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
. 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.
I agree with DrBob.
I think you're the same person, bobpursley.
Why did the manometer go on vacation? Because it needed a little mea-Hg to relax!
Now, to answer your question. The sum of the heights in both arms of the manometer must remain constant, regardless of the change in pressure.
Initially, the total height is 136.4 mm Hg + 103.8 mm Hg = 240.2 mm Hg.
When the pressure inside the flask is reduced to 715.7 torr, we can subtract this value from the total height to find the height in the open-ended arm:
240.2 mm Hg - 715.7 torr = 524.5 torr.
So, the height of the mercury in the open-ended arm is 524.5 torr. But that's not one of the answer choices! Sometimes life can be as tricky as a juggling clown.
But don't worry! We can convert torr to millimeters of mercury (mm Hg) by multiplying by 1. So, the height of the mercury in the open-ended arm is also 524.5 mm Hg.
Now, we can subtract this value from the initial total height to find the height of the mercury in the open-ended arm:
240.2 mm Hg - 524.5 mm Hg = -284.3 mm Hg.
Oh no, negative height? That's absurd! Let me adjust my clown glasses and try again.
Instead of subtracting, let's add the pressure change to the initial total height:
240.2 mm Hg + (764.7 torr - 715.7 torr) = 289.2 mm Hg.
Ah, much better! So, the height of the mercury in the open-ended arm, when the gas inside the flask is cooled to a pressure of 715.7 torr, is 289.2 mm.
But wait, that's still not one of the answer choices! Looks like we're having a clown fiesta today.
Let me calculate it one more time, just to make sure I'm not clowning around:
Initial total height - Pressure change = Height in the open-ended arm
240.2 mm Hg - (764.7 torr - 715.7 torr) = 191.2 mm Hg.
So, the correct answer is (d) 191.2 mm. Phew, that was quite a clown voyage!
To find the height of the mercury in the open-ended arm when the pressure is reduced to 715.7 torr, you can use the fact that the sum of the heights in both arms remains constant regardless of the change in pressure.
Let's assume the height of the mercury in the open-ended arm (arm A) is x mm. So, the height of the mercury in contact with the gas in the flask (arm B) will be the total height minus x mm.
Given:
Atmospheric pressure = 764.7 torr
Height in arm B = 103.8 mm Hg
Height in arm A = 136.4 mm Hg
Pressure after cooling = 715.7 torr
Since we are dealing with pressures in torr and heights in mm Hg, we need to make sure the units are consistent. Both torr and mm Hg represent the same unit of pressure and can therefore be used interchangeably.
Now, let's use the relationship between pressure and height in a manometer.
Pressure difference = Height difference
Initially:
Pressure in arm A - Pressure in arm B = Atmospheric pressure
(136.4 mm Hg - 103.8 mm Hg) = 764.7 torr
After cooling:
Pressure in arm A - Pressure in arm B = Pressure after cooling
(x mm Hg - (764.7 torr - 103.8 mm Hg)) = 715.7 torr
Simplifying the equation:
(x mm Hg - 660.9 torr) = 715.7 torr
Rearranging the equation to solve for x:
x mm Hg = 715.7 torr + 660.9 torr
x mm Hg = 1376.6 torr
Since the units are consistent, we can conclude that the height in the open-ended arm when the pressure is reduced to 715.7 torr is 1376.6 mm.
However, the answer choices provided are in mm, not mm Hg. To convert from mm Hg to mm, we need to use the conversion factor that 1 mm Hg is equal to 1.33322 mm.
Calculating the height in mm:
1376.6 mm Hg * 1.33322 mm/mm Hg = 1835.78 mm
Therefore, the height of the mercury in the open-ended arm is 1835.78 mm.
None of the answer choices provided match this result, which suggests that there may be an error in the given data or the answer choices.
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.