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.

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

COMEDK 2025Set 2025-A1 markMCQ
$T_1$
›Reveal solutionSolution
For an adiabatic process in a monoatomic ideal gas, the relation $TV^{\gamma-1} = \text{constant}$ applies. Here volume doubles, so $\frac{T_1}{T_2} = 2^{2/3}$ , which corresponds to option (B).

The key concept is the adiabatic process for an ideal gas. “Adiabatic” means no heat exchange with the surroundings. When the piston is released suddenly, the gas expands quickly, so there’s no time for heat to flow — the process is adiabatic. For an ideal gas undergoing a reversible adiabatic process (and here the sudden release approximates a free expansion, but the problem treats it as a quasi-static adiabatic expansion because the piston is frictionless and the gas does work), the relation between temperature and volume is:
$TV^{\gamma-1} = \text{constant}$
where $\gamma = \frac{C_p}{C_v}$ . For a monoatomic ideal gas, $\gamma = \frac{5}{3}$ .

Now, step by step:

Identify the volume change. The initial length of the gas column is $L$ , and the final length is $2L$ . Since the cylinder has a constant cross-sectional area, volume is proportional to length. So:

$V_1 \propto L, \quad V_2 \propto 2L \quad \Rightarrow \quad \frac{V_2}{V_1} = 2$

Apply the adiabatic relation. For an adiabatic process:

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

Substitute the values. $\gamma = \frac{5}{3}$ , so $\gamma - 1 = \frac{2}{3}$ . And $\frac{V_2}{V_1} = 2$ . Thus:

$\frac{T_1}{T_2} = 2^{2/3}$

Match with the options. This is exactly option (B).

Watch out

A common mistake is to use $PV^\gamma = \text{constant}$ and then try to relate temperature via the ideal gas law, but that leads to the same result if done correctly. Another pitfall is confusing $\gamma$ for diatomic gases (7/5) — here it’s monoatomic, so $\gamma = 5/3$ .

Tip

Remember the shortcut: For an adiabatic process, $T \propto V^{-(\gamma-1)}$ . If volume doubles, temperature drops by a factor of $2^{\gamma-1}$ . For monoatomic gas, that’s $2^{2/3}$ .

✓Final answer
$The correct option is (B) . ANSWER: B$
COMEDK 2025Set 2025-E1 markMCQ
$\frac{c_p}{c_v}=\gamma=1.4$
›Reveal solutionSolution
For a gas expanding to six times its volume, the adiabatic final pressure is much lower than the isothermal final pressure because adiabatic expansion also cools the gas, reducing pressure further. The correct option is (D).

Concept & Intuition

The key difference between isothermal and adiabatic expansions lies in temperature.

Isothermal: Temperature stays constant. Pressure drops only because volume increases (Boyle’s Law: $P \propto 1/V$ ).

Adiabatic: No heat exchange. The gas does work using its internal energy, so its temperature falls. Pressure drops for two reasons: volume increase and temperature decrease.

Thus, for the same volume increase, the final pressure after adiabatic expansion is always lower than after isothermal expansion. This is a classic result for an ideal gas.

Step-by-step reasoning

Isothermal expansion For an ideal gas at constant temperature:

$P_1 V_1 = P_2 V_2$
Initial volume $V_1 = V$ , final volume $V_2 = 6V$ .

Initial pressure at NTP is $P_1 = 1$ atm (say).
$P_{\text{iso}} = \frac{P_1 V_1}{V_2} = \frac{1 \cdot V}{6V} = \frac{1}{6} \text{ atm}$

Adiabatic expansion For a reversible adiabatic process:

$P_1 V_1^\gamma = P_2 V_2^\gamma$
with $\gamma = 1.4$ .
$P_{\text{adia}} = P_1 \left( \frac{V_1}{V_2} \right)^\gamma = 1 \cdot \left( \frac{1}{6} \right)^{1.4}$
Compute: $\left( \frac{1}{6} \right)^{1.4} = 6^{-1.4}$ .

Since $6^{1.4} > 6^1 = 6$ , we have $6^{-1.4} < \frac{1}{6}$ .

Numerically: $6^{1.4} = e^{1.4 \ln 6} \approx e^{1.4 \times 1.7918} = e^{2.5085} \approx 12.3$ .

So $P_{\text{adia}} \approx \frac{1}{12.3} \approx 0.0813$ atm, while $P_{\text{iso}} = 0.1667$ atm.

Clearly, $P_{\text{adia}} < P_{\text{iso}}$ .

Check the options

(A) says adiabatic pressure is 1.4 times greater — false (it’s smaller).

(B) says final temperature after adiabatic is 1.4 times less — temperature does drop, but not by exactly a factor of 1.4; also the question asks about pressure comparison.

(C) says pressure remains same — false.

(D) says final pressure after adiabatic is less than after isothermal — true.

Watch out

A common mistake is to forget that in adiabatic expansion the temperature drops, so the pressure falls faster than in isothermal expansion. Many students incorrectly think the pressures are equal because they only consider the volume change.

Tip

For any expansion ratio $r > 1$ , since $\gamma > 1$ , we always have $r^{-\gamma} < r^{-1}$ . Hence adiabatic final pressure is always less than isothermal final pressure for the same volume increase.

