JEE Mains Physics · Thermodynamics

Thermodynamic Processes and Work from P–V Graphs

Each standard process holds one quantity fixed and so zeroes one term of the first law; the work done by the gas along any path is the signed area under it on a P–V diagram.

All Thermodynamics notes JEE Mains Physics strategy

Why this matters

Twenty-five PYQs, twenty-three of them multiple choice, two from 2026, and twelve carry a figure. Eleven ask which process is which, often as a match list of W, Q and ΔU or as a graph to recognise. Ten find work from a P–V path: an area under straight legs, an isothermal logarithm, or a comparison of paths. Four give the process as an equation, such as PV^(3/2) = constant, and ask for the work or a temperature.

Concept 1 of 3: The standard processes and their W, Q and ΔU

Each standard process holds one thing fixed, and that one fact zeroes one term of the first law. Fix T and ΔU = 0. Fix V and W = 0. Allow no heat and Q = 0. Fix P and nothing vanishes, but the three terms keep a fixed ratio. Learn one row per process and most match-list questions answer themselves. W below is the work done BY the gas.

Definition

Isothermal: $PV = \text{const}$ , a hyperbola on a P–V graph; a higher temperature lies farther from the axes.
Isochoric: $P \propto T$ , a straight line through the origin of a P–T graph.
Isobaric: $V \propto T$ . On a V–T graph the slope is nR/P, so the steeper line has the lower pressure.
Adiabatic: $PV^{\gamma} = \text{const}$ , steeper on a P–V graph than the isotherm through the same point.
On any graph with T along one axis, an isotherm is a straight line at right angles to that axis.
"Sudden", "insulated walls" or "constant entropy" mean adiabatic; "slow, in contact with a large reservoir" means isothermal.

Process	Held fixed	Work by the gas W	Heat into the gas Q	Change ΔU
Isothermal	Temperature	$nRT\ln\dfrac{V_2}{V_1}$	Q = W	0 Heat flows in, yet the temperature does not rise: all of it leaves as work.
Isochoric	Volume	0	$nC_V\Delta T$ , so Q = ΔU	$nC_V\Delta T$
Isobaric	Pressure	$P\Delta V = nR\Delta T$	$nC_P\Delta T$	$nC_V\Delta T$
Adiabatic	No heat exchanged	$\dfrac{nR(T_1 - T_2)}{\gamma - 1}$	0	−W
Cyclic	Start and end state the same	Area enclosed on the P–V diagram	Q = W	0
Free expansion into a vacuum	Insulated, no outside pressure	0	0	0, so an ideal gas keeps its temperature

Physics sign convention: W is work done BY the gas and Q is heat INTO the gas, so Q = ΔU + W in every row.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 30 January 2023 · Q8Moderate

Example 1 · Thermodynamics · Thermodynamic Processes and Work from P-V Graphs

Heat is given to an ideal gas in an isothermal process. A. Internal energy of the gas will decrease. B. Internal energy of the gas will increase. C. Internal energy of the gas will not change. D. The gas will do positive work. E. The gas will do negative work. Choose the correct answer from the options given below:

Isothermal does not mean no heat

In an isothermal process the temperature is fixed, but heat still flows: Q = W. The process with no heat is the adiabatic one.

An isochoric line must pass through the origin

P ∝ T describes an isochoric process only when the P–T line passes through absolute zero. A straight line with an intercept, P = a + bT, is not isochoric.

Drill 10 more on the standard processes and their w, q and δu

Concept 2 of 3: Work as the area under a P–V path

W = ∫P dV, so on a P–V diagram the work done by the gas is the area between the path and the volume axis. Moving to the right (expansion) the area counts positive; moving to the left (compression) it counts negative. Break a path into straight pieces and add their signed areas. An isotherm is a curve, so there you use its formula instead.

Definition

Horizontal leg (isobaric): $W = P(V_2 - V_1)$ .
Vertical leg (isochoric): $W = 0$ .
Straight sloping leg: a trapezium, $W = \tfrac{1}{2}(P_1 + P_2)(V_2 - V_1)$ .
Isothermal: $W = nRT\ln\dfrac{V_2}{V_1} = P_1V_1\ln\dfrac{P_1}{P_2}$ .
From the same start to the same final volume in an expansion: $W_{\text{adiabatic}} < W_{\text{isothermal}} < W_{\text{isobaric}}$ , because the adiabat falls fastest and the isobar not at all.
An ideal gas held at constant temperature has bulk modulus $B = -V\dfrac{dP}{dV} = P$ .
A curve that starts and ends at the same volume: the size of W is the area between the curve and that vertical line. W is positive if the gas expands at the higher pressure and is compressed at the lower, negative the other way round. Read the two semi-axes of an arc separately, on the V scale and the P scale.
Units: Pa × m³ = J; kPa × L = J; $1\ \text{dyne cm}^{-2} = 0.1\ \text{Pa}$ ; $1\ \text{cm}^{3} = 10^{-6}\ \text{m}^{3}$ .

