Animated Solution for Physics - Thermodynamics: Comprehension Passage

Imagine a perfectly insulated container divided into two compartments by a thermally conducting partition.

The lower compartment holds 2 moles of a hot monoatomic gas at 700 K. The upper compartment holds 2 moles of a cooler diatomic gas at 400 K, and it is topped with a movable, frictionless piston.

Because the container is perfectly insulated, no heat can escape to the surroundings. Any heat lost by the hot gas must be entirely absorbed by the cooler gas.

The Rigid Partition

In the first scenario, the partition is rigidly fixed. This is a crucial constraint!

A fixed partition means the lower gas is trapped in a constant volume. Therefore, its heat exchange process is isochoric, and we must use the molar heat capacity at constant volume, $C_V$.

On the other hand, the upper gas is under a movable piston exposed to the atmosphere. The piston is free to move, which means the pressure of the upper gas remains constant. Its heat exchange process is isobaric, so we use $C_p$.

Let's set up the heat exchange equation. The heat lost by the lower gas equals the heat gained by the upper gas:

Substituting the given values for the monoatomic and diatomic gases:

Notice how elegantly the $2$ and the gas constant $R$ cancel out on both sides. We are left with a simple linear equation:

Bringing the temperature terms to one side, we get:

The Free Partition

Now, the rules of the game change. The partition is no longer fixed; it is free to move without friction.

If the partition can move freely, it will adjust its position until the pressures in both compartments equalize.

Furthermore, since the top piston is also free to move, the pressure of the entire system is dictated by the constant atmospheric pressure and the weight of the piston. This means both gases now undergo a constant pressure process.

Once again, the net heat exchange with the surroundings is zero. But this time, we use $C_p$ for both gases:

Substituting the values, where $C_{p1}=\frac{5}{2}R$ for the monoatomic gas:

Canceling the common terms, we simplify the equation:

The new equilibrium temperature is 525 K.

First Law of Thermodynamics

We need to find the total work done by the gases. We could calculate the work done by each gas individually, but there is a brilliant shortcut!

Let's apply the First Law of Thermodynamics to the entire system:

Since the container is perfectly insulated, $Q_{\text{total}}=0$. Therefore, the total work done is simply the negative of the total change in internal energy:

Remember, the change in internal energy always depends on $C_V$, regardless of the thermodynamic process. Let's calculate it for each gas:

Adding these together gives the total change in internal energy:

Finally, we find the total work done:

The total work done by the gases is -100R. The negative sign indicates that work is done on the system by the surroundings as it contracts.