Animated Solution for Physics - Thermodynamics: An ideal gas having initial pressure $p$, volume $V$ and temperature $T$ is allowed to expand adiabatically until its volume becomes $5.66V$ while its temperature falls to $T/2$. (a) How many degrees of freedom do gas molecules have? (b) Obtain the work done by the gas during the expansion as a function of the initial pressure $p$ and volume $V$.

Imagine a gas trapped in a perfectly insulated cylinder. Suddenly, the piston is released, and the gas expands. Because the cylinder is insulated, no heat can enter or escape. This is the essence of an adiabatic process. In this problem, we are going to act as thermodynamic detectives, using just the initial and final states of the gas to uncover its fundamental nature—its degrees of freedom—and the exact amount of work it performed during its expansion.

We are given that the gas expands from volume $V$ to $5.66V$, and its temperature drops from $T$ to $T/2$. Because the process is adiabatic, the pressure, volume, and temperature are intimately linked. To find the degrees of freedom, we first need the adiabatic index, $\gamma$.

The relationship between temperature and volume in an adiabatic process is given by:

Let's plug in our initial and final states:

The $T$ and $V^{\gamma -1}$ terms beautifully cancel out from both sides, leaving us with a purely numerical equation:

To bring the exponent down, we take the natural logarithm on both sides:

Solving for $\gamma -1$, we get:

This gives us our adiabatic index: $\gamma =1.4$.

Now that we have $\gamma$, finding the degrees of freedom ($f$) is straightforward. The adiabatic index is related to the degrees of freedom by the formula:

Substituting $\gamma =1.4$:

A degree of freedom of 5 tells us a lot about the gas. It means the gas molecules can translate in 3 directions and rotate in 2. This is the classic signature of a diatomic gas like Oxygen or Nitrogen at room temperature!

Next, we need to find the work done by the gas. The formula for work done in an adiabatic process is:

We know the initial pressure $p_i=p$ and volume $V_i=V$. We also know the final volume $V_f=5.66V$. But we need the final pressure $p_f$. We can find this using the adiabatic relation between pressure and volume:

Substituting our values:

Solving for $p_f$:

Now, we have all the pieces of the puzzle. Let's substitute them into the work formula:

The gas did $1.23pV$ of work on its surroundings. But where did the energy for this work come from? Since the process was adiabatic ($Q=0$), the First Law of Thermodynamics ($\Delta U=Q-W$) tells us that $\Delta U=-W$. The gas had to sacrifice its own internal energy to push the piston outward. This loss of internal energy is exactly why we observed the temperature drop to $T/2$ in the first place! Physics is beautifully consistent.