An ideal gas undergoes four different processes starting from the same initial state, drawn on a $P$-$V$ diagram (pressure $P$ vertical, volume $V$ horizontal). The four processes are, in some order, adiabatic, isothermal, isobaric and isochoric. Starting from the common initial point (upper-left region): curve 4 is the flat, topmost curve along which the pressure barely changes as the volume increases; curve 3 falls off below it as the volume increases; curve 2 is the lowest of the three expansion curves, falling most steeply as the volume increases; and curve 1 is a vertical line dropping straight down from the initial state (volume unchanged, pressure decreasing). Which numbered curve is the adiabatic process? (a) 4 (b) 3 (c) 2 (d) 1

> [!TLDR] > The slope of a process on a $P$-$V$ diagram tells you what it is: horizontal (slope 0) is isobaric, vertical is isochoric, and between them the adiabatic is steeper than the isothermal. Curve 4 is isobaric, curve 1 isochoric, curve 3 isothermal and curve 2 — the steepest of the expansion curves — is adiabatic. ### Concept For each process the slope $\dfrac{dP}{dV}$ at the common starting state is: - **Isobaric** ($P=$const): $\dfrac{dP}{dV}=0$ → horizontal line. - **Isothermal** ($PV=$const): $\dfrac{dP}{dV}=-\dfrac{P}{V}$. - **Adiabatic** ($PV^{\gamma}=$const): $\dfrac{dP}{dV}=-\gamma\dfrac{P}{V}$, which is $\gamma$ times steeper than the isothermal (since $\gamma>1$). - **Isochoric** ($V=$const): vertical line, slope $-\infty$. ### Ranking the curves Starting from the same point, order of steepness (least to most negative) is: isobaric (flat) $<$ isothermal $<$ adiabatic $ [!ANSWER] > Option (C) — curve **2** is the adiabatic process.

An ideal gas undergoes four different processes starting from the same initial state, drawn on a P-V diagram (

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(a) 4

(b) 3

(d) 1

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Concept understanding — Adiabatic Compression Factor

Adiabatic Compression Factor: From Intuition to Precision

Imagine you pump air into a bicycle tyre. The pump gets noticeably warm. That warmth isn't coming from outside — it's generated inside the air you're compressing. Why? Because you're doing work on the gas, and since the compression happens too fast for heat to escape, all that work stays inside as internal energy, raising the temperature.

This is the core idea: adiabatic means "no heat exchange with the surroundings." When you compress a gas adiabatically, its temperature rises. The adiabatic compression factor is the ratio that tells you how much the temperature rises for a given compression.

The Intuition First

Think of a gas as a swarm of tiny, fast-moving particles. When you push a piston in, you're moving the wall toward the particles. Each time a particle bounces off the approaching wall, it rebounds with a higher speed than it had — like a tennis ball hit by a moving racket. Faster particles mean higher temperature.

If the compression is slow enough that heat can leak out (isothermal), the temperature stays constant. But if it's fast (adiabatic), the temperature climbs. The adiabatic compression factor captures exactly this: the ratio of final temperature to initial temperature when a gas is compressed without heat loss.

The Precise Statement

For an ideal gas undergoing a reversible adiabatic process, the relationship between temperature ( $T$ ) and volume ( $V$ ) is:

T V^{\gamma - 1} = \text{constant}

where $\gamma$ (gamma) is the adiabatic index — the ratio of specific heats: $\gamma = \dfrac{C_p}{C_v}$ .

If you compress from volume $V_1$ to $V_2$ (so $V_2 < V_1$ ), the temperature changes from $T_1$ to $T_2$ according to:

T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1}

The factor $\left( \dfrac{V_1}{V_2} \right)^{\gamma - 1}$ is the adiabatic compression factor for temperature. Since $V_1/V_2 > 1$ and $\gamma - 1 > 0$ , this factor is always greater than 1 — confirming that temperature rises.

\text{Adiabatic compression factor (temperature)} = \left( \frac{V_1}{V_2} \right)^{\gamma - 1}

You can also express it in terms of pressure. Using $PV^\gamma = \text{constant}$ , you get:

T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}

Here $\left( \dfrac{P_2}{P_1} \right)^{\frac{\gamma - 1}{\gamma}}$ is the pressure-based version.