✓Final answer
$The correct option is (D). ANSWER: D$
COMEDK 2025Set 2025-M1 markMCQ
$\frac{5}{2}$
›Reveal solutionSolution
For an adiabatic process, $PV^\gamma = \text{constant}$ . Given $\gamma = \frac{5}{2}$ and a $\frac{3}{2}\%$ increase in volume, the percentage decrease in pressure is found via logarithmic differentiation to be $\frac{15}{4}\%$ , so the correct option is (A).

The key concept here is the adiabatic relation for an ideal gas: $PV^\gamma = \text{constant}$ , where $\gamma = C_p/C_v$ . When a gas changes adiabatically, pressure and volume are linked so that no heat enters or leaves the system. The problem gives us the percentage change in volume and asks for the corresponding percentage change in pressure — a perfect setup for using logarithmic differentiation to convert small relative changes into each other.

Why this works:

If $PV^\gamma = k$ , taking natural logs gives $\ln P + \gamma \ln V = \ln k$ . Differentiating (for small changes) yields $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$ , so the relative change in pressure is directly proportional to the relative change in volume, with $\gamma$ as the factor. Percentages are just relative changes multiplied by 100.

Step-by-step solution:

Write the adiabatic condition For an adiabatic process:

$PV^\gamma = \text{constant}$
Here $\gamma = \frac{5}{2}$ .

Take natural logarithms

$\ln P + \gamma \ln V = \text{constant}$

Differentiate (for small changes)

$\frac{dP}{P} + \gamma \frac{dV}{V} = 0$
This gives the relation between infinitesimal relative changes.

Solve for the relative change in pressure

$\frac{dP}{P} = -\gamma \frac{dV}{V}$

Plug in the given percentage change in volume The volume increases by $\frac{3}{2}\%$ , so

$\frac{dV}{V} = +\frac{3}{2} \times \frac{1}{100} = \frac{3}{200}$
(We keep it as a fraction to avoid decimals.)

Compute the relative change in pressure

$\frac{dP}{P} = -\frac{5}{2} \times \frac{3}{200} = -\frac{15}{400} = -\frac{3}{80}$

Convert to percentage Percentage change in pressure = $\frac{dP}{P} \times 100\% = -\frac{3}{80} \times 100\% = -\frac{300}{80}\% = -\frac{15}{4}\%$ The negative sign indicates a decrease.

Thus the percentage decrease in pressure is $\frac{15}{4}\%$ .

Tip

Notice that the numbers work out neatly: $\gamma = 5/2$ and the volume increase is $3/2\%$ , so the pressure decrease is $(5/2) \times (3/2) = 15/4\%$ . Always check if the product simplifies before converting to decimals.

Watch out

A common mistake is to forget that the percentage is already given as a pure number (e.g., $3/2$ means $1.5\%$ ), and then incorrectly multiply by 100 again. Keep everything in fractional form to avoid confusion.

✓Final answer
$The correct option is (A) . ANSWER: A$
COMEDK 2023Set 2023-E1 markMCQ
$6.2 \%$
›Reveal solutionSolution

So the pressure DECREASES by 8.68 % (the expansion cools and depressurises the gas); the magnitude asked for is 8.68.

Concept: adiabatic process P V^gamma = constant. Taking logs and differentiating gives the fractional-change relation

dP/P = -gamma (dV/V), i.e. % change in P = -gamma x (% change in V).

Given dV/V = +6.2 % and gamma = 1.4:

% change in P = -1.4 x 6.2 = -8.68 %.

So the pressure DECREASES by 8.68 % (the expansion cools and depressurises the gas); the magnitude asked for is 8.68.

✓Final answer
$The correct option is (A) — 8.68$

ANSWER: A
COMEDK 2022Set 20221 markMCQ
$_1$
›Reveal solutionSolution

Ratio: P1 : P2 = P 2^gamma : 2P = 2^gamma / 2 : 1 = 2^(gamma - 1) : 1

Concept: compare an adiabatic and an isothermal compression of identical gases from the same initial state (P, V) to V/2.

Container C1 (adiabatic): P V^gamma = P1 (V/2)^gamma

P1 = P x 2^gamma

Container C2 (isothermal): P V = P2 (V/2)

P2 = 2P

Ratio:

P1 : P2 = P 2^gamma : 2P = 2^gamma / 2 : 1 = 2^(gamma - 1) : 1

✓Final answer
$2^{\gamma-1}:1$

ANSWER: D
COMEDK 2021Set 20211 markMCQ
$\gamma=\frac{3}{2}$
›Reveal solutionSolution

So the volume decreases by 4/9 %.

Concept: for an adiabatic (reversible) process, P V^gamma = constant. Take logarithms and differentiate to relate the fractional changes.

ln P + gamma ln V = constant

=> dP/P + gamma (dV/V) = 0

=> dV/V = -(1/gamma) (dP/P).

Given gamma = 3/2 and dP/P = +2/3 % :

dV/V = -(2/3) * (2/3) % = -(4/9) %.

So the volume decreases by 4/9 %.

✓Final answer
$\frac{4}{9}$

ANSWER: A

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