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Physics · Thermodynamics

Thermodynamic Processes — Isothermal, Adiabatic, Isobaric & Isochoric

A thermodynamic process is any route by which a system passes from one equilibrium state to another. NCERT §11.8 names four special routes — isothermal, adiabatic, isochoric and isobaric — plus the cyclic process that strings them into a loop. Each is defined by one quantity held fixed, and that single constraint collapses the first law $\Delta Q = \Delta U + \Delta W$ into a clean, examinable form. This deep-dive sets the four processes side by side, derives every work expression NCERT gives, and dissects the slope question that NEET asks almost every other year.

The quasi-static idealisation

Before any process can be drawn as a curve, it must be quasi-static. Suppose the weight on a movable piston is suddenly lifted. The piston accelerates outward and, for an instant, the gas passes through non-equilibrium states that have no well-defined single pressure or temperature. A non-equilibrium state cannot be plotted as one point on a P-V diagram, so a sudden change cannot be drawn as a continuous curve at all.

NCERT therefore introduces the quasi-static process: an idealised, infinitely slow change in which the system remains in thermal and mechanical equilibrium with its surroundings at every stage. At each instant the system's pressure differs from the external pressure only infinitesimally, and the same holds for temperature. Only such a process traces a continuous, well-defined path that can be integrated for work. Every formula in this article assumes the process is quasi-static — this is the silent assumption NCERT states it will keep "from now on... except when stated otherwise."

The first law as a sorting engine

The whole subject of processes is one equation read four ways. The first law of thermodynamics states

$$\Delta Q = \Delta U + \Delta W, \qquad \Delta W = P\,\Delta V .$$

Here $\Delta Q$ is heat supplied to the system, $\Delta U$ the change in internal energy (a state variable, depending only on temperature for an ideal gas), and $\Delta W$ the work done by the system. Each special process fixes one quantity, and that constraint kills one term in the equation. Read the table below as the spine of everything that follows.

ProcessHeld fixedTerm that vanishesFirst law reduces to
IsothermalTemperature $T$$\Delta U = 0$$\Delta Q = \Delta W$
AdiabaticHeat $\Delta Q = 0$ (insulated)$\Delta Q = 0$$\Delta U = -\Delta W$
IsochoricVolume $V$$\Delta W = 0$$\Delta Q = \Delta U$
IsobaricPressure $P$none$\Delta Q = \Delta U + P\,\Delta V$

Notice the symmetry: isothermal and adiabatic kill different terms ($\Delta U$ versus $\Delta Q$) yet are constantly confused; isochoric and isobaric are the constant-volume and constant-pressure twins that pair with $C_v$ and $C_p$ respectively. We now take each in turn, then assemble the master grid.

Isothermal process

An isothermal process holds the temperature fixed throughout — for example, a gas expanding slowly in a metallic cylinder immersed in a large constant-temperature reservoir, whose heat capacity is so large that the heat it exchanges does not alter its own temperature. With $T$ constant, the ideal gas equation gives

$$PV = \text{constant} \qquad (\text{Boyle's Law}),$$

so pressure varies inversely with volume. To find the work, sum $\Delta W = P\,\Delta V$ over the path, substituting $P = \mu RT / V$ and taking the constants out of the integral:

$$W = \int_{V_1}^{V_2} P\,dV = \mu RT \int_{V_1}^{V_2}\frac{dV}{V} = \mu RT \,\ln\!\frac{V_2}{V_1} = 2.303\,\mu RT \,\log_{10}\!\frac{V_2}{V_1}.$$

Because the internal energy of an ideal gas depends only on temperature, $\Delta U = 0$, and the first law gives the defining result $\Delta Q = \Delta W$: the heat absorbed is spent entirely on work. For $V_2 > V_1$ (expansion) $W > 0$ — the gas absorbs heat and does work; for $V_2 < V_1$ (compression) $W < 0$ — work is done on the gas and heat is released.

Adiabatic process

An adiabatic process is the opposite bargain: the system is thermally insulated from its surroundings so that no heat is exchanged, $\Delta Q = 0$. The first law then reads $\Delta U = -\Delta W$. When the gas expands and does positive work, that work is paid for out of internal energy, so the gas cools; when work is done on the gas, its internal energy and temperature rise. The governing relations, quoted by NCERT without proof, are

$$PV^{\gamma} = \text{constant}, \qquad TV^{\gamma-1} = \text{constant}, \qquad \gamma = \frac{C_p}{C_v}.$$

For a change from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$ this gives $P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$. The work done is obtained by integrating $\int P\,dV$ with $P = (\text{const})V^{-\gamma}$:

$$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{\mu R\,(T_1 - T_2)}{\gamma - 1}.$$

Consistency check: if the gas does work ($W > 0$) then $T_2 < T_1$ — the gas cools, exactly as the first law demands; if work is done on the gas ($W < 0$) then $T_2 > T_1$.

