Animated Solution for Chemistry - States of Matter: Two closed bulbs of equal volume ($V$) containing an ideal gas initially at pressure $p_i$ and temperature $T_1$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $T_2$. The final pressure $p_f$ is

A Tale of Two Bulbs

Imagine a perfectly balanced system: two identical glass bulbs, each with a volume $V$, connected by a very thin, negligible tube. Initially, both bulbs are filled with an ideal gas at the exact same pressure $p_i$ and the exact same temperature $T_1$.

Everything is in perfect thermodynamic equilibrium. The gas molecules are happily bouncing around, completely unaware of the disturbance that is about to happen.

Conservation of Mass

To understand what happens next, we need to rely on our most trusted tool in thermodynamics: the Ideal Gas Equation, $pV=nRT$.

Because our two bulbs form a closed system—meaning no gas can leak out and no new gas can enter—the total amount of gas must remain absolutely constant. This is the principle of conservation of mass, or in our case, the conservation of moles.

We can rearrange the ideal gas law to find the number of moles:

Let's calculate the total number of moles initially present in the entire system. We simply add the moles in the left bulb to the moles in the right bulb.

For the left bulb, the number of moles is $\frac{p_{i}V}{R{T}_1}$. Since the right bulb is identical and under the exact same conditions, its number of moles is also $\frac{p_{i}V}{R{T}_1}$.

Adding them together gives us the total initial moles:

Keep this expression safe; it is the foundation of our solution.

Heating the System

Now, we introduce a disturbance. We heat the right bulb, raising its temperature to a new value, $T_2$.

What happens physically? The gas molecules in the right bulb gain kinetic energy. They start moving faster and hitting the walls harder, which temporarily spikes the pressure in the right bulb. Because the bulbs are connected, this pressure difference forces gas molecules to flow through the narrow tube from the right bulb into the left bulb.

This flow continues until the pressure in both bulbs equalizes at a new, final value, $p_f$.

Let's write down the total number of moles for this new final state.

The left bulb is still at its original temperature $T_1$, but its pressure has changed to $p_f$. So, the number of moles in the left bulb is now $\frac{p_{f}V}{R{T}_1}$.

The right bulb is at the new temperature $T_2$ and the new pressure $p_f$. Its number of moles is $\frac{p_{f}V}{R{T}_2}$.

Adding these gives us the total final moles:

Bringing It All Together

Here is the crucial logical step. Because the system is closed, the initial total moles must equal the final total moles.

Let's equate our two expressions:

This is our master equation. Now, the magic of algebra takes over. Look closely at the right side. We can factor out the common terms, $\frac{p_{f}V}{R}$:

Notice that the term $\frac{V}{R}$ appears on both sides of the equation. Since the volume and the gas constant are non-zero, we can beautifully cancel them out:

We are almost at the finish line. Let's simplify the terms inside the bracket by taking a common denominator:

To find the final pressure $p_f$, we just cross-multiply. The $T_1$ in the denominator on the left side cancels perfectly with the $T_1$ in the numerator on the right side.

Isolating $p_f$, we arrive at our final, elegant result:

This beautiful equation tells us exactly how the final pressure depends on the initial pressure and the two temperatures, perfectly capturing the physics of the redistributing gas!