- #1
decentfellow
- 130
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Homework Statement
Two glass bulbs of equal volume are connected by a narrow tube and are filled with a gas at ##0^{\circ}C## and a pressure of ##76\text{ cm}## of mercury. One of the bulbs is then placed in melting ice and the other is placed in water at ##62^{\circ}C##. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible.
Homework Equations
$$PV=nRT$$
The Attempt at a Solution
I am not sure how to attempt this question a little hint would be helpful. But, still I will show my work.
Since the bulbs are immersed in different reservoirs having different temperatures, so the temperature in both the bulbs can be simply found out by the ideal gas equation. Let the bulb immersed in the melting ice reservoir be bulb 1 and the other one be bulb 2.
So, the pressure in the bulb 1 remains unchanged since the bulb was initially at ##0^{\circ}C## and the reservoir in which it is kept is also at the same temperature.
While for bulb 2 it is kept in a reservoir which has a temperature higher than its initial temperature. As the volume of the bulb remains constant so the pressure in bulb 2 is
$$\dfrac{{(P_2)}_{i}}{{(P_2)}_{f}}=\dfrac{{(T_2)}_{i}}{{(T_2)}_{f}}\implies {(P_2)}_{f}\approx 93.2\text{ cm of Hg}$$.
But I don't feel that this is correct (not only due to the fact that the book gives a different answer) as there will be some heat conduction through the rod connecting the two bulbs. But again due to the reservoirs supplying the constant temperature indefinitely(that's what I think reservoirs do) then...I am really confused.
The book tells that the gaseous molecules rearrange themselves in such a manner so that there is mechanical equilibrium. Hence the temperature in the bulbs is not the same throughout but gradually decreases from ##T_2## (temperature of hot reservoir) to ##T_1## (temperature of cold reservoir).
Now, how was I supposed to take the hint that the system was supposed to be in mechanical equilibrium.