Work done by the gas

W = \int_{V_1}^{V_2} P\,dV \qquad W_{\text{iso}} = nRT\ln\frac{V_2}{V_1}

Worked example

A gas goes in a straight line on a P–V diagram from A (1 L, 500 kPa) to B (5 L, 200 kPa), then at constant pressure back to C (1 L, 200 kPa). Find the work done by the gas on each leg and in total.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 23 January 2025 · Q93Moderate

Example 2 · Thermodynamics · Thermodynamic Processes and Work from P-V Graphs

Using the given P-V diagram, the work done by an ideal gas along the path ABCD is-

A leg to the left is negative work

Any part of a path that runs towards smaller volume has negative work by the gas, however high the pressure. Add the areas with their signs.

Read each axis on its own scale

For an elliptical or circular arc on a P–V diagram, read the semi-axis along V in m³ and the semi-axis along P in Pa, each from its own axis. A circle on the page is usually an ellipse in physical units.

Drill 9 more on work as the area under a p–v path

Concept 3 of 3: Processes given by an equation such as PV^x = constant

Some questions describe the process by a rule linking P and V, or V and T. Turn it into the form PV^x = constant, where x is a number fixed by the process. Then one formula gives the work, and PV = nRT turns it into a temperature rule. For a rule of any other shape, find P, V and T at each end separately from PV = nRT.

Definition

$PV^{x} = \text{const}$ (x ≠ 1): $W = \dfrac{P_1V_1 - P_2V_2}{x - 1} = \dfrac{nR(T_1 - T_2)}{x - 1}$ .
With PV = nRT: $TV^{x - 1} = \text{const}$ , so $T \propto V^{1 - x}$ .
Given $V \propto T^{m}$ : then $T \propto V^{1/m}$ , so $1 - x = \dfrac{1}{m}$ .
x = 0 is isobaric, x = 1 isothermal, x = γ adiabatic, and x very large is isochoric.
Any other rule P(V): find each end's temperature from $T = \dfrac{PV}{nR}$ .

Polytropic process PV^x = constant

W = \frac{P_1V_1 - P_2V_2}{x - 1} = \frac{nR(T_1 - T_2)}{x - 1} \qquad T \propto V^{1 - x}

Worked example

1 mol of an ideal gas at 400 K follows

PV^{4/3} = \text{constant}

while its volume grows 8 times. Find its final temperature and the work it does. (

R = 8.3\ \text{J mol}^{-1}\text{K}^{-1}

)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q9Moderate

Example 3 · Thermodynamics · Thermodynamic Processes and Work from P-V Graphs

The pressure and volume of an ideal gas are related as

PV^{3/2}= K

(Constant). The work done when the gas is taken from state

A\left( P_{1},{\text{ }V}_{1},{\text{ }T}_{1} \right)

to state

B\left( P_{2},{\text{ }V}_{2},{\text{ }T}_{2} \right)

is:

x = 1 needs the logarithm

The polytropic work formula divides by x − 1, so it fails for an isothermal process. There use W = nRT ln(V₂/V₁).

The index belongs to the process, γ to the gas

x in PV^x = constant is set by the process; γ = Cp/Cv is set by the gas. They are equal only in an adiabatic process, so do not call x by the name γ.

Drill 3 more on processes given by an equation such as pv^x = constant

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Work as the area under a P–V path
Work done by the gas
$W = \int_{V_1}^{V_2} P\,dV \qquad W_{\text{iso}} = nRT\ln\frac{V_2}{V_1}$
Processes given by an equation such as PV^x = constant
Polytropic process PV^x = constant
$W = \frac{P_1V_1 - P_2V_2}{x - 1} = \frac{nR(T_1 - T_2)}{x - 1} \qquad T \propto V^{1 - x}$

Reference tables (1)

The standard processes and their W, Q and ΔU6 rows

Process	Held fixed	Work by the gas W	Heat into the gas Q	Change ΔU
Isothermal	Temperature	$nRT\ln\dfrac{V_2}{V_1}$	Q = W	0 Heat flows in, yet the temperature does not rise: all of it leaves as work.
Isochoric	Volume	0	$nC_V\Delta T$ , so Q = ΔU	$nC_V\Delta T$
Isobaric	Pressure	$P\Delta V = nR\Delta T$	$nC_P\Delta T$	$nC_V\Delta T$
Adiabatic	No heat exchanged	$\dfrac{nR(T_1 - T_2)}{\gamma - 1}$	0	−W
Cyclic	Start and end state the same	Area enclosed on the P–V diagram	Q = W	0
Free expansion into a vacuum	Insulated, no outside pressure	0	0	0, so an ideal gas keeps its temperature

Physics sign convention: W is work done BY the gas and Q is heat INTO the gas, so Q = ΔU + W in every row.

Watch out for (6)

Isothermal does not mean no heat
→ The standard processes and their W, Q and ΔU
An isochoric line must pass through the origin
→ The standard processes and their W, Q and ΔU
A leg to the left is negative work
→ Work as the area under a P–V path
Read each axis on its own scale
→ Work as the area under a P–V path
x = 1 needs the logarithm
→ Processes given by an equation such as PV^x = constant
The index belongs to the process, γ to the gas
→ Processes given by an equation such as PV^x = constant

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