To solve this problem, we will use the properties of an isothermal, reversible expansion of an ideal gas.

Given Data

• Initial volume, $V_1 = 5 \, \text{dm}^3$
• Final volume, $V_2 = 10 \, \text{dm}^3$
• Temperature, $T = 300 \, \text{K}$
• Number of moles, $n = 1 \, \text{mol}$
• For an ideal gas, $R = 8.314 \, \text{J K}^{-1} \text{mol}^{-1}$

Key Formulas and Concepts

1. Isothermal Process: Since the temperature is constant, the internal energy change $\Delta U = 0$ and the enthalpy change $\Delta H = 0$ for an ideal gas.

2. Heat and Work:

   • Heat absorbed by the system in an isothermal, reversible expansion $q = nRT \ln \frac{V_2}{V_1}$.
   • The work done by the gas $w = -q$ in an isothermal process, as $\Delta U = 0$.

3. Gibbs Free Energy Change (∆G):

   • For an isothermal, reversible process, $\Delta G = nRT \ln \frac{V_2}{V_1}$.

4. Helmholtz Free Energy Change (∆A):

   • For an isothermal process, $\Delta A = w = -nRT \ln \frac{V_2}{V_1}$.

Solution Steps

1. Calculate $q$ (Heat Absorbed)

$q = nRT \ln \frac{V_2}{V_1}$ Substitute the values: $q = (1 \, \text{mol})(8.314 \, \text{J K}^{-1} \text{mol}^{-1})(300 \, \text{K}) \ln \frac{10}{5}$ $q = 2494.2 \, \text{J} \times \ln 2$ $q \approx 2494.2 \, \text{J} \times 0.693$ $q \approx 1729.6 \, \text{J}$

2. Calculate $w$ (Work Done)

Since $w = -q$ for an isothermal process: $w = -1729.6 \, \text{J}$

3. Calculate $\Delta U$ (Change in Internal Energy)

For an ideal gas undergoing an isothermal process: $\Delta U = 0$

4. Calculate $\Delta H$ (Change in Enthalpy)

For an ideal gas in an isothermal process: $\Delta H = 0$

5. Calculate $\Delta G$ (Change in Gibbs Free Energy)

$\Delta G = -q = -1729.6 \, \text{J}$

6. Calculate $\Delta A$ (Change in Helmholtz Free Energy)

$\Delta A = w = -1729.6 \, \text{J}$

Summary of Results

• $q = 1729.6 \, \text{J}$
• $w = -1729.6 \, \text{J}$
• $\Delta U = 0$
• $\Delta H = 0$
• $\Delta G = -1729.6 \, \text{J}$
• $\Delta A = -1729.6 \, \text{J}$

Would you like a deeper explanation of any specific step?

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Here are some related questions to explore further:

1. How would the calculations change if the expansion were adiabatic instead of isothermal?
2. What would happen to $\Delta G$ and $\Delta A$ if the process were not reversible?
3. Can you calculate the entropy change for this process?
4. How does the ideal gas assumption affect these calculations?
5. How would these values change if the temperature were different?

Tip: In isothermal processes for ideal gases, $\Delta U$ and $\Delta H$ are always zero since they depend only on temperature changes.