Q.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 $P_2$.

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:

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:

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.

You can also express it in terms of pressure. Using $PV^\gamma = \text{constant}$, you get:

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:

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. …