1.If the atmosphere can support a column of mercury 762 mm high at sea level, what height of water could be supported? The density of water is approximately 114 that of mercury.Answer in units of mm
2.If the atmosphere can support a column of mercury 762 mm high at sea level, what height of a hypothetical liquid with a density 1.40 times that of mercury could be supported?
Answer in units of mm
density x gravity x height = pressure
g/cc.......9.8.......m.......kPa
#1.
1.0 x 9.8 x h = (762/760)*101.325
2. You substitute.
To answer both questions, we need to apply the principle of hydrostatic pressure. The equation for hydrostatic pressure is:
P = ρgh
Where:
P is the pressure exerted by the column of liquid,
ρ is the density of the liquid,
g is the acceleration due to gravity (approximately 9.8 m/s²),
h is the height of the column of liquid.
Now let's solve each question step by step.
1. If the atmosphere can support a column of mercury 762 mm high at sea level, what height of water could be supported?
First, we need to find the pressure exerted by the mercury column:
P_mercury = ρ_mercury * g * h_mercury
Given that the density of mercury is approximately 114 times that of water, we can write:
ρ_mercury = 114 * ρ_water
The pressure exerted by the water column can be expressed as:
P_water = ρ_water * g * h_water
To determine the height of the water column, we need to equate the pressures of the mercury and water columns:
P_mercury = P_water
Substituting the values, we get:
114 * ρ_water * g * h_mercury = ρ_water * g * h_water
The density of water cancels out, simplifying the equation to:
114 * h_mercury = h_water
To find the height of the water column, we multiply the height of the mercury column by 114:
h_water = 114 * 762 mm
h_water ≈ 86968 mm
So, the height of water that could be supported is approximately 86968 mm.
2. If the atmosphere can support a column of mercury 762 mm high at sea level, what height of a hypothetical liquid with a density 1.40 times that of mercury could be supported?
Following the same method as in the previous question, the equation becomes:
1.40 * ρ_mercury * g * h_hypothetical = ρ_mercury * g * h_mercury
The density of mercury cancels out, simplifying the equation to:
1.40 * h_hypothetical = h_mercury
To find the height of the hypothetical liquid column, we divide the height of the mercury column by 1.40:
h_hypothetical = 762 mm / 1.40
h_hypothetical ≈ 544.29 mm
So, the height of a hypothetical liquid with a density 1.40 times that of mercury that could be supported is approximately 544.29 mm.