Enthalpy Change in an Isothermal and Reversible Expansion of a Gas

Introduction:
In an isothermal and reversible expansion, the enthalpy change (ΔH) of a gas can be determined using the equation ΔH = q + w, where q represents the heat transferred to the system and w represents the work done by the system. In this case, the expansion is isothermal, meaning the temperature remains constant throughout the process.

Given Information:
- Initial volume (V₁) = 5 L
- Final volume (V₂) = 10 L
- Moles of gas (n) = 1

Calculating the Work Done (w):
During an isothermal expansion, the work done by the system can be calculated using the equation w = -nRT ln(V₂/V₁), where R is the gas constant and T is the temperature. Since the temperature remains constant in this case, the equation simplifies to w = -nRT ln(V₂/V₁).

Calculating the Enthalpy Change (ΔH):
Since the process is isothermal, the heat transferred to the system (q) can be calculated using the equation q = -w. Therefore, ΔH = q + w = -w + w = 0.

Explanation:
In an isothermal and reversible expansion, the temperature remains constant. As a result, the internal energy of the gas does not change, and therefore, there is no change in enthalpy (ΔH = 0). This means that the heat transferred to the system (q) is equal in magnitude but opposite in sign to the work done by the system (w), resulting in a net enthalpy change of zero.

Conclusion:
The enthalpy change (ΔH) for the isothermal and reversible expansion of 1 mole of gas from 5L to 10L is zero. This is due to the constant temperature, which results in no change in internal energy and, consequently, no net enthalpy change.