# Homework Help: Gas Manometer

1. Apr 12, 2012

### samantha.

1. The problem statement, all variables and given/known data

Consider a manometer with a glass bulb (Pgas=88kPa) at one end and which is open to the atmosphere (P =101 kPa) at the other. The manometer has two liquids oil and water in it, as shown. If the oil has a density of 823 kg/m3 and water a density of 1.00x10^3 kg/m3 find the height of the water.

http://s16.postimage.org/9ihj14zir/Screen_shot_2012_04_12_at_3_22_43_PM.png

2. Relevant equations

P=pgh

3. The attempt at a solution

I keep deducing the equation to be;

Patm + ρwater*g*H = Pgas + ρoil*g*0.699m

Solving for H, I get 0.750m. The answer though, is 0.269m.

Where am I going wrong?

Last edited: Apr 12, 2012
2. Apr 12, 2012

### collinsmark

With the values you have (and the figure as it is), the problem is not solvable. No amount of water would ever cause the height of the oil to be 69.9 cm, as depicted in the figure.

I'm guessing the pressure of the gas should be 98 kPa, not 88. It was probably a typo of some sort.

(And just to be nit-picky about this problem, what is the water doing "floating" on oil? Oil is lighter that water. The water would sink to the bottom, and the oil would slurp up replacing the water at the top. Well, unless there's some sort of hidden barrier separating them. Anyway, good luck with it. )

3. Apr 12, 2012

### samantha.

88kPa is the value given. If it was a typo it wasn't on my part. I noticed the water floating on top of oil.. strange yes. Haha.
I spent half of my night panicking over my exam tomorrow, trying to figure out where I was going wrong in all of these problems that my professor gave me when in fact there's an issue with almost every question.

Thanks for the reply though. :)

4. Apr 12, 2012

### collinsmark

By the way, I think your approach to this problem was valid. Try replacing 88 kPa with 98 kPa, and see what happens.

Good luck on your exam tomorrow!

5. Apr 12, 2012

### samantha.

Thank you, Sir. :)