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Visualized Solution
The Sigma Insight: Kinetic Theory of Gases
The problem of two connected glass bulbs at different temperatures is a classic exploration of the Kinetic Theory of Gases and the Ideal Gas Law. It beautifully demonstrates how macroscopic properties like pressure and temperature are governed by the conservation of microscopic particles.
Analyzing the Setup
Imagine two identical glass bulbs connected by a thin tube. Initially, both bulbs are submerged in an environment at (which is ). They are filled with a gas at a uniform pressure of .
Because the bulbs are identical and at the same temperature and pressure, they hold the exact same amount of gas.
Suddenly, the environment changes. The left bulb is kept in melting ice at , while the right bulb is plunged into a hot water bath at (which is ).
The Master Equation
Conservation of Moles
When the right bulb is heated, the kinetic energy of its gas molecules increases. This causes the gas to expand and flow through the connecting tube into the colder left bulb until the pressure equalizes across the entire system. Let's call this new equilibrium pressure .
The crucial physical principle here is the Conservation of Mass. Since the volume of the connecting tube is negligible, no gas is "lost" in the transit. The total number of moles of gas before heating must equal the total number of moles after heating.
Using the Ideal Gas Law, , we can express the number of moles as .
Initially, the total number of moles in both bulbs is:
After the heating process, the new moles in the left and right bulbs are:
Equating the initial and final states gives us our master equation:
The Mathematical Execution
Notice the elegance of this equation. The volume and the universal gas constant are present in every single term. We can completely factor them out and cancel them, leaving a pure relationship between pressure and temperature:
Now, we substitute our known values. Remember, thermodynamics demands absolute temperatures in Kelvin! We plug in , , , and :
The Final Calculation and a Subtle Typo
Let's simplify the algebraic fraction on the left side. The common denominator is , and the numerator becomes :
We can cancel the from the denominators on both sides:
Solving for yields:
A Note on Textbooks: You might notice that some reference materials provide an answer of . This stems from a classic arithmetic typo in older solution manuals, where was mistakenly calculated as instead of . However, the mathematically precise and physically accurate answer is indeed .
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