Question

A sample containing 2.50 mol of an ideal gas at \(325 \mathrm{K}\) is expanded from an initial volume of \(10.5 \mathrm{L}\) to a final volume of 60.0 L. Calculate the final pressure. Calculate \(\Delta G\) and \(\Delta A\) for this process for (a) an isothermal reversible path and (b) an isothermal expansion against a constant external pressure equal to the final pressure. Explain why \(\Delta G\) and \(\Delta A\) do or do not differ from one another.

Short answer

The final pressure after the isothermal expansion of the ideal gas is approximately 107.4 J/L. For the isothermal reversible process, both ΔG and ΔA are approximately -13462.6 J. For the isothermal expansion against constant external pressure equal to the final pressure, ΔG remains the same, approximately -13462.6 J, while ΔA is approximately -5305.4 J. ΔG and ΔA differ in the irreversible process because the work done is not the maximum work that could have been done.

Step-by-step solution

Find the initial pressure

First, we need to find the initial pressure of the gas. We can use the ideal gas law to do this: \[PV = nRT\] Where: P = Pressure V = Volume n = Number of moles R = Ideal gas constant (\(8.314 \frac{J}{mol \cdot K}\)) T = Temperature in Kelvin Plugging in the given values: \[\textnormal{P} \times 10.5 \ \textnormal{L} = 2.50 \ \textnormal{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K}\] Now, we can solve for the initial pressure: \[P = \frac{2.50 \ \mathrm{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K}}{10.5\ \textnormal{L}} \approx 644.5 \frac{\textnormal{J}}{\textnormal{L}}\]

Find the final pressure

Now that we have the initial pressure, we can find the final pressure after the gas has expanded to the final volume. Since the expansion is isothermal, the temperature remains constant. We can use the ideal gas law again to find the final pressure: \[PV = nRT\] Plugging in the values, with the new volume: \[\textnormal{P} \times 60.0 \ \textnormal{L} = 2.50 \ \textnormal{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K}\] Now, we can solve for the final pressure: \[P = \frac{2.50 \ \mathrm{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K}}{60.0\ \textnormal{L}} \approx 107.4 \frac{\textnormal{J}}{\textnormal{L}}\]

Calculate ΔG and ΔA for the isothermal reversible path

For an isothermal reversible process, we can use the following formulas to calculate the change in Gibbs free energy (ΔG) and Helmholtz free energy (ΔA): \[\Delta G = -nRT \ln \frac{P_{\textnormal{final}}}{P_{\textnormal{initial}}}\] \[\Delta A = -nRT \ln \frac{V_{\textnormal{final}}}{V_{\textnormal{initial}}}\] Using the calculated final and initial pressures, and the given final and initial volumes: \[\Delta G = -2.50 \ \textnormal{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K} \times \ln \frac{107.4 \frac{\textnormal{J}}{\textnormal{L}}}{644.5 \frac{\textnormal{J}}{\textnormal{L}}} \approx -13462.6 \ \textnormal{J}\] \[\Delta A = -2.50 \ \textnormal{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K} \times \ln \frac{60.0 \ \textnormal{L}}{10.5 \ \textnormal{L}} \approx -13462.6 \ \textnormal{J}\] For the isothermal reversible process, ΔG and ΔA are the same.

Calculate ΔG and ΔA for the isothermal expansion against constant external pressure

For an isothermal expansion against a constant external pressure equal to the final pressure, we can use the following formulas: \[\Delta G = -nRT \ln \frac{P_{\textnormal{final}}}{P_{\textnormal{initial}}}\] \[\Delta A = -P_{\textnormal{ext}}\Delta V\] Where ΔV is the change in volume from the initial state to the final state (V_final - V_initial). Using the values calculated above: \[\Delta G = -2.50 \ \textnormal{mol} \times 8.314 \ \frac{\textnormal{J}}{\textnormal{mol} \cdot \textnormal{K}} \times 325 \ \mathrm{K} \times \ln \frac{107.4 \frac{\textnormal{J}}{\textnormal{L}}}{644.5 \frac{\textnormal{J}}{\textnormal{L}}} \approx -13462.6 \ \textnormal{J}\] \[\Delta A = -107.4 \frac{\textnormal{J}}{\textnormal{L}} \times (60.0 \ \textnormal{L} - 10.5 \ \textnormal{L}) \approx -5305.4\ \textnormal{J}\] In this case, ΔG and ΔA are not the same, because work done in an irreversible process is not the maximum work that could have been done. So, ΔG and ΔA differ from one another depending on whether the process is reversible or irreversible. They are the same for a reversible process and differ for an irreversible process, as shown in the calculations above.

