Osmotic pressure

Storyboard

Osmotic pressure is generated in a solution when a semipermeable membrane is present. This membrane allows the solvent to pass through while retaining the solute on one side, creating a pressure imbalance. As a result, there is a reduction in pressure on the solvent side, driving the solvent to move through the membrane toward the side containing the solute.

This process continues until the pressure on the solute side increases enough to balance the initial pressure reduction or until the solute becomes diluted enough that the pressure difference is eliminated, reaching osmotic equilibrium.

>Model

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Mechanisms

Iframe

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Code
Concept

Mechanisms

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Osmotic pressure and U tube

Image

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When a semipermeable membrane is placed at the bottom of a U-shaped tube and water is added, it can be observed that adding dissolved material causes the column with the solute to rise:

This phenomenon is due to the negative pressure generated by osmotic pressure.

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$p_0$
p_0
Atmospheric pressure
Pa
$N_A$
N_A
Avogadro's number
-
$g$
g
Gravitational Acceleration
m/s^2
$\rho_w$
rho_w
Liquid density
kg/m^3
$M$
M
Mass
kg
$M_m$
M_m
Molar Mass
kg/mol
$N_s$
N_s
Number of ions
-
$n$
n
Número de Moles
mol
$\Psi$
Psi
Osmotic pressure
Pa
$R$
R
Universal gas constant
J/mol K
$\Delta p$
Dp
Variación de la Presión
Pa

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$T$
T
Absolute temperature
K
$\Delta h$
Dh
Height of liquid column
m
$h_1$
h_1
Height or depth 1
m
$h_2$
h_2
Height or depth 2
m
$p_1$
p_1
Pressure in column 1
Pa
$p_2$
p_2
Pressure in column 2
Pa
$V$
V
Volume
m^3

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ \Delta h = h_2 - h_1 $

Dh = h_2 - h_1


$ \Delta p = p_2 - p_1 $

Dp = p_2 - p_1


$ \Delta p = \rho_w g \Delta h $

Dp = rho_w * g * Dh


$ n = \displaystyle\frac{ M }{ M_m }$

n = M / M_m


$ n \equiv\displaystyle\frac{ N_s }{ N_A }$

n = N / N_A


$ \Psi =\displaystyle\frac{ N_s }{ V } R T $

Psi = N_s * R * T / V


$ p_1 = p_2 - \Psi $

p_1 = p_2 - Psi


$ p_1 = p_0 + \rho_w g h_1 $

p_t = p_0 + rho_w * g * h


$ p_2 = p_0 + \rho_w g h_2 $

p_t = p_0 + rho_w * g * h

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Behavior of the Solute as Gas Ideal

Equation

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The osmotic pressure ($\Psi$) behaves like the pressure of an ideal gas of the number of ions ($N_s$) in the volume ($V$) at the absolute temperature ($T$), using the universal gas constant ($R$) as described by:

$ \Psi =\displaystyle\frac{ N_s }{ V } R T $

$T$
Absolute temperature
$K$
5177
$N_s$
Number of ions
$-$
9850
$\Psi$
Osmotic pressure
$Pa$
6608
$R$
Universal gas constant
8.4135
$J/mol K$
4957
$V$
Volume
$m^3$
5226

Como la energía molar libre de Gibbs es

$ dg = - s dT + v dp + \mu dN $



se tiene que para el equilibrio entre un sistema con y sin material disuelto (dg=0) e igual temperatura (dT=0) que

$\displaystyle\frac{V}{N_A}dp=\displaystyle\frac{V}{N_A}(p - \Phi)=\mu dN=\mu (N-N_s)$



Como sin material disuelto se debe asumir que el vapor satisface la ecuación de los gases se tiene que

$\mu\sim \displaystyle\frac{R}{N_A} T$



con lo que se obtiene que

$ \Psi =\displaystyle\frac{ N_s }{ V } R T $

ID:(12820, 0)



Osmotic pressure and water column

Equation

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If two columns of water are separated at their base by a semipermeable membrane that allows water to pass through but blocks the solute present in one of them, the columns will exhibit different heights. This is because the presence of a solute reduces the osmotic pressure, leading to an adjustment in the height of the column to balance the pressure difference.

If the pressure in the first column is the pressure in column 1 ($p_1$), the pressure in the second column (without solute) is the pressure in column 2 ($p_2$), and the osmotic pressure is the osmotic pressure ($\Psi$), we can express the relationship as follows:

$ p_1 = p_2 - \Psi $

$\Psi$
Osmotic pressure
$Pa$
6608
$p_1$
Pressure in column 1
$Pa$
6261
$p_2$
Pressure in column 2
$Pa$
6262

ID:(12827, 0)



Pressure difference between columns

Equation

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The height difference, denoted by the height difference ($\Delta h$), implies that the pressure in both columns is distinct. In particular, the pressure difference ($\Delta p$) is a function of the liquid density ($\rho_w$), the gravitational Acceleration ($g$), and the height difference ($\Delta h$), as follows:

