As shown in the figure below, a monometer is attached to a tank of gas in which the pressure is 104.0 kPa. The manometer liquid is mercury, with a density of 13.59 g/cm 3 . If g = 9.81 m/s 2 and the atmospheric pressure is 101.33 kPa, calculate (a) the difference in mercury levels in the manometer, in cm, and (b) the gage pressure of the gas, in kPa.

Answer:

a) The difference in mercury levels in the manometer is 2 centimeters.

b) The gage of the gas is 2.670 kilopascals.  

Explanation:

a) Pressure in gases is absolute. A manometer helps to determine the hydrostatic difference between pressure of the gas ([tex]P_{g}[/tex]) and atmospheric pressure ([tex]P_{atm}[/tex]), both measured in pascals. A kilopascal equals 1000 pascals and 1 meter equals 100 centimeters. That is:

[tex]P_{g}-P_{atm} = \rho \cdot g \cdot L[/tex] (1)

Where:

[tex]\rho[/tex] - Density of mercury, measured in kilograms per cubic meter.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]L[/tex] - Difference in mercury levels, measured in meters.

If we know that [tex]P_{g} = 104000\,Pa[/tex], [tex]P_{atm} = 101330\,Pa[/tex], [tex]\rho = 13590\,\frac{kg}{m^{3}}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the difference in mercury levels in the manometer is:

[tex]L = \frac{P_{g}-P_{atm}}{\rho\cdot g}[/tex]

[tex]L = \frac{104000\,Pa-101330\,Pa}{\left(13590\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]L = 0.020\,m[/tex]

[tex]L = 2\,cm[/tex]

The difference in mercury levels in the manometer is 2 centimeters.

b) The gage pressure is the difference between gas pressure and atmospheric pressure: ([tex]P_{g} = 104000\,Pa[/tex], [tex]P_{atm} = 101330\,Pa[/tex])

[tex]P_{gage} = P_{g}-P_{atm}[/tex] (2)

[tex]P_{gage} = 104000\,Pa-101330\,Pa[/tex]

[tex]P_{gage} = 2670\,Pa[/tex]

[tex]P_{gage} = 2.670\,kPa[/tex]

The gage of the gas is 2.670 kilopascals.

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