Isochoric process

An isochoric (isovolumetric) process holds the volume constant — heating a gas sealed in a rigid vessel is the standard example. Since $\Delta V = 0$, the work term vanishes outright:

$$\Delta W = P\,\Delta V = 0.$$

The first law then collapses to $\Delta Q = \Delta U$. Every joule of heat supplied goes into raising the internal energy and hence the temperature; none escapes as work. The temperature change for a given heat input is set by the molar specific heat at constant volume:

$$\Delta Q = \Delta U = \mu C_v\,\Delta T.$$

On a P-V diagram this is a vertical line — and a vertical line encloses no area, the geometric statement that no work is done.

Isobaric process

An isobaric process holds the pressure constant — boiling water in an open vessel at atmospheric pressure is the textbook case. No term vanishes, so this is the one process where the full first law stays in play. With $P$ constant the work is simply

$$W = P\,(V_2 - V_1) = P\,\Delta V = \mu R\,(T_2 - T_1) = \mu R\,\Delta T,$$

using $PV = \mu RT$ for the last step. The heat supplied splits between raising the internal energy and doing this work, and the temperature change is governed by the molar specific heat at constant pressure:

$$\Delta Q = \mu C_p\,\Delta T.$$

This is why $C_p > C_v$: at constant pressure the gas must be fed extra heat to cover the work $\mu R\,\Delta T$ that it does in expanding, on top of the $\mu C_v\,\Delta T$ that goes into internal energy — the origin of the relation $C_p - C_v = R$.

i
Foundation

Every reduction above starts from one equation. If $\Delta Q$, $\Delta U$ and $\Delta W$ feel slippery, lock them down in the first law of thermodynamics first.

The master comparison table

The four processes form a perfect parallel structure. Memorise this single grid — condition, the first-law reduction, the work expression, and the shape of the P-V curve — and the bulk of NEET thermodynamics is mechanical recall.

ProcessDefining conditionFirst-law formWork done $W$P-V curve
Isothermal $T$ const; $PV=$ const $\Delta U=0,\ \Delta Q=\Delta W$ $\mu RT\ln\!\dfrac{V_2}{V_1}=2.303\,\mu RT\log\!\dfrac{V_2}{V_1}$ Rectangular hyperbola
Adiabatic $\Delta Q=0$; $PV^{\gamma}=$ const $\Delta U=-\Delta W$ $\dfrac{P_1V_1-P_2V_2}{\gamma-1}=\dfrac{\mu R(T_1-T_2)}{\gamma-1}$ Steeper hyperbola (slope $\times\gamma$)
Isochoric $V$ const; $\dfrac{P}{T}=$ const $\Delta Q=\Delta U=\mu C_v\Delta T$ $W=0$ Vertical line
Isobaric $P$ const; $\dfrac{V}{T}=$ const $\Delta Q=\mu C_p\Delta T=\Delta U+P\Delta V$ $P\Delta V=\mu R\Delta T$ Horizontal line
Figure 1 V P start Isobaric (P const) Isochoric (V const) Isothermal Adiabatic
Figure 1 — The four processes leaving a single initial state. Isobaric runs horizontally (constant P), isochoric runs vertically (constant V), and the two curves fall away to the right: the adiabatic (coral) drops more steeply than the isothermal (teal).

Isothermal vs adiabatic slope

The most heavily tested single fact in this subtopic is which curve is steeper. Differentiate each constraint at a common point. For the isothermal $PV = \text{const}$,

$$P\,dV + V\,dP = 0 \;\Rightarrow\; \left(\frac{dP}{dV}\right)_{\text{iso}} = -\frac{P}{V}.$$

For the adiabatic $PV^{\gamma} = \text{const}$,

$$V^{\gamma}\,dP + \gamma P V^{\gamma-1}\,dV = 0 \;\Rightarrow\; \left(\frac{dP}{dV}\right)_{\text{adi}} = -\gamma\,\frac{P}{V}.$$

Both slopes are negative, but the adiabatic slope is larger in magnitude by exactly the factor $\gamma = C_p/C_v > 1$. The adiabatic curve is steeper than the isothermal by the factor $\gamma$. Physically, an adiabatic compression also heats the gas, so its pressure climbs faster than in an isothermal compression where the extra heat is bled off to the reservoir.

