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.

TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
$Q.The gas that gives the highest fractional conversion of heat to work in an isobaric process is (A) monatomic gas (B) diatomic gas (C) polyatomic gas (D) all types of gases give the same fractional conversion$
›Reveal solutionSolution
The fractional conversion of heat to work in an isobaric process depends on the heat capacity ratio $\gamma$ . Monatomic gases, with the smallest $\gamma$ , give the highest fractional conversion.

The fractional conversion of heat to work tells us what fraction of the heat supplied actually becomes useful work, rather than just raising the internal energy of the gas. In an isobaric (constant pressure) process, this fraction is determined entirely by the type of gas through its heat capacity ratio.

For any isobaric process, the first law gives us:
$Q = \Delta U + W$
where $Q$ is heat supplied, $\Delta U$ is the change in internal energy, and $W$ is work done. The fractional conversion is:
$\eta = \frac{W}{Q}$
Let me work through how this depends on the gas type.

Express work in an isobaric process

At constant pressure, $W = P\Delta V = nR\Delta T$ for an ideal gas.

Express heat supplied

For an isobaric process, $Q = nC_P\Delta T$ , where $C_P$ is the molar heat capacity at constant pressure.

Calculate the fractional conversion

$\eta = \frac{W}{Q} = \frac{nR\Delta T}{nC_P\Delta T} = \frac{R}{C_P}$

Relate to the heat capacity ratio

We know that $C_P - C_V = R$ (Mayer's relation) and $\gamma = \frac{C_P}{C_V}$ .

From these: $C_P = \frac{\gamma R}{\gamma - 1}$

Substitute to find the fractional conversion

$\eta = \frac{R}{C_P} = \frac{R(\gamma - 1)}{\gamma R} = \frac{\gamma - 1}{\gamma} = 1 - \frac{1}{\gamma}$ $\eta_{\text{isobaric}} = 1 - \frac{1}{\gamma}$
Now the key insight: this fraction increases as $\gamma$ decreases. The values of $\gamma$ for different gas types are: …
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TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
$P_1 V_1$
›Reveal solutionSolution
Work in a cycle = area enclosed on the $P$ – $V$ diagram. The triangle has base $V_1$ and height $2P_1$ , so the net work has magnitude $\tfrac12 (V_1)(2P_1) = P_1V_1$ — option (B).

The concept first

For any quasi-static process, $W = \int P\,dV$ — literally the area under the curve on a $P$ – $V$ plot. Go round a closed loop and the "areas under" the outward and return paths partly cancel; what survives is exactly the area enclosed:
$W_{\text{net}} = \oint P\,dV = \pm\,(\text{enclosed area}).$
The sign is set by the sense of travel: clockwise $\Rightarrow$ net work done BY the gas (positive), anticlockwise $\Rightarrow$ net work done ON the gas (negative). Also worth remembering: over a cycle $\Delta U = 0$ , so $Q_{\text{net}} = W_{\text{net}}$ — a cycle simply shuttles heat into work (or vice-versa).

Step-by-step

Write the vertices.

$A = (V_1,\,P_1),\qquad B = (2V_1,\,P_1),\qquad C = (V_1,\,3P_1).$

Work on each leg.

$C \to A$ (vertical, $V = V_1$ constant): isochoric, $dV = 0 \Rightarrow W_{CA} = 0$ .

$A \to B$ (horizontal, $P = P_1$ constant): isobaric expansion,

$W_{AB} = P_1 (2V_1 - V_1) = +P_1V_1.$

$B \to C$ (straight slanted line, compression from $2V_1$ to $V_1$ ): the area under this line is that of a trapezium of parallel sides $P_1$ and $3P_1$ and width $V_1$ :

$|W_{BC}| = \frac{P_1 + 3P_1}{2}\,(V_1) = 2P_1V_1,$
and because the volume decreases, $W_{BC} = -2P_1V_1$ .

Add them up. …
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TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$P'$
›Reveal solutionSolution
The pressure change when one ball is removed is proportional to the weight of one ball divided by the piston area, and the original pressure is proportional to the weight of all n balls; thus the fractional change is $1/n$ .

Concept & Intuition

The gas supports the weight of chamber B plus the metal balls inside it. The pressure of the gas is simply the total weight supported divided by the area of the piston (chamber B’s base). Removing one ball reduces the total weight, so the pressure drops. The fractional change in pressure equals the fractional change in the supported weight, because the area stays constant. This is a direct application of $P = F/A$ for a gas in equilibrium with a movable piston.

Identify the forces in equilibrium Chamber B (with its n balls) rests on the gas. The gas exerts an upward force $P A$ (where $A$ is the cross‑sectional area of chamber B). This balances the total weight of chamber B plus the balls. Let $W_B$ be the weight of chamber B itself, and let $w$ be the weight of a single ball. Then

$P A = W_B + n w.$

Pressure after removing one ball If one ball is taken away, the total weight becomes $W_B + (n-1)w$ . The new pressure $P'$ satisfies

$P' A = W_B + (n-1)w.$

Find the difference $P - P'$ Subtract the two equations:

$(P - P') A = [W_B + n w] - [W_B + (n-1)w] = w.$
Hence
$P - P' = \frac{w}{A}.$

Express the original pressure $P$ in terms of $w$ and $A$ From the first equation,

$P = \frac{W_B + n w}{A}.$
But note that the weight of chamber B itself is also supported by the gas. However, the problem asks for $(P - P')/P$ . We can compute this ratio without knowing $W_B$ separately if we realize that the change in pressure depends only on the weight of one ball, while the original pressure depends on the total weight. The ratio is
$\frac{P - P'}{P} = \frac{w/A}{(W_B + n w)/A} = \frac{w}{W_B + n w}.$
This does not simplify to a constant unless we know $W_B$ . But wait — the problem statement says “the weight of chamber B is supported by the gas.” It does not say chamber B itself is weightless. So is the answer not simply $1/n$ ? Let’s re‑examine. …
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$\rightarrow$
›Reveal solutionSolution

For an adiabatic process the work done by an ideal gas is $W = \dfrac{P_1 V_1 - P_2 V_2}{\gamma - 1}$ . Evaluating this for the two states in the figure gives the expression in option (B).

For an adiabatic process no heat is exchanged ( $Q = 0$ ), so by the first law of thermodynamics the work done by the gas equals the drop in its internal energy, $W = -\Delta U$ . For an ideal gas this can be written entirely in terms of the end-state pressures and volumes:
$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}, \qquad \gamma = \frac{C_P}{C_V}.$
Take state A as $(P_0, V_0)$ . Using the adiabatic relation

$P_A V_A^\gamma = P_B V_B^\gamma$ …
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