Animated Solution for Chemistry - Chemical Thermodynamics: An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If $T_i$ is the initial temperature and $T_f$ is the final temperature, then which of the following statements is correct ?

The problem of comparing reversible and irreversible expansions is a classic in thermodynamics. It tests your deep understanding of the First Law, work, and internal energy. Let's break down the physics step-by-step.

The question states that an ideal gas expands in an isolated system. In strict thermodynamic terms, an isolated system cannot exchange heat or matter with its surroundings. This immediately tells us that the process is adiabatic, meaning there is no heat transfer:

Now, let's bring in the First Law of Thermodynamics, which relates internal energy ($\Delta U$), heat ($q$), and work ($W$):

Since $q=0$, the equation simplifies beautifully:

This is a profound statement. It means that any work done by the gas comes entirely at the expense of its own internal energy.

For an ideal gas, the internal energy is a direct measure of its temperature. The relationship is given by:

Equating this to the work done, we get our master equation for this process:

During an expansion, the gas does work on the surroundings. By convention, work done by the system is negative ($W<0$). Because the work is negative, the change in internal energy must also be negative ($\Delta U<0$). This mathematically proves that the gas cools down during an adiabatic expansion:

Irreversible Work

Here is where the magic happens. We need to compare the work done in a reversible process versus an irreversible process.

A reversible expansion is an idealized process where the internal pressure of the gas is always infinitesimally greater than the external pressure. This delicate balance ensures that the gas does the maximum possible work. In contrast, an irreversible expansion happens against a much lower external pressure, resulting in less work done.

In terms of magnitude:

However, we must be extremely careful with signs! Since the work is negative (expansion), a larger magnitude means a more negative value. For example, if reversible work is $-100\text{ J}$ and irreversible work is $-50\text{ J}$, then $-100<-50$. Therefore:

Since the work done is directly equal to the change in internal energy ($\Delta U=W$), the reversible process will experience a more negative change in internal energy:

Substituting our temperature relation:

Dividing both sides by the positive constant $n{C}_v$ and adding $T_i$ to both sides, we arrive at our final, elegant conclusion:

Or, written another way:

The irreversible process ends up at a higher final temperature because it didn't expend as much of its internal energy doing work. The correct option is (a).