Pressure (P) is directly proportional to height difference (h). Therefore …
P ∝ h => P = k∙h => k = P/h
k₁ = k₂ => P₁/h₁ = P₂/h₂ => h₂ = h₁(P₂/P₁) = 32.6mmHg(715.7 Torr/764.7 Torr) = 30.5mmHg height differential after pressure change. So, the change in height differential = 32.6 mm - 20.5 mm = 2.1 mm => Open end Hg height decreases from 136.4 mm to 134.3 mmHg. (None of the answers listed are correct)
Rational => Lowering the pressure reduces pressure on the mercury-gas interface in the closed end thereby allowing weight of mercury column from open end side of manometer to push the closed end up by 2.1 mmHg while decreasing the mercury height by 2.1 mmHg in the open end. New height in closed height is 105.9 mmHg and open end 134.3 mmHg => 105.9 mmHg + 134.3 mmHg = 103.8 mmHg + 136.4 mmHg = 240.2 mmHg which is the sum of heights before and after pressure change.
The atmospheric pressure is 764.7 torr. A gas sample is placed in a flask attached to an open-end mercury manometer. The height of the mercury in the open-ended arm is 136.4 mm Hg, and the height in the arm in contact with the gas in the flask is 103.8 mm Hg
If the gas inside the flask is cooled so
that its pressure is reduced to a value of 715.7 torr, what will
be the height of the mercury in the open-ended arm?
(Hint: The sum of the heights in both arms must remain
constant regardless of the change in pressure.)
(a) 49.0 mm (b) 95.6 mm (c) 144.6 mm (d) 120.1 mm
4 answers
. My solution follows:
First, determine the pressure of the flask initially. That is
Pgas = Po + h (where h is height difference of Hg on open end side)
h = 136.4 - 103.8 = 32.6 mm
Pgas = 764.7 + 32.6 = 797.3 mm Hg in the flask initially.
Reducing the pressure in the flask to 715.7 is LOWER THAN atmospheric pressure; therefore, the Hg column on the flask side will be HIGHER then the right side and
Pgas = Po - h (where h is the height difference of Hg on flask side)
715.7 = 764.7 - h
h = 764.7-715.7= 49 mm
So we divide the 49 mm (24.5 each side) + and - the mid-point of
(136.4 + 103.8) = 240.2 and half of that is 120.1 (that is if Hg level was 120.1 on each side P in flask = Po.
So on flask side we have 120.1 + 24.5 = 144.6 mm.
On the open end side we have 120.1 - 24.5 = 95.6 mm.
95.6 mm is one of the answers AND the column on the left is higher than that on the right which it must be to have Po of 764.7 and Pgas of 715.7.
Finally, taking your numbers, with the open ended side higher than the flask side, does not give a pressure of 715.7 in the flask as follows:
Pgas = Po + h
h = 134.3-105.8 = 28.4
Pgas = 764.7 + 28.4 = 793.1 and although this ls less than 764.7 it isn't the 715.7 mm stated in the problem and it isn't lower than Po either.
First, determine the pressure of the flask initially. That is
Pgas = Po + h (where h is height difference of Hg on open end side)
h = 136.4 - 103.8 = 32.6 mm
Pgas = 764.7 + 32.6 = 797.3 mm Hg in the flask initially.
Reducing the pressure in the flask to 715.7 is LOWER THAN atmospheric pressure; therefore, the Hg column on the flask side will be HIGHER then the right side and
Pgas = Po - h (where h is the height difference of Hg on flask side)
715.7 = 764.7 - h
h = 764.7-715.7= 49 mm
So we divide the 49 mm (24.5 each side) + and - the mid-point of
(136.4 + 103.8) = 240.2 and half of that is 120.1 (that is if Hg level was 120.1 on each side P in flask = Po.
So on flask side we have 120.1 + 24.5 = 144.6 mm.
On the open end side we have 120.1 - 24.5 = 95.6 mm.
95.6 mm is one of the answers AND the column on the left is higher than that on the right which it must be to have Po of 764.7 and Pgas of 715.7.
Finally, taking your numbers, with the open ended side higher than the flask side, does not give a pressure of 715.7 in the flask as follows:
Pgas = Po + h
h = 134.3-105.8 = 28.4
Pgas = 764.7 + 28.4 = 793.1 and although this ls less than 764.7 it isn't the 715.7 mm stated in the problem and it isn't lower than Po either.
I agree with DrBob.
I think you're the same person, bobpursley.
1 answer