Consider a cycle tyre being filled with air by a pump. Let $V$ be the fixed volume of the tyre, and at each stroke of the pump a small volume $\Delta V$ (with $\Delta V \ll V$) of air is transferred into the tube adiabatically. Find the work done when the pressure in the tube is increased from $P_1$ to $P_2$.

Consider a cycle tyre being filled with air by a pump. Let V be the fixed volume of the tyre, and at each stro

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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. …

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Each pump stroke pushes a small volume $\Delta V$ of air into the fixed-volume tube against the current pressure $P$ , doing work $P\,\Delta V$ . Using the adiabatic condition $PV^{\gamma}=$ const to relate $\Delta V$ to the pressure rise $dP$ gives $dW=\dfrac{V}{\gamma}dP$ , and integrating from $P_1$ to $P_2$ yields $W=\dfrac{(P_2-P_1)V}{\gamma}$ .

Concept

The air already in the tube of fixed volume $V$ is compressed adiabatically when a further $\Delta V$ is forced in. Treating the addition as an adiabatic compression of gas from $V+\Delta V$ to $V$ :

P\,(V+\Delta V)^{\gamma}=(P+dP)\,V^{\gamma}.

Derivation

Expand to first order in the small quantities ( $\Delta V\ll V$ ):

$P\,V^{\gamma}\left(1+\frac{\Delta V}{V}\right)^{\gamma}\approx P\,V^{\gamma}\left(1+\gamma\frac{\Delta V}{V}\right)=(P+dP)V^{\gamma},$ …

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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. …
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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,

…

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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:

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

…

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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 ( $Q \approx 0$ ).

…

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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). …
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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:

$Q = 0$

…

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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:

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

…

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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)] …

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