# Pressure (Barometer)

[SOLVED] Pressure (Barometer)

## Homework Statement

A mercury barometer of height 1.000m has some ideal gas trapped inside. When the tube is vertical, the height of the mercury column is 0.700m. When the tube is inclined at an angle of 30 degrees to the horizontal, the length of the mercury is 0.950m.
Assuming temperature is constant, the atmospheric pressure, in mm Hg, is ?
(a picture of the question is attached)

Pressure = hpg

## The Attempt at a Solution

I tried working it along this line whereby Atmospheric Pressure = Pressure of the ideal gas(hpg) + Pressure of vertical length of Mercury .
Formed 2 simultaneous equations with the 2 diagrams as attached. But no luck in getting the answer so far. Please tell me whether the direction that im going at is correct.

#### Attachments

• quest.GIF
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learningphysics
Homework Helper
Yes, looks right. You need to take into account that the ideal gas compresses, hence its pressure changes.

From what i have. It should be something like this.

Equation 1(Diagram 1) : Atmospheric pressure = Pressure from Ideal Gas(hpg) + Pressure in the mercury lever.

A x Density of Mercury x Gravity = 0.3 x Density of Gas x Gravity + 0.7 x Density of Mercury x Gravity

For Diagram 2:(heres the part.. Whats the new density of the ideal gas , in this case, now that volume has changed?. and how do i find it?)

A x Density of Mercury x Gravity = (sin30 x 1 - sin30 x 0.950) x Density of Gas x Gravity + sin30 x 0.95 x Density of Mercury x Gravity

An Update
I seem to got the answer.
I let P be the initial Pressure that the ideal gas exert, and the volume that the ideal gas occupy in diagram 1 seem to be approximately 6 times more than that in diagram 2 ( (1-0.7)/(1-0.95) = 6). Using PV=CONSTANT, the pressure that ideal gas exert in diagram shld be 6 times bigger than that in Diagram 1)

So

A x Density of Mercury x Gravity = P + 0.7 x Density of Mercury x Gravity---- (1)
A x Density of Mercury x Gravity = 6P + sin30 x 0.95 x Density of Mercury x Gravity---(2)

Solving simultaneously, i got the answer. ! cheers.

Please let me know if there are other possible solutions that u can think of. Thanks !!

Last edited:
learningphysics
Homework Helper
Do they give the width of the barometer? Seems like we need that...

I got the answer already. as explained in my earlier edited message. Theres no need for the width value as when we calculate the volume ratios, they cancel out I guess.

Thanks a lot buddy!

learningphysics
Homework Helper
I got the answer already. as explained in my earlier edited message. Theres no need for the width value as when we calculate the volume ratios, they cancel out I guess.

Thanks a lot buddy!

Yeah, looks good... it's an approximation it seems (valid for very small width) but that seems to be the only way to answer the question.

you approximated the trapezoidal area of the gas in the second picture as a rectangular area... which is what they wanted I think.

Last edited:
Yeap. Thanks for the enlightenment. It was your reply that gave me the idea of how to do it.

Code:
You need to take into account that the ideal gas compresses, hence its pressure changes.

Thanks buddy . This thread can be closed now.