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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P

(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.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In an adiabatic process, which of the following quantities remains constant?$
(a) Temperature
(b) Pressure
(c) Volume
(d) None of these
›Reveal solutionSolution
"Adiabatic" specifically means no heat transfer into or out of the system (Q = 0) -- it says nothing about P, V, or T individually staying constant; in fact all three typically DO change in an adiabatic process.

By definition, an adiabatic process is one in which no heat enters or leaves the system: Q = 0. This is different from:

Isothermal process: temperature (T) stays constant.

Isobaric process: pressure (P) stays constant.

Isochoric (isovolumetric) process: volume (V) stays constant.

In an adiabatic process, the system can still do work on (or have work done on it by) the surroundings, changing its internal energy, and consequently P, V, and T all generally change together, following the adiabatic relation:

P V^gamma = constant

(where gamma = Cp/Cv). Since none of pressure, volume, or temperature is individually held fixed in an adiabatic process, the correct answer is that none of the listed quantities remains constant.

✓Final answer
$(d) None of these.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The relation between pressure (P) and volume (V) of an ideal gas in an adiabatic process is (where γ = Cp/Cv)$
(a) (PV)^γ = constant
(b) P^γ V = constant
(c) PV = constant
(d) PV^γ = constant
›Reveal solutionSolution

Adiabatic process: P V^gamma = constant. Answer (D).

In an adiabatic process no heat is exchanged (Q = 0). Combining the first law with the ideal-gas law leads to the relation:

P V^gamma = constant,

where gamma = Cp/Cv is the ratio of specific heats. (For an isothermal process, by contrast, PV = constant.)

✓Final answer
$(D) PV^gamma = constant.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. For adiabatic process, the relation between pressure and volume is (A) PV^γ = constant (B) P^(1-γ)V^γ = constant (C) PV^(γ-1) = constant (D) P^γV^γ = constant$
›Reveal solutionSolution

For an adiabatic process on an ideal gas, $PV^\gamma = \text{constant}$ .

In an adiabatic process, $dQ = 0$ . Applying the first law of thermodynamics ( $dQ = dU + PdV$ ) with $dU = nC_VdT$ , combined with the ideal gas law $PV = nRT$ , and eliminating $T$ using Mayer's relation $C_P - C_V = R$ , leads (after integration) to the relation:

$PV^\gamma = \text{constant}$

where $\gamma = C_P/C_V$ is the ratio of specific heats of the gas. This is steeper than the isothermal curve $PV = \text{constant}$ on a P-V diagram, since $\gamma > 1$ .

✓Final answer
$(A) PV^γ = constant.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): Air quickly leaking out of a balloon becomes cooler. Reason (R): The leaking air undergoes adiabatic expansion. Select the correct answer from the codes below.$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) and (R) both are false.
›Reveal solutionSolution

The leaking air's rapid, essentially heat-free expansion is a textbook case of adiabatic cooling — the reason given is exactly why the observation is true.

Assertion (A): Air leaking quickly out of a balloon becomes cooler — this is true and can be felt directly (a deflating balloon's nozzle feels cold).

Reason (R): The leaking air undergoes adiabatic expansion — also true. The process happens so fast that essentially no heat is exchanged with the surroundings ( $Q \approx 0$ ).

Why R explains A: In an adiabatic expansion, the gas does work on its surroundings (pushing outward as it expands) at the expense of its own internal energy, since no heat flows in to replace that energy ( $\Delta U = -W$ , with $Q=0$ from the first law of thermodynamics). A drop in internal energy for an ideal gas means a drop in temperature — hence the escaping air cools. This is precisely the mechanism behind (A), so (R) is the correct explanation of (A).

✓Final answer
$Both (A) and (R) are true, and (R) is the correct explanation of (A) — option (a) .$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The thermodynamic process in which no heat exchange takes place is called -$
(a) Adiabatic
(b) Isochoric
(c) Isobaric
(d) Isothermal
›Reveal solutionSolution
A process with no heat exchange with the surroundings is called an adiabatic process.

In thermodynamics, processes are classified by what is held constant or restricted:

Isochoric: volume is held constant.

Isobaric: pressure is held constant.

Isothermal: temperature is held constant (system in thermal contact with a reservoir, heat CAN flow).

