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**1. Homework Statement**

A cylinder of height H = 1.2 m high rests on a table. The bottom half of the cylinder is filled with an ideal gas; on top of that is a frictionless, movable disk of negligible mass and thickness. Above the disk, the top half of the cylinder is completely filled with liquid mercury, and the top is open to the air. Initially, the gas is at temperature [itex]T_{1}[/itex] = 319 K; air pressure is [itex]P_{0}[/itex] = 1.013 x 10^5 Pa, and the density of mercury is [itex]\rho_{Hg}[/itex] = 13600 kg/m3. Now the temperature of the gas is raised until one-half of the mercury spills out of the cylinder. Assume you can ignore the expansion of the mercury and the cylinder due to change in temperature, and evaporation of mercury is negligible. Find [itex]T_{2}[/itex], the new temperature of the system.

**2. Homework Equations**

→ Ideal Gas Law is the start.

→ [itex]P_{1} = P_{0} + \rho hg[/itex]

**3. The Attempt at a Solution**

I answered the question incorrectly by following this method..

[itex](P_{0} + \rho_{Hg}gh/2) * V_{1}/T_{1} = (P_{0} + \rho_{Hg}gh * 3/4) * V_{2} /T_{2}[/itex]

[itex](P_{0} + \rho_{Hg}gh/2) * A * h/2 /T_{1} = (P_{0} + \rho_{Hg}gh * 3/4) * A * 3h/4 /T_{2}[/itex]

I only considered the volume of the mercury on the top of the disk. I thought that the cylinder is filled with half mercury and half gas, so I included h/2. That is the height of the mercury in the cylinder. Then, since the temperature is raised until more than a half of the mercury is spilled out of the cylinder, I believe I include 3h/4. I solved for [itex]T_{2}[/itex], but I have the wrong answer.

Don't know what goes wrong here.