Key concepts

Isothermal Expansion

When a gas expands while maintaining a constant temperature, the process is known as isothermal expansion. It's a key concept in thermodynamics and involves the ideal gas law, which relates the pressure, volume, and temperature of an ideal gas. In isothermal conditions, the temperature (T) remains fixed, meaning the product of pressure (P) and volume (V) stays constant.

An example of this comes from an exercise where a gas at a certain temperature expands from one volume to another. Using the ideal gas law, one could determine the gas's final pressure after expansion. Isothermal expansion can be visualized on a P-V diagram, where it is represented by a hyperbolic curve, indicating that as volume increases, pressure decreases proportionally to maintain the constant temperature.

In terms of energy, isothermal expansion can do work on the surroundings. Since the internal energy for an ideal gas in an isothermal process doesn't change (due to constant temperature), the work done by the gas comes from the heat absorbed from the surroundings.

Gibbs Free Energy

Gibbs free energy (G) is a thermodynamic potential that measures the maximum amount of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. It encompasses both the enthalpy (heat content) of the system and its entropy (degree of disorder), thus serving as a criterion for spontaneous processes.

The change in Gibbs free energy, denoted as \(\Delta G\), helps predict whether a process will occur spontaneously. A negative \(\Delta G\) implies a process can occur without the input of energy, while a positive \(\Delta G\) suggests energy is needed for the process to proceed. In the context of isothermal expansion, \(\Delta G\) can be calculated using the formula \(\Delta G = -nRT \ln \frac{P_{\textnormal{final}}}{P_{\textnormal{initial}}}\), which considers the change in the system's pressure from an initial to a final state.

Helmholtz Free Energy

Helmholtz free energy (A) is another important thermodynamic potential, similar to Gibbs free energy, but it is used under the condition of constant volume and temperature. It signifies the amount of useful work obtainable from a closed thermodynamic system.

The change in Helmholtz free energy, denoted as \(\Delta A\), is found using the formula \(\Delta A = -nRT \ln \frac{V_{\textnormal{final}}}{V_{\textnormal{initial}}}\) for reversible processes or \(\Delta A = -P_{\textnormal{ext}}\Delta V\) for irreversible processes against an external pressure, \(P_{\textnormal{ext}}\). The choice of calculations depends on whether the process is reversible or irreversible. The concept of Helmholtz free energy is crucial when dealing with processes at constant volume, as it reveals the energy available to do work other than expansion or compression work.

Reversible Process

A reversible process in thermodynamics is an idealized transformation that occurs in infinite small steps, allowing the system to remain in equilibrium with its surroundings at all times. Reversible processes are extremely efficient and represent the limit of no energy waste; there's no increase in entropy in the universe due to the process.

In the context of isothermal expansion, a reversible process allows for calculations of changes in both Gibbs and Helmholtz free energies using related formulas. These changes reflect the maximum work that can be done by a system because they are executed so slowly that the system can theoretically be returned to its initial state without any net energy change in the universe. In our gas expansion example, the values of \(\Delta G\) and \(\Delta A\) are identical, highlighting that for a reversible isothermal process, both quantities measure the same maximum work that can be extracted.

Irreversible Process

In contrast to the reversible process, an irreversible process happens spontaneously and cannot be reversed without changing the surroundings. It's a real-world process that involves friction, inelastic deformations, and other non-equilibrium changes, resulting in an increase in the entropy of the universe.

In practical terms, irreversible processes are more common than reversible ones. For example, the isothermal expansion of a gas against a constant external pressure that is different from the gas's internal pressure is considered irreversible. Such a process involves energy dissipation and, consequently, less work is performed by the system compared to a reversible process. \(\Delta G\) for the irreversible process remains the same as that for the reversible process, but \(\Delta A\) would differ because it now depends on the external pressure and the change in volume \(\Delta V\), thus making the energy extracted as work from the process lower than the maximum possible outlined by a reversible process.