The situation obtained when a perfect gas expands into a vacuum is

Analysis

This question tests your understanding of Thermodynamics, specifically the expansion of an ideal (perfect) gas into a vacuum. This process is known as Joule expansion or free expansion.

Concept Explanation

To solve this, we need to look at the First Law of Thermodynamics and the nature of an ideal gas:

1. Work Done ($w$): When a gas expands into a vacuum, there is no external pressure ($P_{ext} = 0$). Since work done $w = -P_{ext} \Delta V$, it follows that $w = 0$.
2. Internal Energy ($\Delta U$): For an ideal gas, internal energy depends only on temperature. Since there is no work done and no heat exchange ($q = 0$) with the surroundings (the system is isolated), the temperature remains constant. Therefore, $\Delta U = 0$.
3. Enthalpy ($\Delta H$): Enthalpy is defined as $H = U + PV$. For an ideal gas, $PV = nRT$. Since $T$ is constant and $U$ is constant, $\Delta H$ must also be zero.
4. Entropy ($\Delta S$): Entropy is a measure of disorder. When a gas expands into a larger volume, the number of possible microstates increases, meaning the system becomes more disordered. Therefore, the entropy change ($\Delta S$) is positive. Consequently, the term $T\Delta S$ is positive.

Step-by-Step Solution

• Step 1: Identify the conditions of free expansion. Because the gas expands into a vacuum, no work is done ($w = 0$) and no heat is exchanged ($q = 0$).
• Step 2: Apply the First Law of Thermodynamics ($\Delta U = q + w$). Since $q=0$ and $w=0$, then $\Delta U = 0$.
• Step 3: Relate $\Delta U$ to $\Delta H$. For an ideal gas, $\Delta H = \Delta U + \Delta(PV)$. Since $\Delta U = 0$ and $\Delta(PV) = \Delta(nRT) = 0$ (because $T$ is constant), then $\Delta H = 0$.
• Step 4: Determine the entropy change. Expansion into a vacuum is a spontaneous, irreversible process that increases the volume available to the gas molecules. This leads to an increase in entropy ($\Delta S > 0$). Thus, $T\Delta S > 0$.

Conclusion

Comparing our findings ($\Delta H = 0$ and $T\Delta S > 0$) with the given options, we find that they match option D.

Exam Tip: Always remember that for an ideal gas, the internal energy and enthalpy are functions of temperature only. If the temperature doesn't change during an expansion, $\Delta U$ and $\Delta H$ will always be zero!