Adiabatic Expansion and Compression of Gases

An adiabatic process is one in which no heat is transferred between the system and its surroundings ($\Delta Q = 0$). Adiabatic processes play a significant role in various natural phenomena, such as cloud formation in meteorology, and are key to understanding thermodynamics in gases.

Cloud Formation through Adiabatic Cooling

Cloud formation is driven by the adiabatic cooling of air as it rises through the atmosphere. When air moves upward, the external pressure decreases, causing the air to expand and cool. This cooling can cause the air to reach its dew point, at which point water vapor condenses into tiny droplets, forming clouds. Since the motion of the air is faster than heat transfer, this cooling is considered an adiabatic process.

\[ \text{Lifting of air} \quad \Rightarrow \quad \text{Adiabatic cooling} \]

As the air rises and expands, its internal energy ($U$) decreases, leading to a reduction in temperature, and eventually reaching saturation, forming water droplets.

 

Adiabatic Expansion of an Ideal Gas

In an adiabatic process, there is no heat exchange between the system and its surroundings, meaning:

\[ \Delta Q = 0 \]

From the First Law of Thermodynamics:

\[ \Delta U = \Delta Q + \Delta W = \Delta Q - p \, dV \]

Since $ \Delta Q = 0 $ in an adiabatic process, this simplifies to:

\[ \Delta U = - p \, dV \]

For an ideal gas, the internal energy change is $\Delta U = C_v \, dT$, and with the ideal gas equation $p = \frac{nRT}{V}$ we get:

\[ C_v \, dT = - nR \frac{dV}{V} \]

We then solve the integral between the initial and final states, assuming $C_v$ is constant:

\[ \int_{T_i}^{T_f} C_v \frac{dT}{T} = - nR \int_{V_i}^{V_f} \frac{dV}{V} \implies C_v \ln \frac{T_f}{T_i} = - nR \ln \frac{V_f}{V_i}\]

This equation shows the relationship between temperature change and volume change during an adiabatic process. In an adiabatic process, no heat is exchanged, and the process is driven purely by the work done on or by the gas, resulting in changes to internal energy and temperature $ \Delta U = - W.

 

Relationships between the macroscopic variables for adiabatic processes

From the previous expression:

\[ C_v \ln \frac{T_f}{T_i} = - nR \ln \frac{V_f}{V_i} \]

We can now explore how temperature changes during adiabatic expansion. Using the fact that $C_p - C_v = nR$, we define the adiabatic exponent ($\gamma$) as:

\[ \gamma = \frac{C_p}{C_v} \]

Substituting this into our earlier equation, we arrive at the following expression for temperature-volume relationship for adiabatic processes:

\[ \ln \frac{T_f}{T_i} = - \frac{nR}{C_v} \ln \frac{V_f}{V_i} = (\gamma - 1) \ln \frac{V_f}{V_i} \]

Taking the exponential of both sides, we find:

\[ \frac{T_f}{T_i} = \left( \frac{V_f}{V_i} \right)^{1 - \gamma} \implies  TV^{\gamma - 1} = \text{constant} \]

This equation represents the temperature change during a reversible adiabatic expansion. As the volume increases, the temperature decreases, and vice versa.

We can further derive a similar pressure-volume relationship:

From the ideal gas law:

\[ \frac{T_f}{T_i} = \frac{p_f V_f}{p_i V_i} \]

Substituting this into the previous expression for temperature change gives us:

\[ \frac{p_f V_f}{p_i V_i} = \left( \frac{V_f}{V_i} \right)^{1 - \gamma} \]

Rearranging this, we derive the following relationship between pressure and volume during an adiabatic process:

\[ p_i V_i^\gamma = p_f V_f^\gamma \implies  pV^\gamma = \text{constant} \]

This equation represents the pressure change during reversible adiabatic expansion. The pressure decreases as the volume increases, according to the adiabatic exponent $\gamma$.

