Respuesta :
Answer : The height of the water will be, 348.16 cm.
Explanation :
As we know that,
[tex]P=\rho \times h\times g[/tex]
where,
P = pressure of the column
[tex]\rho[/tex] = density of a gas or a liquid
h = height of the column of a gas or a liquid
g = acceleration due to gravity (constant)
As per question we conclude that the pressure is same in both the cases. So, the density is inversely proportional to the height of the column.
Thus the formula becomes :
[tex]\frac{\rho_{Hg}}{\rho_{water}}=\frac{h_{water}}{h_{Hg}}[/tex]
where,
[tex]\rho_{Hg}[/tex] = density of Hg (mercury) = [tex]13.6g/cm^3[/tex]
[tex]\rho_{water}[/tex] = density of water = [tex]1.00g/cm^3[/tex]
[tex]h_{Hg}[/tex] = height of Hg (mercury) = 256 mm = 25.6 cm
conversion used : (1 mm = 0.1 cm)
[tex]h_{water}[/tex] = height of water = ?
Now put all the given values in the above formula, we get height of water.
[tex]\frac{13.6g/cm^3}{1.00g/cm^3}=\frac{h_{water}}{25.6cm}[/tex]
[tex]h_{water}=348.16cm[/tex]
Therefore, the height of the water will be, 348.16 cm.
The height of the column of water is [tex]\boxed{{\text{348}}{\text{.16 cm}}}[/tex] .
Further explanation:
The property is a unique feature of the substance that differentiates it from the other substances. It is classified into two types:
1. Intensive properties:
These are the properties that depend on the nature of the substance. These don't depend on the size of the system. Their values remain unaltered even if the system is further divided into a number of subsystems. Temperature, refractive index, concentration, pressure, and density are some of the examples of intensive properties.
2. Extensive properties:
These are the properties that depend on the amount of the substance. These are additive in nature when a single system is divided into many subsystems. Mass, enthalpy, volume, energy, size, weight, and length are some of the examples of extensive properties.
The expression to calculate the pressure of any liquid at depth is as follows:
[tex]${\text{P}}={\text{h\rho g}}$[/tex] …… (1)
Here,
P is the pressure of the liquid column.
h is the height of the column.
[tex]${\text{\rho }}$[/tex] is the density of the liquid.
g is the acceleration due to gravity.
Since the pressure is the same in the given problem, the left-hand side of equation (1) becomes constant, and the equation becomes as follows:
[tex]${\text{h\rho g}}={\text{k}}$[/tex] …… (2)
Here, k is a constant.
Rearrange equation (2) for the density of the liquid.
[tex]${\text{\rho }}=\frac{{\text{k}}}{{{\text{hg}}}}$[/tex] …… (3)
The value of k and g are constants so equation (3) becomes,
[tex]${\text{\rho }}\propto\frac{1}{{\text{h}}}$[/tex] …… (4)
We have two liquids in the given problem, one is water, and the other is mercury. So equation (4) becomes,
[tex]\dfrac{\rho_{\text{Hg}}}{\rho_{\text{water}}}=\dfrac{\text{h}_{\text{water}}}{\text{h}_{\text{Hg}}}[/tex] …… (5)
Here,
[tex]${{\text{\rho }}_{{\text{Hg}}}}$[/tex] is the density of mercury.
[tex]${{\text{\rho }}_{{\text{water}}}}$[/tex] is the density of water.
[tex]{{\text{h}}_{{\text{water}}}}[/tex] is the height of the water column.
[tex]{{\text{h}}_{{\text{Hg}}}}[/tex] is the height of the mercury column.
Rearrange equation (5) for the height of water.
[tex]\text{h}_{\text{water}}=\dfrac{(\rho_{\text{Hg}})(\text{h}_{\text{Hg}})}{\rho_{\text{water}}}[/tex] …… (6)
The height of the mercury column is first converted to cm. The conversion factor for this is,
[tex]{\text{1 mm}}={10^{-1}}\;{\text{cm}}[/tex]
So the height of the mercury column is calculated as follows:
[tex]\begin{aligned}{\text{Height ofmercury column}}&=\left({{\text{256 mm}}}\right)\left({\frac{{{{10}^{-1}}\;{\text{cm}}}}{{{\text{1 mm}}}}}\right)\\&={\text{25}{\text{.6 cm}}\\\end{aligned}[/tex]
The value of [tex]${{\text{\rho }}_{{\text{Hg}}}}$[/tex] is [tex]13.6\;{\text{g/c}}{{\text{m}}^{\text{3}}}[/tex] .
The value of [tex]${{\text{\rho }}_{{\text{water}}}}$[/tex] is [tex]1\;{\text{g/c}}{{\text{m}}^{\text{3}}}[/tex] .
The value of [tex]{{\text{h}}_{{\text{Hg}}}}[/tex] is 25.6 cm.
Substitute these values in equation (6).
[tex]\begin{aligned}{{\text{h}}_{{\text{water}}}}&=\frac{{\left( {{\text{13}}{\text{.6 g/c}}{{\text{m}}^3}}\right)\left({{\text{25}}{\text{.6 cm}}}\right)}}{{\left({{\text{1 g/c}}{{\text{m}}^3}}\right)}}\\&={\mathbf{348}}{\mathbf{.16 cm}}\\\end{aligned}[/tex]
Learn more:
1. Find the final pressure of the system: https://brainly.com/question/6340739
2. Why is density an important property of matter? https://brainly.com/question/1593730
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Gases and the kinetic-molecular theory
Keywords: height, mercury column, water column, height, pressure, acceleration due to gravity, h, g, P, 348.16 cm, conversion factor, k, constant.
