A container of volume 1 m^3 is divided into two equal parts by a partition. One part has an ideal ga

Q: Why is the work done zero if the volume is increasing?

Work done by a gas is calculated against an external pressure ($W = \int P_{\text{ext}} dV$). In a vacuum, $P_{\text{ext}} = 0$, so the gas does no work while expanding.

Q: Doesn't expansion always cause cooling?

Not always! Adiabatic reversible expansion causes cooling because the gas does positive work at the expense of its internal energy. In free expansion (into a vacuum), no work is done, so an ideal gas doesn't cool down.

Q: What if the gas was not ideal?

For a real gas, internal energy depends on both temperature and volume due to intermolecular forces. During free expansion, the volume increases, increasing the potential energy. To keep total internal energy constant, kinetic energy (and thus temperature) must decrease.

Q: Is this process reversible or irreversible?

Free expansion is a highly irreversible process. You cannot spontaneously compress the gas back into half the container without doing external work.

Q: Does the entropy of the gas change?

Yes! Even though the temperature is constant and $Q=0$, the volume increases, giving the molecules more accessible microstates. The entropy of the gas increases ($\Delta S = nR \ln(V_f/V_i)$).

Question

A container of volume $1	ext{ m}^3$ is divided into two equal parts by a partition. One part has an ideal gas at $300	ext{ K}$ and the other part is vacuum. The whole system is thermally isolated from the surroundings. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be ......

Sigma Percentile · Accepted Answer

The problem of free expansion is one of the most classic and conceptually rich scenarios in thermodynamics. It tests our fundamental understanding of the First Law of Thermodynamics and the behavior of ideal gases.

### Analyzing the Setup

Imagine a rigid container divided into two equal halves by a partition. One half is filled with an ideal gas at a temperature of $300	ext{ K}$, while the other half is a perfect vacuum. The entire container is thermally isolated from its surroundings.

The phrase **"thermally isolated"** is our first major clue. It implies that the boundaries of the container are perfectly insulated, meaning no heat can flow into or out of the system. Mathematically, this gives us our first constraint:
$$Q = 0$$

### The Master Equation: First Law of Thermodynamics

Now, the partition is suddenly removed. The gas, previously confined to one half, rushes into the vacuum to occupy the entire volume of the container. This process is known as **free expansion**.

To understand what happens to the temperature, we must evaluate the work done by the gas. Work done by an expanding gas is calculated against the external pressure it pushes against:
$$W = \int P_{	ext{ext}} dV$$

Since the gas is expanding into a vacuum, the external pressure $P_{	ext{ext}}$ is exactly zero. The gas doesn't have to exert any force to push the vacuum away. Consequently, the work done by the gas is zero:
$$W = 0$$

With both heat and work determined, we invoke the First Law of Thermodynamics, which states that the change in internal energy ($\Delta U$) of a system is equal to the heat added to the system minus the work done by the system:
$$\Delta U = Q - W$$

Substituting our values:
$$\Delta U = 0 - 0 = 0$$

### Final Calculation and Conclusion

The internal energy of the gas remains completely unchanged during the free expansion.

For an **ideal gas**, the internal energy is a function of temperature alone ($U = \frac{f}{2}nRT$). It does not depend on the volume or pressure. Since the internal energy does not change ($\Delta U = 0$), the temperature must also remain constant ($\Delta T = 0$).

Therefore, the final temperature of the gas is exactly equal to its initial temperature:
$$T_f = T_i = 300	ext{ K}$$

*A word of caution:* This result holds strictly for an ideal gas. If the gas were real, the expansion would increase the average distance between molecules. Because real gas molecules exert attractive intermolecular forces, pulling them apart requires energy, which would be drawn from their kinetic energy, leading to a slight drop in temperature (the Joule-Thomson effect). However, for an ideal gas, free expansion is always an isothermal process!

Visualized Solution

$Initial Setup$

$Heat Exchange (Q)$

$Work Done (W)$

$First Law of Thermodynamics$

$Final Temperature$

$Conclusion & Extension$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Final Calculation and Conclusion

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