In an adiabatic process, there is no heat exchange between the system and its surroundings (Q = 0). For such a process, the first law gives ΔE = W.

This means that the internal energy increases if work is done on the system, and this usually leads to a temperature rise. If work is done by the system, the internal energy decreases, which is usually accompanied by a temperature drop.

**Examples of adiabatic processes:**

– Reactions or transformations taking place in a thermal-insulated chamber.

– Fast processes for which there is not enough time for heat exchange to take place.

- E.g. Operating a spray can, bicycle pump or opening a carbonated beverage can.
- Air masses ascending or descending in the Earth’s atmosphere.

### Derivation of the PV relation for an adiabat of an ideal gas.

$$\begin{aligned} dE &= \delta Q + \delta W \\ &= 0 \, – P \, dV \end{aligned}$$

Since dE = nCvdT, we have

$$n C_{v} \, dT = \, – P \, dV$$

This means that as the gas is compressed adiabatically (dV < 0), its temperature rises (dT > 0), which is thus called adiabatic heating.

Conversely, if the gas is expanded adiabatically, its temperature falls, i.e., adiabatic cooling.

We now need to eliminated dT to find the relationship between P and V during the transformation.

By differentiating the ideal gas equation PV = nRT, we get:

$$V \, dP + P \, dV = nR \, dT$$

By substituting the adiabatic relation (Equation 1) into the right hand side of the equation above, we get:

$$V \, dP + P \, dV = \, – nR \frac{P \, dV}{n C_{v}}$$

Rearranging,

$$\begin{aligned} V \, dP + \left( 1 + \frac{R}{C_{v}} \right) P \, dV \, &= 0 \\ V \, dP + \left( \frac{C_{p}}{C_{v}} \right) P \, dV \, &= 0 \\ V \, dP + \gamma P \, dV \, &= 0 \end{aligned}$$

Integrating both sides of the equation, we obtain:

$$\int\limits_{P_{1}}^{P_{2}} \frac{1}{P} \, dP + \gamma \int\limits_{V_{1}}^{V_{2}} \frac{1}{V} \, dV = 0$$

Hence,

$$P_{1} V_{1}^{\gamma} = P_{2} V_{2}^{\gamma}$$

Substitute in the ideal gas law and you get the other forms,

$$\begin{aligned}T_{1} V_{1}^{\gamma \, – 1} \, &= T_{2} V_{2}^{\gamma \, -1} \\ T_{1}^{\gamma} P_{1}^{1 \, – \gamma} \, &= T_{2}^{\gamma}P_{2}^{1 \, – \gamma} \end{aligned}$$

**Adiabatic ideal gas process**

Equation of the adiabat (line of constant $PV^{\gamma}$)

$$P_{1} V_{1}^{\gamma} = P_{2} V_{2}^{\gamma} = K$$

Adiabatic work done by the gas,

$$ \begin{aligned} W_{out} &= \int\limits_{V_{i}}^{V_{f}} P \, dV \\ &= \int\limits_{V_{i}}^{V_{f}} \frac{K}{V^{\gamma}} \, dV \\ &= K \frac{ \left( V_{f}^{1 \, – \gamma} \, – V_{i}^{1 \, – \gamma} \right)}{1 \, – \gamma} \end{aligned}$$