Figure 2 V P Isothermal: slope = −P/V Adiabatic: slope = −γP/V Adiabatic is steeper by the factor γ > 1
Figure 2 — Through the same state, the adiabatic tangent (coral) is steeper than the isothermal tangent (teal). The ratio of their slopes is exactly $\gamma = C_p/C_v$.

A direct corollary, asked in NEET 2016: compressing a gas to the same final volume takes more work adiabatically than isothermally. The steeper adiabatic curve sits above the isothermal one during compression, so the area under it — the work — is larger.

Cyclic process and loop area

A cyclic process returns the system to its initial state after a sequence of steps. Since internal energy is a state variable depending only on the state, the net change over a complete cycle is zero:

$$\Delta U_{\text{cycle}} = 0 \;\Rightarrow\; \Delta Q_{\text{net}} = W_{\text{net}}.$$

All the heat absorbed over the cycle emerges as net work. Geometrically, that net work equals the area enclosed by the closed loop on the P-V diagram: positive for a clockwise loop (the gas does net work, as in a heat engine) and negative for an anticlockwise loop (net work is done on the gas, as in a refrigerator). The straight-segment legs of a rectangular loop are simply isobaric (horizontal) and isochoric (vertical) processes back to back.

Figure 3 V P a b c d W_net = enclosed area Clockwise loop ⇒ net work done BY the gas
Figure 3 — A cyclic process abcda. Over the full loop $\Delta U = 0$, so the net heat absorbed equals the net work, which is the shaded enclosed area. A clockwise traversal gives positive net work.
Quick recap

Four processes in one breath

  • Isothermal: $T$ fixed, $PV=$ const, $\Delta U=0$, so $\Delta Q=\Delta W=\mu RT\ln(V_2/V_1)=2.303\,\mu RT\log(V_2/V_1)$.
  • Adiabatic: $\Delta Q=0$, $PV^{\gamma}=$ const and $TV^{\gamma-1}=$ const, $\Delta U=-\Delta W$, $W=(P_1V_1-P_2V_2)/(\gamma-1)=\mu R(T_1-T_2)/(\gamma-1)$.
  • Isochoric: $V$ fixed, $W=0$, so $\Delta Q=\Delta U=\mu C_v\Delta T$; vertical P-V line.
  • Isobaric: $P$ fixed, $W=P\Delta V=\mu R\Delta T$, $\Delta Q=\mu C_p\Delta T$; horizontal P-V line.
  • Slopes: isothermal $-P/V$, adiabatic $-\gamma P/V$; the adiabatic is steeper by $\gamma$, so adiabatic compression takes more work.
  • Cyclic: $\Delta U=0$ over a loop, net $W=$ enclosed area; clockwise positive, anticlockwise negative.
  • Two zeros to never swap: isothermal is $\Delta U=0$; adiabatic is $\Delta Q=0$.

NEET PYQ Snapshot — Thermodynamic Processes

Processes are tested almost every year. The same three handles recur: identify the curve, kill the right term, read the loop.

NEET 2022

An ideal gas undergoes four different processes from the same initial state, shown as curves 1, 2, 3 and 4. The processes are adiabatic, isothermal, isobaric and isochoric. The curve representing the adiabatic process is:

  1. 2
  2. 3
  3. 4
  4. 1
Answer: (1) curve 2

Slope test. $\left(\tfrac{dP}{dV}\right)_{\text{adi}}=-\gamma P/V$ while $\left(\tfrac{dP}{dV}\right)_{\text{iso}}=-P/V$. Since $\gamma>1$, the adiabatic is the steeper of the two falling curves. The vertical curve is isochoric, the horizontal one isobaric, the gentler curve isothermal, and the steepest curve (2) is the adiabatic.

NEET 2016

A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half. Then:

  1. Compressing the gas through the adiabatic process will require more work to be done.
  2. Compressing the gas isothermally or adiabatically will require the same amount of work.
  3. Which case requires more work depends on the atomicity of the gas.
  4. Compressing the gas isothermally will require more work to be done.
Answer: (1) adiabatic requires more work

Steeper curve, larger area. Isothermal compression to $V/2$ raises pressure to $2P_1$ (Boyle); adiabatic compression to $V/2$ raises it to $2^{\gamma}P_1$, which is higher. The adiabatic path lies above the isothermal during compression, so the area under it — the work done — is greater.

