# Finding pressure P of air held in between two columns of Mercury

1. Nov 2, 2015

### Janiceleong26

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

2. Relevant equations
P=hρg

3. The attempt at a solution
The length of mercury column ...
(1) on the left side (exposed to atmosphere) would be ( 760 + x/2 )mm
(2) on the right side (exposed to atmosphere) would be ( 760 + (50-y)/2 )mm
(3) on the left (exposed to air at pressure P) would be ( 760 -x/2) mm
(4) on the right (exposed to air at pressure P) would be ( 760- (50-y)/2 ) mm

By equating (1)=(2) or (3)=(4) ,
x=50-y
A and D are out as x and y does not satisfy this equation.

I chose B as (by assuming from the diagram) x is shorter than y, but I do not know how to calculate the Air pressure P.. Any hints? Thanks in advance

2. Nov 2, 2015

### Staff: Mentor

The results you obtained under items 1 and 2 are both correct expressions for the pressure P (although the factor of 2 in the denominator should not be in there. Where did that come from?). The results you obtained under items 3 and 4 make no sense to me.
So,

P = 760 + x

and

P = 760 + (50-y)

So which combination of values is consistent with these two equations?

3. Nov 2, 2015

### Janiceleong26

Oh I see.. B.
Oh I thought that the column (1) exposed to air would increase by x/2, and column (2) exposed to air pressure would decrease by x/2. But I guess it's wrong.
Thanks for the help!