In classical thermodynamics, during the free expansion of an ideal gas into vacuum, the work done by the gas on the surroundings has what value?

Introduction / Context:
Free expansion is a classic thought experiment in thermodynamics used to illustrate the concepts of work, internal energy and irreversibility. In a free expansion, a gas expands into a vacuum without any opposing external pressure. Understanding how much work is done in such a process is essential for correctly applying the first law of thermodynamics. This question focuses on the value of work done by an ideal gas during free expansion.

Given Data / Assumptions:

• We are considering an ideal gas obeying the ideal gas equation.
• The gas initially occupies part of a container and then expands into a vacuum region when a partition is removed.
• There is no external pressure resisting the expansion in the vacuum region.
• The system is thermally insulated, so heat exchange may be negligible in the ideal case, although the question focuses on work.

Concept / Approach:
In thermodynamics, the work done by a gas during expansion is usually calculated as W = integral of P_external dV, where P_external is the external pressure opposing the expansion. In a free expansion into vacuum, the external pressure is effectively zero because there is no gas or mechanism pushing back. Since P_external is zero throughout the expansion, the integral of zero with respect to volume is zero. Therefore, the work done by the gas on the surroundings is zero. The gas volume increases and its molecules spread out, but no mechanical work is transferred to the surroundings because there is nothing to push against.

Step-by-Step Solution:
Step 1: Recall that thermodynamic work for a quasi static expansion is W = integral P_external dV.Step 2: Recognise that in a free expansion, the gas expands into a vacuum, so the external pressure on the boundary is zero.Step 3: Substitute P_external = 0 into the integral, which gives W = integral 0 dV.Step 4: Evaluate the integral of zero over any change in volume; the result is W = 0.Step 5: Understand physically that although the gas occupies more space, there is no opposing force to do work against.Step 6: Conclude that the work done in a free expansion of an ideal gas is exactly zero.

Verification / Alternative check:
Consider the first law of thermodynamics, which states that delta U = Q - W, where delta U is the change in internal energy, Q is heat added and W is work done by the system. For an ideal gas, internal energy depends only on temperature. In a free expansion of an ideal gas, the temperature remains constant if the container is insulated and the process is ideal, so delta U equals zero. If no heat is exchanged (Q = 0), then from delta U = Q - W we get 0 = 0 - W, which implies W = 0. This reinforces the conclusion that no work is done during an ideal free expansion.

Why Other Options Are Wrong:
Stating that the work is minimum but not zero is incorrect, because the calculated work is exactly zero due to zero external pressure. Claiming that the work is maximum would apply to a reversible expansion against a finite pressure, not to free expansion into vacuum. Saying that the work is indeterminate without more data ignores the fact that the definition of free expansion already fixes the external pressure at zero, which is enough to determine that W is zero.

Common Pitfalls:
Students sometimes think that any expansion of a gas must involve positive work because the gas volume increases. However, work in thermodynamics is not simply change in volume, it depends on both pressure and volume change. Others confuse internal energy changes with work and assume that if the gas occupancy changes, energy must have been transferred as work. To avoid these mistakes, always refer to the formal definition W = integral P_external dV and remember that in free expansion P_external is zero throughout.

Final Answer:
In a free expansion of an ideal gas into vacuum, the work done by the gas is zero because there is no external pressure opposing the expansion.