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AKJ1

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1. Homework Statement

1. Homework Statement

A U-tube of uniform cross section contains mercury. The initial temperature of the system is 301K and the barometric pressure is 749 torr. Suppose the left side of the tube is closed at the top and the temperature is raised to a temperature such that the air column on the left is 60cm high. Determine the temperature at which this occurs.

Initially, the mercury level is 50 cm on both sides. The height of each tube, as measured from the bottom, is 100 cm.

## Homework Equations

Ideal gas law PV=nRT

Pascals principle P = P

_{0}+ρgh

## The Attempt at a Solution

I know what my initial pressure and temperature is. In addition I can write my initial volume on the left tube as A(.50) .

My goal is to find the temperature when the the column of mercury on the left tube drops by 10cm (air column now 60cm).

So now I want to develop an expression to describe the final state of the left side of the tube.

Im assuming the number of moles will not change & R is a constant of course.

P

_{li}V

_{li}/(T

_{i}) = P

_{lf}V

_{lf}/(T

_{f})

So now I have one equation and two unknowns ( I don't know the final pressure or temperature, the volume can be written in terms of AH). So I need to develop another expression. Here is where I can take full advantage of pascals principle.

When the mercury goes down on the left side & goes up on the right side, the pressure on the left side of the tube, at the "surface" of the mercury, must be equal to the pressure on the right tube at the same point.

The pressure on the right side of the tube can be written as

P = P

_{air}+ρ

_{hg}gh

Where, h= 0.20m . I choose 0.20m because when the left side goes down by .10m the right side goes up by .10m. Thus if I want to look at the pressure on the right side, I must acknowledge that the point is 0.20m below the surface of the liquid.

So now I can say P

_{lf}= P = P

_{air}+ρ

_{hg}gh

∴ I am now ready to solve for T

_{f}.

Issue is, when I went to solve for T

_{f}, I got an answer that did not correspond to one of the solutions.

here are the values I used

ρ

_{hg}= 13500 kg/m^3

h = 0.20m

P

_{air}= 101325 pa

P

_{li}= 749 torr = 99858 pa

V

_{li[/SUB = A(0.50m) Vlf[/SUB = A(0.40m) And of course the initial temperature is 301 K.}