What $\gamma$ Means

$\gamma$ depends on the number of degrees of freedom of the gas molecule:

Gas type	Degrees of freedom	$\gamma$	Example
Monatomic	3 (translation only)	5/3 ≈ 1.67	He, Ar
Diatomic / linear triatomic (rigid)	5 (3 translation + 2 rotation)	7/5 = 1.40	N₂, O₂; CO₂ (theoretical)
Non-linear triatomic	6 (3 translation + 3 rotation)	4/3 ≈ 1.33	H₂O vapour

A higher $\gamma$ means the temperature rises more sharply for the same compression. Monatomic gases heat up the most — they have only translational motion to store energy, so all the work of compression goes into raising temperature.

Watch out

CO₂ is a linear triatomic molecule (O=C=O), so the rigid-rotor kinetic-theory model actually predicts the same $\gamma$ as a diatomic gas (7/5 = 1.40), not 4/3. The 4/3 value belongs to non-linear triatomic molecules such as water vapour, which have a genuine third rotational degree of freedom. In practice, real CO₂ shows $\gamma \approx 1.30$ because its bending vibrational mode is active at ordinary temperatures — a correction beyond this simple rigid-molecule model. Don't memorise "triatomic = 4/3" as a blanket rule; it only holds for non-linear triatomics.

Also, do not confuse the adiabatic compression factor with the plain compression ratio of an engine. The compression ratio is $V_1/V_2$ — a purely geometric ratio. The adiabatic compression factor includes $\gamma$ and tells you the thermal effect of that compression.

A Quick Example

Air at 300 K is compressed adiabatically to one-tenth its volume. For air (mostly diatomic), $\gamma = 1.4$ .

T_2 = 300 \times (10)^{1.4 - 1} = 300 \times 10^{0.4}

$10^{0.4} \approx 2.51$ , so $T_2 \approx 753\ \text{K}$ (about 480°C). That's why diesel engines don't need spark plugs — the adiabatic compression of air alone raises its temperature enough to ignite the injected fuel.

The Big Picture

The adiabatic compression factor isn't a separate formula to memorise — it's the temperature multiplier that emerges naturally from the adiabatic condition $TV^{\gamma-1} = \text{constant}$ . Whenever you see "adiabatic compression" in an exam, immediately think: temperature rises, and the rise is governed by $\gamma$ (fixed by the gas's degrees of freedom) and the volume (or pressure) ratio. That's the entire concept.

This topic is commonly searched as "Adiabatic Compression Factor 11 physics important questions" or "Adiabatic Compression Factor formula and examples", and it maps cleanly onto the Class 11 Physics portion of the NCERT/CBSE syllabus. Because adiabatic compression factor shows up repeatedly in JEE Main, NEET and state engineering/medical entrance exams, mastering the underlying idea (not just the formula) is genuinely worth the extra time.

The slope of a process on a $P$ - $V$ diagram tells you what it is: horizontal (slope 0) is isobaric, vertical is isochoric, and between them the adiabatic is steeper than the isothermal. Curve 4 is isobaric, curve 1 isochoric, curve 3 isothermal and curve 2 — the steepest of the expansion curves — is adiabatic.

Concept

For each process the slope $\dfrac{dP}{dV}$ at the common starting state is:

Isobaric ( $P=$ const): $\dfrac{dP}{dV}=0$ → horizontal line.
Isothermal ( $PV=$ const): $\dfrac{dP}{dV}=-\dfrac{P}{V}$ .
Adiabatic ( $PV^{\gamma}=$ const): $\dfrac{dP}{dV}=-\gamma\dfrac{P}{V}$ , which is $\gamma$ times steeper than the isothermal (since $\gamma>1$ ).
Isochoric ( $V=$ const): vertical line, slope $-\infty$ .

Ranking the curves

Starting from the same point, order of steepness (least to most negative) is: isobaric (flat) $<$ isothermal $<$ adiabatic $<$ isochoric (vertical). Matching to the diagram:

Curve 4 (flat, top) $=$ isobaric.
Curve 3 (moderate fall) $=$ isothermal.
Curve 2 (steepest fall) $=$ adiabatic.
Curve 1 (vertical) $=$ isochoric.

Distractors: (A) 4 is isobaric, (D) 1 is isochoric, (B) 3 is isothermal — all wrong because the adiabatic must lie between the isothermal and the vertical isochoric.

✓Final answer

Option (C) — curve 2 is the adiabatic process.

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