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

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

b) The gage of the gas is 2.670 kilopascals.

Step-by-step 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\,(kg)/(m^(3))` and
`g = 9.807\,(m)/(s^(2))`, the difference in mercury levels in the manometer is:

`L = (P_(g)-P_(atm))/(\rho\cdot g)`

`L = (104000\,Pa-101330\,Pa)/(\left(13590\,(kg)/(m^(3)) \right)\cdot \left(9.807\,(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.