Question

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0,\) and \(w=0 .\) (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) What is the "driving force" for the expansion of the gas: enthalpy or entropy?

Short answer

(a) Yes, the process is spontaneous because the change in entropy (\(\Delta S\)) is positive, resulting in a negative change in Gibbs Free Energy (\(\Delta G < 0\)). (b) No work is done during this process because the external pressure (\(P_{ext}\)) is zero, as the gas is expanding into a vacuum. Therefore, the work done (\(w\)) is also zero. (c) The driving force for the expansion of the gas is entropy, as the enthalpy (\(\Delta H\)) remains constant (equal to zero) during the isothermal process, while the entropy (\(\Delta S\)) increases.

Step-by-step solution

(a) Evaluate the spontaneity of the process

To determine if a process is spontaneous, we need to consider the change in Gibbs free energy, which is given by the following equation: \(\Delta G = \Delta H - T \Delta S\) where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, \(T\) is the temperature, and \(\Delta S\) is the change in entropy. A process is considered spontaneous if \(\Delta G < 0\). In the given problem, since the process is isothermal, we can re-write the equation as: \(\Delta G = q + w - T \Delta S\) We are given that \(q = 0\) and \(w = 0\), so the equation simplifies to: \(\Delta G = - T \Delta S\) If \(\Delta S > 0\), the process will be spontaneous since \(\Delta G\) will be less than zero. An expansion of a gas results in an increase in entropy because the particles of the gas have more space to move around, which means \(\Delta S > 0\). Thus, the process is spontaneous.

(b) Explain why no work is done during the process

During the isothermal expansion of a gas into a vacuum, the gas expands freely with no pressure exerted externally (since it is in a vacuum). Work is defined as the energy transferred due to the force acting on an object over a certain distance, and in this case, the force would be exerted by the pressure of the gas. The formula for work done in a thermodynamic system involving a gas is given by: \(w = - \int_{V_1}^{V_2} P_{ext} dV\) where \(w\) is the work done by the gas, \(V_1\) and \(V_2\) are the initial and final volumes of the gas, and \(P_{ext}\) is the external pressure. In this case, since the gas is expanding into a vacuum, \(P_{ext} = 0\), and hence, the work done is equal to \(0\).

(c) Identify the driving force for expansion: enthalpy or entropy

In order to determine the driving force for the expansion of the gas, we need to consider whether the change in enthalpy (\(\Delta H\)) or the change in entropy (\(\Delta S\)) is the dominant factor in the Gibbs free energy equation (\(\Delta G = \Delta H - T \Delta S\)). Since the process is isothermal, \(\Delta E = 0\), and the internal energy of a gas depends only on its temperature, the enthalpy of the gas also remains constant during the expansion process because \(\Delta H = \Delta E + P_\Delta V\). Therefore, \(\Delta H = 0\). On the other hand, as mentioned earlier, the entropy of the gas increases during expansion, which means \(\Delta S > 0\). Since \(\Delta H = 0\) and \(\Delta S > 0\), the driving force for the gas expansion is the entropy change. The expansion of the gas results in an increase in disorder or randomness in the system, which is consistent with an increase in entropy.

Key concepts

Spontaneity of Process

The spontaneity of a process indicates whether it can occur naturally without any external influence. In thermodynamics, we determine this using the Gibbs free energy change, represented by the equation: \[ \Delta G = \Delta H - T \Delta S \] Here, \( \Delta G \) denotes the Gibbs free energy change, \( \Delta H \) is the enthalpy change, \( T \) is the temperature, and \( \Delta S \) is the entropy change. For a process to be spontaneous, \( \Delta G \) must be less than zero.
In the case of isothermal expansion into a vacuum:

• The equation simplifies to \( \Delta G = -T \Delta S \) because both heat \( q \) and work \( w \) are zero.
• Since the entropy (\( \Delta S \)) increases during gas expansion, \( \Delta G \) becomes negative.
• This means the process occurs spontaneously due to the natural tendency of entropy to increase.

Work in Thermodynamics

In thermodynamics, work refers to the energy transferred due to a force acting over a distance. For a gas, this is often equated to the pressure and volume change involved, using the formula: \[ w = - \int_{V_1}^{V_2} P_{ext} \, dV \] Where \( w \) is the work done by the gas, \( V_1 \) and \( V_2 \) are initial and final volumes, and \( P_{ext} \) is the external pressure.
For isothermal expansion into a vacuum:

• The external pressure \( P_{ext} \) is zero.
• Because there's no opposing force, no work is done (\( w = 0 \)).

This kind of expansion where no work is done is known as "free expansion."

Entropy Change

Entropy measures the disorder or randomness in a system. In thermodynamic processes, the change in entropy (\( \Delta S \)) can dictate the spontaneity of reactions. For the isothermal expansion of a gas:

• The gas particles move from a confined space to a larger volume.
• This increase in space allows particles more freedom, raising the disorder in the system.

This results in a positive change in entropy (\( \Delta S > 0 \)).
In our scenario:

• Since the enthalpy change \( \Delta H \) is zero during the process, the driving force is the increase in entropy.
• The natural tendency toward greater entropy is what propels the spontaneous expansion of the gas.

Understanding how entropy relates to spontaneity helps unravel why some processes, like expansion into a vacuum, occur without any input of energy.