Question

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 1

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 ($P_{g}$) and atmospheric pressure ($P_{atm}$), both measured in pascals. A kilopascal equals 1000 pascals and 1 meter equals 100 centimeters. That is:

$P_{g}-P_{atm} = \rho \cdot g \cdot L$ (1)

Where:

$\rho$ - Density of mercury, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$L$ - Difference in mercury levels, measured in meters.

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

$L = \frac{P_{g}-P_{atm}}{\rho\cdot g}$

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

$L = 0.020\,m$

$L = 2\,cm$

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

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

$P_{gage} = P_{g}-P_{atm}$ (2)

$P_{gage} = 104000\,Pa-101330\,Pa$

$P_{gage} = 2670\,Pa$

$P_{gage} = 2.670\,kPa$

The gage of the gas is 2.670 kilopascals.