8.3 Adiabatic Process and Equation

An adiabatic process is a thermodynamic process in which there is no heat transfer into or out of the system ($Q=0$). This condition is achieved if the system is perfectly insulated from its surroundings or if the process occurs so rapidly that there is no time for significant heat exchange.

For an adiabatic process, the relationship between the pressure ($P$) and volume ($V$) of an ideal gas is described by the adiabatic equation:

$P{V}^{\gamma }=\text{constant}$

1. The First Law of Thermodynamics in an Adiabatic Process

The First Law of Thermodynamics is given by $Q=\Delta U+W$. In an adiabatic process, the heat transfer ($Q$) is zero.

• Condition: $Q=0$
• First Law Equation: The law simplifies to: $0=\Delta U+W⟹\Delta U=-W$
• Implication: This expresses the conservation of energy — any work done by the system comes entirely at the expense of its own internal energy.
  • If the system expands (does work on surroundings, $W>0$): internal energy decreases ($\Delta U<0$) and temperature drops.
  • If the system is compressed (work done on it, $W<0$): internal energy increases ($\Delta U>0$) and temperature rises.

2. The Adiabatic Index ($\gamma$)

The exponent $\gamma$ (gamma) in the adiabatic equation is known as the adiabatic index or the heat capacity ratio.

• Definition: $\gamma =\frac{C_p}{C_v}$ where $C_p$ is the molar specific heat at constant pressure and $C_v$ is the molar specific heat at constant volume.
• Significance: Since $C_p>C_v$ always (by Mayer's relation, $C_p-C_v=R$), we have $\gamma >1$.
  • Monatomic ideal gas: $\gamma \approx 1.67$
  • Diatomic ideal gas (e.g., air): $\gamma \approx 1.4$

3. Mathematical Derivation of the Adiabatic Equation

The equation $P{V}^{\gamma }=\text{constant}$ is derived by combining the First Law of Thermodynamics with the Ideal Gas Law.

Step 1: Start with the First Law

For an infinitesimal adiabatic process, $dQ=0$. The First Law gives: $dU=-dW$

Step 2: Substitute expressions for $dU$ and $dW$

• Change in internal energy: $dU=n{C}_{v}dT$
• Work done by the gas: $dW=PdV$

Substituting: $n{C}_{v}dT=-PdV$

Step 3: Use the Ideal Gas Law

From $PV=nRT$, differentiating: $PdV+VdP=nRdT⟹dT=\frac{PdV+VdP}{nR}$

Step 4: Combine the Equations

Substitute $dT$ into the equation from Step 2: $n{C}_v\left(\frac{PdV+VdP}{nR}\right)=-PdV$

Simplify (cancel $n$, multiply through by $R$): $C_v(PdV+VdP)=-R(PdV)$

Step 5: Apply Mayer's Relation

Using $R=C_p-C_v$: $C_v(PdV+VdP)=-(C_p-C_v)(PdV)$

Expanding: $C_{v}PdV+C_{v}VdP=-C_{p}PdV+C_{v}PdV$ $C_{v}VdP=-C_{p}PdV$

Step 6: Separate Variables and Integrate

$\frac{dP}{P}=-\frac{C_p}{C_v}\frac{dV}{V}=-\gamma \frac{dV}{V}$

Integrating both sides: $\ln P=-\gamma \ln V+\text{constant}$ $\ln P+\gamma \ln V=\text{constant}⟹\ln (P{V}^{\gamma })=\text{constant}$

Therefore: $\boxed{P{V}^{\gamma }=\text{constant}}$

4. Temperature–Volume and Temperature–Pressure Relations

Using the Ideal Gas Law $PV=nRT$ alongside $P{V}^{\gamma }=\text{constant}$, two additional adiabatic relations can be derived:

Temperature–Volume: $T{V}^{\gamma -1}=\text{constant}$

Temperature–Pressure: $P^{1-\gamma }{T}^{\gamma }=\text{constant}$

These confirm that in an adiabatic process, temperature, pressure, and volume all change simultaneously.

5. Adiabatic vs. Isothermal Process

The adiabatic curve is steeper than the isothermal curve on a P-V diagram because $\gamma >1$.

Real-World Example

Why does pumping a bicycle tire make the pump feel hot?

When you pump the tire, you rapidly compress the air inside the pump. This compression is nearly adiabatic (too fast for heat to escape). By the First Law for an adiabatic process ($\Delta U=-W$), the work done on the gas increases its internal energy, raising its temperature significantly.

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