Question: An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg
ID: 1828368 • Letter: Q
Question
Question:
An open test tube at 293 K is filled at the bottom with 12.1 cm of Hg and 5.6 cm of water is placed above the mercury. Calculate the pressure at the bottom of the test tube if the atmospheric pressure is 756 mm Hg. Use a density of 1355 kg/m^3 for Hg and 998 kg/m^3 for water. Give the answer in the units of dynes/cm^2, psia, and Pa.
The correct answer is supposed to be about 1.175 x 10^6 dyn/cm^2.
my solution:
Converted my atmosphere pressure,Pa to be Pa = 756mm Hg to be 100791.73 Pa
using p as rho
Pa - Pb = pgh(water) - pgh(mecury)
Pb = 100791.73 - (998*9.8*.056) - (1355*9.8*.121)
Detailed explanation please! Thank you!
Explanation / Answer
In the question density of mercury should be 13550 kg/m^3 instead of 1355 kg/m^3.
Atmospheric pressure = 756 mm Hg = 0.756*9.81*13550 Pascals = 100491.67 Pascals
Pressure due to 12.1 cm Hg is 0.121*9.81*13550 Pascals = 16084 Pa
Pressure due to 5.6 cm of water = 0.056*9.81*998 Pascals = 548.26 Pascals
Total pressure at bottom = 100491.67 + 16084 + 548.26 = 117123.9 Pa
Unit conversion: 1 psi = 6894.76 Pa
Thus, Pressure at bottom = 117123.9 / 6894.76 psi = 16.987 psi
0.1 Pa = 1 dyn/cm^2
Thus, pressure at bottom = 117123.9/0.1 dyn/cm^2 = 1171239 dyn/cm^2 = 1.175*10^6 dyn/cm^2
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