Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The g

NCERT Exemplar · Q5

Q. Consider two containers A and B containing identical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is

(a)

2^{\gamma - 1}

(b)

\left(\dfrac{1}{2}\right)^{\gamma - 1}

(c)

\left(\dfrac{1}{1-\gamma}\right)^{2}

(d)

\left(\dfrac{1}{\gamma - 1}\right)^{2}

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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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Isothermal compression follows $PV = \text{const}$ , while adiabatic compression follows $PV^\gamma = \text{const}$ . When both gases are compressed to half their volume, the adiabatic process produces a higher pressure by a factor of $2^{\gamma-1}$ .

The heart of this problem lies in understanding how pressure responds differently to compression depending on whether heat can escape. In an isothermal process, the gas stays at constant temperature because it exchanges heat with its surroundings—compress it and it wants to heat up, but that heat flows out, keeping $T$ fixed. In an adiabatic process, no heat escapes, so all the compression work goes into raising the internal energy and temperature, which drives the pressure up more steeply.

The mathematical signature of these processes:

Isothermal: $PV = \text{constant}$ (Boyle's law at fixed $T$ )
Adiabatic: $PV^\gamma = \text{constant}$ (where $\gamma = C_p/C_v > 1$ )

Let both gases start with pressure $P_0$ , volume $V_0$ , and temperature $T_0$ . Both are compressed to volume $V_f = V_0/2$ .

Step-by-step solution

Find the final pressure for gas A (isothermal compression)

For an isothermal process:

P_0 V_0 = P_A \cdot \frac{V_0}{2}

Solving for $P_A$ :

P_A = \frac{P_0 V_0}{V_0/2} = 2P_0

Find the final pressure for gas B (adiabatic compression)

For an adiabatic process:

P_0 V_0^\gamma = P_B \left(\frac{V_0}{2}\right)^\gamma

Solving for $P_B$ :

$P_B = P_0 \cdot \frac{V_0^\gamma}{(V_0/2)^\gamma} = P_0 \cdot \frac{V_0^\gamma}{V_0^\gamma / 2^\gamma} = P_0 \cdot 2^\gamma$ …

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