[manjui]

Hello, could someone help me with some homework and explain this to me?

"If the atmospheric pressure is 0.960 atm in a manometer, what is the pressure of the enclosed gas if the if the liquid is mercury and has a hight of 10.3 cm?"
I know I have to keep all the units the same...here's what I am doing, please let me know where I am going wrong! 

I convert 10.3 cm to mm (103 mm) and 0.960 atm to mm Hg (729.6 mm Hg) then I add the two together and get 832.6 mm Hg and convert that to atm: 1.095 atm. But I know that's not the right answer!

[ARGOS++]

Dear manjui;

Which side of the manometer shows you the higher mercury level?:
            http://en.wikipedia.org/wiki/Pressure_measurement#Liquid_column 

Good Luck!
                    ARGOS⁺⁺

[manjui]

The right side. I also forgot to mention that it has a closed end.

[ARGOS++]

Dear manjui;

In this case:  Which side is closed and on which side is the unknown pressure applied?

Good Luck!
                    ARGOS⁺⁺

[manjui]

http://session.masteringchemistry.com/problemAsset/1098123/1/BLB10.10.21.jpg

[ARGOS++]

Dear manjui;

The right most picture makes only sense, if the pressure above the mercury on the right side is equal zero (p₀ = 0 mm Hg).

So calculate from the balance:  p_left + p_right = 0.

Good Luck!
                    ARGOS⁺⁺

[manjui]

The right most picture (iii) is the one I need help with. If the atmospheric pressure is 0.960 atm, what is the pressure of the enclosed gas in sample (iii)?

[ARGOS++]

Dear manjui;

Yes I know that!
But what is the "total" pressure on the right side of this manometer (iii)?

Good Luck!
                    ARGOS⁺⁺

[manjui]

I have no idea. All it says is: "if the atmospheric pressure is 0.960 atm, what is the pressure of the enclosed gas of (iii)?"

[ARGOS++]

Dear manjui;

In picture (iii) the surrounding pressure has no effect, as I told you, and according to:
     http://www.chm.davidson.edu/Chemistryapplets/GasLaws/Pressure.html  

So the pressure on the right side is: p_right = p₀  + p_Hg with p₀ = 0 mm.

And as the sample pressure must be equal p_right,  so p_left = ?

Good Luck!
                    ARGOS⁺⁺

[manjui]

I'm sorry but that makes no sense. 

[ARGOS++]

Dear manjui;

NO - that makes very much sense!

It follows from: http://www.chm.davidson.edu/Chemistryapplets/GasLaws/Pressure.html 

              --  that simply: p_Sample = p_Hg!

You may study the given link once more and very carefully!

Good Luck!
                    ARGOS⁺⁺