$ \Delta p = \rho_w g \Delta h $

$g$
Gravitational Acceleration
9.8
$m/s^2$
5310
$\Delta h$
Height of liquid column
$m$
5819
$\rho_w$
Liquid density
$kg/m^3$
5407
$\Delta p$
Variación de la Presión
$Pa$
6673

If there is the pressure difference ($\Delta p$) between two points, as determined by the equation:

$ \Delta p = p_2 - p_1 $



we can utilize the water column pressure ($p$), which is defined as:

$ p_t = p_0 + \rho_w g h $



This results in:

$\Delta p=p_2-p_1=p_0+\rho_wh_2g-p_0-\rho_wh_1g=\rho_w(h_2-h_1)g$



As the height difference ($\Delta h$) is:

$ \Delta h = h_2 - h_1 $



the pressure difference ($\Delta p$) can be expressed as:

$ \Delta p = \rho_w g \Delta h $

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Height difference

Equation

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When two liquid columns are connected with the height of liquid column 1 ($h_1$) and the height of liquid column 2 ($h_2$), a the height difference ($\Delta h$) is formed, which is calculated as follows:

$ \Delta h = h_2 - h_1 $

$\Delta h$
Height of liquid column
$m$
5819
$h_1$
Height or depth 1
$m$
6259
$h_2$
Height or depth 2
$m$
6260



the height difference ($\Delta h$) will generate the pressure difference that will cause the liquid to flow from the higher column to the lower one.

ID:(4251, 0)



Pressure difference

Equation

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When two liquid columns are connected with the pressure in column 1 ($p_1$) and the pressure in column 2 ($p_2$), a the pressure difference ($\Delta p$) is formed, which is calculated according to the following formula:

$ \Delta p = p_2 - p_1 $

$p_1$
Pressure in column 1
$Pa$
6261
$p_2$
Pressure in column 2
$Pa$
6262
$\Delta p$
Variación de la Presión
$Pa$
6673



the pressure difference ($\Delta p$) represents the pressure difference that will cause the liquid to flow from the taller column to the shorter one.

ID:(4252, 0)



Number of moles

Equation

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The number of moles ($n$) corresponds to the number of particles ($N$) divided by the avogadro's number ($N_A$):

$ n \equiv\displaystyle\frac{ N_s }{ N_A }$

$ n \equiv\displaystyle\frac{ N }{ N_A }$

$N_A$
Avogadro's number
6.02e+23
$-$
9860
$N$
$N_s$
Number of ions
$-$
9850
$n$
Número de Moles
$mol$
6679

ID:(3748, 0)



Number of moles with molar mass

Equation

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The number of moles ($n$) is determined by dividing the mass ($M$) of a substance by its the molar Mass ($M_m$), which corresponds to the weight of one mole of the substance.

Therefore, the following relationship can be established:

$ n = \displaystyle\frac{ M }{ M_m }$

$M$
Mass
$kg$
5183
$M_m$
Molar Mass
$kg/mol$
6212
$n$
Número de Moles
$mol$
6679

The number of moles ($n$) corresponds to the number of particles ($N$) divided by the avogadro's number ($N_A$):

$ n \equiv\displaystyle\frac{ N_s }{ N_A }$



If we multiply both the numerator and the denominator by the particle mass ($m$), we obtain:

$n=\displaystyle\frac{N}{N_A}=\displaystyle\frac{Nm}{N_Am}=\displaystyle\frac{M}{M_m}$



So it is:

$ n = \displaystyle\frac{ M }{ M_m }$

The molar mass is expressed in grams per mole (g/mol).

ID:(4854, 0)



Atmospheric pressure column pressure (1)

Equation

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The water column pressure ($p$) is with the liquid density ($\rho_w$), the column height ($h$), the gravitational Acceleration ($g$) and the atmospheric pressure ($p_0$) equal to:

$ p_1 = p_0 + \rho_w g h_1 $

$ p_t = p_0 + \rho_w g h $

$p_0$
Atmospheric pressure
$Pa$
5817
$h$
$h_1$
Height or depth 1
$m$
6259
$g$
Gravitational Acceleration
9.8
$m/s^2$
5310
$\rho_w$
Liquid density
$kg/m^3$
5407
$p_t$
$p_1$
Pressure in column 1
$Pa$
6261

ID:(4250, 1)



Atmospheric pressure column pressure (2)

Equation

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The water column pressure ($p$) is with the liquid density ($\rho_w$), the column height ($h$), the gravitational Acceleration ($g$) and the atmospheric pressure ($p_0$) equal to:

$ p_2 = p_0 + \rho_w g h_2 $

$ p_t = p_0 + \rho_w g h $

$p_0$
Atmospheric pressure
$Pa$
5817
$h$
$h_2$
Height or depth 2
$m$
6260
$g$
Gravitational Acceleration
9.8
$m/s^2$
5310
$\rho_w$
Liquid density
$kg/m^3$
5407
$p_t$
$p_2$
Pressure in column 2
$Pa$
6262

ID:(4250, 2)