NEET 2024

A system is taken through the cycle abcda: $a(100~\text{cm}^3,100~\text{kPa})\to b(400~\text{cm}^3,100~\text{kPa})\to c(400~\text{cm}^3,300~\text{kPa})\to d(100~\text{cm}^3,300~\text{kPa})\to a$. The work done by the gas along path $bc$ is:

  1. Zero
  2. 30 J
  3. −90 J
  4. −60 J
Answer: (1) Zero

Identify the leg. Along $bc$ the volume is fixed at $400~\text{cm}^3$ while pressure rises from $100$ to $300~\text{kPa}$ — an isochoric process. With $\Delta V=0$, $W=P\Delta V=0$. The pressure rise feeds internal energy, not work. (Do not confuse this single leg with the net cycle work, which is the enclosed loop area.)

FAQs — Thermodynamic Processes

Short answers to the process questions NEET aspirants get wrong most often.

In an isothermal process, is it the heat or the internal energy that is zero?
The internal energy change is zero, not the heat. For an ideal gas the internal energy depends only on temperature, and in an isothermal process the temperature is fixed, so ΔU = 0. The first law ΔQ = ΔU + ΔW then becomes ΔQ = ΔW: the heat supplied is non-zero and equals the work done by the gas. Writing ΔQ = 0 for an isothermal process is the most common NEET error — that is the adiabatic condition, not the isothermal one.
What stays constant in an adiabatic process — the temperature or the heat exchange?
The heat exchange is zero (the system is insulated), so ΔQ = 0. The temperature is NOT constant in an adiabatic process. From the first law, ΔU = −ΔW: when the gas expands it does work at the expense of its own internal energy, so it cools; when it is compressed, work done on it raises its internal energy, so it heats up. The governing relation is PVγ = constant, equivalently TV^(γ−1) = constant.
Why is the adiabatic P-V curve steeper than the isothermal one?
Differentiating PV = constant gives the isothermal slope dP/dV = −P/V. Differentiating PVγ = constant gives the adiabatic slope dP/dV = −γP/V. Since γ = Cp/Cv > 1, the adiabatic slope is steeper by exactly the factor γ. Physically, in an adiabatic compression the gas also heats up, so pressure rises faster than in an isothermal compression where the heat is allowed to escape.
Why is the work done zero in an isochoric process?
Work done by a gas is W = P ΔV. In an isochoric process the volume is held constant, so ΔV = 0 and therefore W = 0 — irrespective of how the pressure changes. The first law then reduces to ΔQ = ΔU = nCv ΔT: every joule of heat supplied goes into raising the internal energy and temperature of the gas. On a P-V diagram an isochoric process is a vertical line, which encloses no area, confirming zero work.
What is a quasi-static process and why do we assume it?
A quasi-static process is an idealised, infinitely slow process in which the system stays in thermal and mechanical equilibrium with its surroundings at every stage — the pressure and temperature of the system differ from the external values only infinitesimally. We assume it because only equilibrium states have well-defined P, V and T, so only a quasi-static process can be drawn as a continuous curve on a P-V diagram and integrated for work. Real processes that are slow and free of large gradients approximate it well.
How is the net work in a cyclic process related to the P-V loop?
In a cyclic process the system returns to its initial state, so the internal energy change over the full cycle is zero (ΔU = 0). The first law gives net ΔQ = net W. Graphically, the net work equals the area enclosed by the closed P-V loop: positive (work done by the gas) for a clockwise loop, negative (work done on the gas) for an anticlockwise loop.
Does compressing a gas to half its volume take more work isothermally or adiabatically?
Adiabatic compression takes more work. Because the adiabatic curve is steeper, reaching the same final volume drives the pressure higher (the trapped heat raises the temperature), so the gas pushes back harder throughout the stroke. The area under the adiabatic curve therefore exceeds the area under the isothermal curve for the same volume change. This is exactly the result tested in NEET 2016.
Related Topics
Thermodynamics — Parent chapter Full chapter map: state variables, laws, processes, engines. First Law of Thermodynamics ΔQ = ΔU + ΔW; the equation every process reduces. Internal Energy, Heat & Work Why U is a state variable while Q and W are path-dependent. Reversible & Irreversible Processes Quasi-static idealisation; what makes a process reversible. Carnot Engine & Efficiency A cycle built from two isothermals and two adiabatics. Heat Engines Cyclic processes converting heat into net work. Second Law of Thermodynamics Direction of processes; why some routes are forbidden.