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

NCERT Exemplar · Q19

V

Karnataka PUCShort· 3mImportance★★★★★est

77% · 27/35 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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 ( $\Delta V\ll V$ ):

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

Start your 14-day free trial to unlock the full solution →

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. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 %. …
Start your 14-day free trial to see the full solution →
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 …
Start your 14-day free trial to see the full solution →
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 % : …
Start your 14-day free trial to see the full solution →

←Previous questionQ18 — A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m …→Next questionQ20 — In a refrigerator one removes heat from a lower temperature and deposi…

Concept

Derivation

🎓Unlock everything free for 14 days

Unlock everything free for 14 days