The pressure-temperature dependence can be derived similarly:

\[ P^{1-\gamma} T^\gamma = \text{constant} \]

 

Applications of Adiabatic Processes

Adiabatic expansion and compression of gases are crucial to various processes, such as:

  • Meteorology: Cloud formation and weather patterns often involve adiabatic cooling of rising air.
  • Heat engines: Many cycles, such as the Otto and Diesel cycles, incorporate adiabatic expansion or compression stages.

 

Joule-Thompson Experiment – Isenthalpic Expansion

The Joule-Thompson experiment demonstrates the behavior of gases when they expand through a porous membrane, which causes a change in pressure without heat exchange (adiabatic process) or external work done on the system. This type of expansion is called isenthalpic expansion, meaning the enthalpy ($H$) remains constant throughout the process.

In the Joule-Thompson experiment, a gas is forced through a porous membrane or valve:

  • The system is adiabatic ($\Delta Q = 0$), meaning no heat is exchanged with the surroundings.
  • Initial state: Pressure $p_1$, Volume $V_1$, Temperature $T_1$.
  • Final state: Pressure $p_2$, Volume $V_2$, Temperature $T_2$.

A piston pushes the gas from volume $V_1$ to $V_f = 0$, and the second piston moves out from $V_i = 0$ to $V_f = V_2$.

Using the first law of thermodynamics:

\[ \Delta U = \Delta Q + \Delta W \]

Since the process is adiabatic ($\Delta Q = 0$), the equation simplifies to:

\[ \Delta U = \Delta W \]

The work done on the left and right pistons is given by:

\[ \Delta W_{net} = \Delta W_{left} + \Delta W_{right} = - \int_{V_1}^{0} p dV - \int_{0}^{V_2} p dV \]

Evaluating the integrals gives:

\[ \Delta W_{net} = p_1 V_1 - p_2 V_2 \]

The change in internal energy, using the first law, is:

\[ U_2 - U_1 = p_1 V_1 - p_2 V_2 \]

Using the definition of enthalpy ($H = U + pV$), we rewrite the internal energy change:

\[ U_2 + p_2 V_2 = U_1 + p_1 V_1 \quad \Rightarrow \quad H_2 = H_1 \]

Therefore, we conclude that the process is isenthalpic ($\Delta H = 0$).

 

Deriving the Joule-Thompson Coefficient

The enthalpy is a state function and can be expressed as:

\[ dH = \left( \frac{\partial H}{\partial p} \right)_T dp + \left( \frac{\partial H}{\partial T} \right)_p dT \]

Since the process is isenthalpic ($dH = 0$), we have:

\[ \left( \frac{\partial H}{\partial p} \right)_T dp = - \left( \frac{\partial H}{\partial T} \right)_p dT =- C_p dT \]

Dividing by $dp$, we get:

\[ \left( \frac{\partial H}{\partial p} \right)_T \frac{d \rho}{d p} = - C_p \left( \frac{\partial T}{\partial p} \right)_H = - C_p \mu \]
 

The Joule-Thompson coefficient ($\mu$) expression follows:

\[ \mu = \left( \frac{\partial T}{\partial p} \right)_H \]

 

Behavior of Gases and Cooling/Heating

The Joule-Thompson coefficient tells us whether a gas cools or heats during an isenthalpic expansion:

  • If $\mu > 0$, the gas cools during expansion.
  • If $\mu < 0$, the gas heats during expansion.

The sign of $\mu$ depends on the type of gas and its temperature. For example, nitrogen ($N_2$) and oxygen ($O_2$) exhibit cooling at room temperature, while hydrogen ($H_2$) and helium ($He$) exhibit heating at the same conditions.

Practical Application – Liquefaction of Gases

The Joule-Thompson effect is used in the liquefaction of gases, where gases are cooled below their inversion temperature (the temperature at which $\mu$ changes sign) and undergo expansion to become liquid.