Question

What happens in a reversible adiabatic expansion process?

• A. Heating takes place
• B. Cooling takes place
• C. Pressure remains constant
• D. Temperature remains constant

Hint:

Remember that in an adiabatic expansion, the system does work without any heat input. This work comes from the internal energy of the gas, causing its temperature to drop. Think of air escaping rapidly from a pressurized tire; the escaping air feels cooler.

Answer 1

The Correct Option is B

Step 1: Understand the terms involved in the question. 
Reversible process: A thermodynamic process that can be reversed by an infinitesimal change in a property of the system without dissipating energy. The system is always in equilibrium with its surroundings.
Adiabatic process: A thermodynamic process in which no heat is exchanged between the system and its surroundings (\( Q = 0 \)).
Expansion process: A process in which the volume of the system increases (\( dV>0 \)). 
Step 2: Apply the first law of thermodynamics to a reversible adiabatic expansion process. 
\[ \Delta U = Q - W \] For an adiabatic process, \( Q = 0 \), so the first law simplifies to: \[ \Delta U = -W \] In an expansion process, the system does work on the surroundings (\( W>0 \)). Therefore, for a reversible adiabatic expansion, \( \Delta U \) must be negative. 
Step 3: Relate the change in internal energy to the change in temperature for an ideal gas. 
For an ideal gas, the internal energy \( U \) is directly proportional to its temperature \( T \): \[ \Delta U = m C_v \Delta T \] where \( m \) is the mass of the gas and \( C_v \) is the specific heat capacity at constant volume (which is always positive). Since \( \Delta U \) is negative in a reversible adiabatic expansion, it follows that \( \Delta T \) must also be negative, indicating a decrease in temperature. 
Step 4: Consider the relationship between pressure, volume, and temperature in a reversible adiabatic process for an ideal gas. 
For a reversible adiabatic process involving an ideal gas, the following relationship holds: \[ PV^\gamma = \text{constant} \] where \( P \) is the pressure, \( V \) is the volume, and \( \gamma = C_p / C_v \) is the adiabatic index (which is always greater than 1). As the gas expands, \( V \) increases. To maintain \( PV^\gamma \) constant, the pressure \( P \) must decrease. Also, using the ideal gas law \( PV = nRT \) (where \( n \) is the number of moles and \( R \) is the ideal gas constant), we can write \( P = nRT/V \). Substituting this into the adiabatic relation: \[ \left( \frac{nRT}{V} \right) V^\gamma = nRT V^{\gamma - 1} = \text{constant} \] Since \( n \) and \( R \) are constant, \( T V^{\gamma - 1} = \text{constant} \). As \( V \) increases during expansion and \( \gamma - 1>0 \), the temperature \( T \) must decrease to keep the product constant. 
Step 5: Evaluate the given options. 
Option 1 (Heating takes place): This is incorrect, as our analysis shows that the temperature decreases.
Option 2 (Cooling takes place): This is correct, as the temperature of the gas decreases during a reversible adiabatic expansion.
Option 3 (Pressure remains constant): This is incorrect; the pressure decreases as the volume increases.
Option 4 (Temperature remains constant): This is incorrect; the temperature decreases as the gas does work on the surroundings at the expense of its internal energy. 
Step 6: Select the correct answer. 
In a reversible adiabatic expansion process, cooling takes place.