approx\; 0.0428\text{ mol}.

Step 2. Work done.

W \;=\; 0.

Why: rigid walls \Rightarrow V constant \Rightarrow dV = 0 \Rightarrow W = \int P\, dV = 0. No exception, no subtlety.

Step 3. Change in internal energy.

\Delta U \;=\; n C_V\, \Delta T \;=\; 0.0428 \times \tfrac{5}{2} \times 8.314 \times (345 - 278) \;=\; 0.0428 \times 20.785 \times 67 \;\approx\; 59.6\text{ J}.

Step 4. Heat absorbed from the first law.

Q \;=\; \Delta U + W \;=\; 59.6 + 0 \;\approx\; 59.6\text{ J}.

Why: with W = 0, the first law says Q = \Delta U. Every joule that crossed the can wall as heat was stored as internal energy. This is why a sealed can of soda gets warm all the way through; the heat has nowhere else to go.

Step 5. Final pressure (Gay-Lussac).

P_2 \;=\; P_1 \cdot \frac{T_2}{T_1} \;=\; 3.0 \times 10^5 \times \frac{345}{278} \;\approx\; 3.72 \times 10^5\text{ Pa} \;\approx\; 3.7\text{ atm}.

Result. Q = +59.6\text{ J}, W = 0, \Delta U = +59.6\text{ J}, P_2 \approx 3.72 \times 10^5 Pa.

What this shows. At constant volume, the first law is at its starkest: heat and internal-energy change are the same number. No energy is "wasted" on pushing a piston; none is recovered as mechanical work. The can heats and pressurises, end of story. That is why aluminium soft-drink cans are engineered to carry several atmospheres of overpressure — the same physics that cooks a pressure cooker's contents can rupture a can that was designed for the fridge but left in the sun.

Common confusions

If you came here to understand the specific heats of gases and the difference between isobaric and isochoric processes, use the formulas in your problems, you have what you need. What follows is for readers who want the deeper picture — enthalpy as the natural state function of isobaric processes, and why the adiabatic index \gamma depends on the temperature through the quantum mechanics of rotational and vibrational modes.

Enthalpy: the state function for constant-pressure processes

Define the enthalpy of a system by

H \;\equiv\; U + PV.

H is a state function because U, P, and V are all state functions. Compute its differential:

dH \;=\; dU + P\, dV + V\, dP.

For an isobaric process (dP = 0), the last term vanishes. Substitute the first law dU = dQ - P\, dV:

dH \;=\; (dQ - P\, dV) + P\, dV \;=\; dQ.

So for a constant-pressure process,

\boxed{\;Q_P \;=\; \Delta H.\;}

Why: in a constant-pressure process, the heat absorbed equals the change in a state function — the enthalpy — rather than a path-dependent quantity. That is why chemists, who work at atmospheric pressure almost always, treat \Delta H as their primary thermodynamic coordinate. It makes their heats of reaction into state-function differences, independent of how the reaction was run.

The molar heat capacity at constant pressure relates to H the way C_V relates to U:

C_P \;=\; \frac{1}{n}\left(\frac{\partial H}{\partial T}\right)_P.

For an ideal gas, H = U + PV = U + nRT = \frac{f}{2} nRT + nRT = \frac{f+2}{2} nRT, so

C_P \;=\; \frac{f+2}{2} R,

which matches the earlier derivation. The two-sided symmetry between (U, C_V) at constant volume and (H, C_P) at constant pressure is the cleanest conceptual statement of Mayer's relation.

Why \gamma depends on temperature — the equipartition cliff

The rule of thumb "\gamma = 7/5 for diatomic gases" is only true in a specific temperature range. Consider hydrogen (H_2). A hydrogen molecule has:

  • three translational degrees of freedom — always active
  • two rotational degrees of freedom — active only above a temperature T_{\text{rot}} \sim 100\text{ K}
  • two vibrational degrees of freedom (one kinetic and one potential) — active only above T_{\text{vib}} \sim 6000\text{ K}

At very low temperatures (T < T_{\text{rot}}), only translational modes contribute and f = 3, giving C_V = \tfrac{3}{2} R, C_P = \tfrac{5}{2} R, \gamma = 5/3 — monatomic-like behaviour, even though H_2 is diatomic. At room temperature, rotations are active but vibrations are not, so f = 5, C_V = \tfrac{5}{2} R, \gamma = 7/5 = 1.4. At very high temperatures (T > T_{\text{vib}}), vibrations also turn on, f = 7, C_V = \tfrac{7}{2} R, \gamma = 9/7 \approx 1.29.

Why don't all modes contribute at all temperatures? Because the energy spacing of rotational and vibrational levels is quantised. Classically, each quadratic degree of freedom has to contribute \frac{1}{2} k_B T to the energy by equipartition; quantum mechanically, a mode whose energy gap \Delta E \gg k_B T simply stays in its ground state and does not take up energy. As you cool below T_{\text{rot}}, the rotational mode "freezes out"; as you cool below T_{\text{vib}}, the vibrational modes freeze out. This is one of the earliest experimental hints that led to the quantum theory of gases — classical physics had no explanation for why diatomic hydrogen behaved like a monatomic gas at low temperatures.

For a class 11 problem, you can safely take \gamma = 5/3 for helium, argon, etc.; \gamma = 7/5 for air, nitrogen, oxygen, hydrogen; and leave the more exotic corrections for college chemistry.

Generalised isobaric and isochoric — beyond ideal gases

For a real gas, a liquid, or a solid, the simple W = P\,\Delta V still holds for isobaric processes (it came purely from the definition of work, not from any gas law), and W = 0 still holds for isochoric processes. What changes is how U and the specific heats relate.

For liquids and solids, U depends on T and (weakly) on P, and C_V \approx C_P because these substances expand so little with heating that the work term is negligible. For a real gas, C_V and C_P depend on both T and the intermolecular forces; Mayer's relation C_P - C_V = R generalises to a slightly more complicated formula involving the thermal expansion coefficient and the isothermal compressibility.

But the isobaric and isochoric idealisations are clean enough to handle most introductory thermodynamics: a real idli cooker is approximately isobaric because the lid can't hold much pressure; a real coke can is approximately isochoric because the walls are essentially rigid on the relevant scale. The formulas in this article are what JEE, NEET, and CBSE exams will ask you to use.

Where this leads next