Adiabatic: the system is thermally insulated from its surroundings (e.g. rapid compression/expansion, or enclosure in a perfectly insulating container), so no heat enters or leaves: $Q = 0$ . Any change in internal energy in an adiabatic process comes entirely from work done on or by the system.

✓Final answer
$The correct option is (a) Adiabatic.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. An adiabatic process occurs at constant :$
(a) Temperature
(b) Pressure
(c) Heat
(d) Temperature and Pressure
›Reveal solutionSolution

"Adiabatic" literally means no heat transfer occurs — the system is thermally insulated from its surroundings during the process.

An adiabatic process is one carried out in a thermally insulated system, or fast enough that there is no time for heat exchange with the surroundings. This means:

$Q = 0$

throughout the process — no heat enters or leaves the system (unlike an isothermal process, where temperature is held constant while heat does flow in or out). In an adiabatic process, temperature and pressure can both change (governed by $PV^\gamma = \text{constant}$ ); it is the heat exchanged, not temperature or pressure, that is held at (zero and) constant.

✓Final answer
$An adiabatic process occurs at constant heat (Q = 0, no heat exchange with surroundings). Option (c) Heat.$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Two statements are given below: one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer using the codes given. Assertion (A) : Air quickly leaking out of a balloon becomes cooler. Reason (R) : The leaking air undergoes adiabatic expansion.$
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true and (R) is not the correct explanation of (A).
(c) (A) is true but (R) is false.
(d) (A) is false and (R) is also false.
›Reveal solutionSolution

Rapidly leaking air expands adiabatically (no time to absorb heat), so by the first law its internal energy — and hence temperature — falls, which is exactly why it feels cooler.

Assertion (A): When air rushes quickly out of a balloon, it expands very fast. Because this happens too quickly for any significant heat to be exchanged with the surroundings, the expansion is essentially adiabatic, and the escaping air does indeed feel cooler.

Reason (R): In an adiabatic expansion, Q = 0. By the first law of thermodynamics:

$\Delta U = Q - W = -W$

Since the gas expands, it does positive work (W > 0) on the surroundings, so $\Delta U < 0$ : the internal energy, and therefore the temperature, of the gas falls. This is precisely the mechanism that explains the cooling described in (A).

So (R) is true, and it correctly and directly explains why (A) is true.

✓Final answer
$Both (A) and (R) are true, and (R) is the correct explanation of (A). Option (a).$
CBSE 2022Set ANNUAL1 markMCQ
$Q. The ratio gamma = Cp/Cv for a gas mixture consisting of 8 g of helium and 16 g of oxygen is :$
(a) 27/17
(b) 23/15
(c) 17/27
(d) 15/23
›Reveal solutionSolution

Find the number of moles of each gas, use the correct degrees of freedom for each (monatomic He: f=3; diatomic O2: f=5) to get each gas's Cv and Cp, mole-average these for the mixture, and take the ratio.

Step 1 — moles of each gas:

Helium: mass = 8 g, molar mass M(He) = 4 g/mol ⇒ n(He) = 8/4 = 2 mol.

Oxygen: mass = 16 g, molar mass M(O2) = 32 g/mol ⇒ n(O2) = 16/32 = 0.5 mol.

Step 2 — molar specific heats of each gas:

Helium is monatomic (f = 3): Cv(He) = (3/2)R, Cp(He) = (5/2)R.

Oxygen is diatomic (f = 5, rigid, room temperature): Cv(O2) = (5/2)R, Cp(O2) = (7/2)R.

Step 3 — mixture's Cv and Cp (mole-weighted average):

Cv(mix) = [n(He)·Cv(He) + n(O2)·Cv(O2)] / [n(He)+n(O2)]

= [2×(3/2)R + 0.5×(5/2)R] / 2.5

= [3R + 1.25R] / 2.5 = 4.25R/2.5 = 1.7R

Cp(mix) = [n(He)·Cp(He) + n(O2)·Cp(O2)] / [n(He)+n(O2)]

= [2×(5/2)R + 0.5×(7/2)R] / 2.5

= [5R + 1.75R] / 2.5 = 6.75R/2.5 = 2.7R

Step 4 — ratio:

γ(mix) = Cp(mix)/Cv(mix) = 2.7R / 1.7R = 27/17.

✓Final answer
$The correct option is (a) 27/17.$

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Concept

Ranking the curves

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