Converting the Unit from cmHg to Pa
Pressure in unit cmHg can be converted to Pa by using the formulaP = hρg
Example 1:
Find the pressure at point A, B, C, D, D, E and F in the unit of cmHg and Pa. [Density of mercury = 13600 kg/m³]
Answer:
Find the pressure at point A, B, C, D, D, E and F in the unit of cmHg and Pa. [Density of mercury = 13600 kg/m³]
Answer:
Pressure in unit cmHg | Pressure in unit Pa |
---|---|
PA = 0 PB = 17 cmHg PC = 17 + 59 = 76 cmHg PD = 76 + 8 = 84 cmHg PE = 76 cmHg PF = 76 cmHg | PA = 0 PB = hρg = (0.17)(13600)(10) = 23,120 Pa PC = hρg = (0.76)(13600)(10) = 103,360 Pa PD = hρg = (0.84)(13600)(10) = 114,240 Pa PE = hρg = (0.76)(13600)(10) = 103,360 Pa PF = hρg = (0.76)(13600)(10) = 103,360 Pa |
Example 4:
Figure above shows a mercury barometer whereby the atmospheric pressure is 760 mm Hg on a particular day. Determine the pressure at point
a. A,
b. B,
c. C.
[Density of Mercury = 13 600 kg/m³]
Answer:
a.
b.
c.
Figure above shows a mercury barometer whereby the atmospheric pressure is 760 mm Hg on a particular day. Determine the pressure at point
a. A,
b. B,
c. C.
[Density of Mercury = 13 600 kg/m³]
Answer:
a.
PA = 0
b.
PB = 50 cmHg
or
PB = hρg = (0.50)(13600)(10) = 68 000 Pa
c.
PC = 76 cmHg
or
PC = hρg = (0.76)(13600)(10) = 103 360 Pa
Example 6:
If the atmospheric pressure in a housing area is 100 000 Pa, what is the magnitude of the force exerted by the atmospheric gas on a flat horizontal roof of dimensions 5m × 4m?
Answer:
Area of the roof = 5 x 4 = 20 m²
Force acted on the roof
F = PA
F = (100 000)(20)
F = 2,000,000 N
If the atmospheric pressure in a housing area is 100 000 Pa, what is the magnitude of the force exerted by the atmospheric gas on a flat horizontal roof of dimensions 5m × 4m?
Answer:
Area of the roof = 5 x 4 = 20 m²
Force acted on the roof
F = PA
F = (100 000)(20)
F = 2,000,000 N
Example 5:
Figure above shows a simple barometer, with some air trapped in the tube. Given that the atmospheric pressure is equal to 101000 Pa, find the pressure of the trapped gas. [Density of Mercury = 13 600 kg/m³]
Answer:
Pressure of the air = Pair
Atmospheric pressure = Patm
Pair + 55 cmHg = Patm
Pair
= Patm – 55 cmHg
= 101 000 – (0.55)(13 600)(10)
= 101 000 – 74 800
= 26 200 Pa
Figure above shows a simple barometer, with some air trapped in the tube. Given that the atmospheric pressure is equal to 101000 Pa, find the pressure of the trapped gas. [Density of Mercury = 13 600 kg/m³]
Answer:
Pressure of the air = Pair
Atmospheric pressure = Patm
Pair + 55 cmHg = Patm
Pair
= Patm – 55 cmHg
= 101 000 – (0.55)(13 600)(10)
= 101 000 – 74 800
= 26 200 Pa
Example 7:
Figure(a) above shows the vertical height of mercury in a mercury barometer in a laboratory. Figure(b) shows the mercury barometer in water at a depth of 2.0 m.
Find the vertical height (h) of the mercury in the barometer in the water. Given that the pressure at a depth of 10 m from the water surface is 75 cmHg. [Density of water = 1000 kg/m³, Density of mercury = 13 600 kg/m³]
Answer:
Atmospheric pressure,
Patm = 75 cmHg
Pressure caused by the water,
Pwater = 2/10 x 75 = 15cmHg
Pressure in 2m under water
= 75 + 15 = 90 cmHg
Vertical height of the mercury = 90cm
Figure(a) above shows the vertical height of mercury in a mercury barometer in a laboratory. Figure(b) shows the mercury barometer in water at a depth of 2.0 m.
Find the vertical height (h) of the mercury in the barometer in the water. Given that the pressure at a depth of 10 m from the water surface is 75 cmHg. [Density of water = 1000 kg/m³, Density of mercury = 13 600 kg/m³]
Answer:
Atmospheric pressure,
Patm = 75 cmHg
Pressure caused by the water,
Pwater = 2/10 x 75 = 15cmHg
Pressure in 2m under water
= 75 + 15 = 90 cmHg
Vertical height of the mercury = 90cm