Osmotic pressure

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Osmotic pressure is the minimum pressure which needs to be applied to a solution to prevent the inward flow of its pure solvent across a semipermeable membrane.^[1] It is also defined as the measure of the tendency of a solution to take in its pure solvent by osmosis. Potential osmotic pressure is the maximum osmotic pressure that could develop in a solution if it were separated from its pure solvent by a semipermeable membrane

Osmosis occurs when two solutions containing different concentrations of solute are separated by a selectively permeable membrane. Solvent molecules pass preferentially through the membrane from the low-concentration solution to the solution with higher solute concentration. The transfer of solvent molecules will continue until equilibrium is attained.^[1][2]

Theory and measurement

Jacobus van 't Hoff found a quantitative relationship between osmotic pressure and solute concentration, expressed in the following equation:

${\displaystyle \Pi =icRT}$

where ${\displaystyle \Pi }$ is osmotic pressure, i is the dimensionless

in the form ${\textstyle P={\frac {n}{V}}RT=c_{\text{gas}}RT}$ where n is the total number of moles of gas molecules in the volume V, and n/V is the molar concentration of gas molecules.

The Pfeffer cell was developed for the measurement of osmotic pressure.

Applications

Osmotic pressure measurement may be used for the determination of

Osmotic pressure is an important factor affecting biological cells.^[4] Osmoregulation is the homeostasis mechanism of an organism to reach balance in osmotic pressure.

• Hypertonicity is the presence of a solution that causes cells to shrink.
• Hypotonicity is the presence of a solution that causes cells to swell.
• Isotonicity is the presence of a solution that produces no change in cell volume.

When a

Consider the system at the point when it has reached equilibrium. The condition for this is that the chemical potential of the solvent (since only it is free to flow toward equilibrium) on both sides of the membrane is equal. The compartment containing the pure solvent has a chemical potential of ${\displaystyle \mu ^{0}(p)}$, where ${\displaystyle p}$ is the pressure. On the other side, in the compartment containing the solute, the chemical potential of the solvent depends on the mole fraction of the solvent, ${\displaystyle 0<x_{v}<1}$. Besides, this compartment can assume a different pressure, ${\displaystyle p'}$. We can therefore write the chemical potential of the solvent as ${\displaystyle \mu _{v}(x_{v},p')}$. If we write ${\displaystyle p'=p+\Pi }$, the balance of the chemical potential is therefore:

${\displaystyle \mu _{v}^{0}(p)=\mu _{v}(x_{v},p+\Pi ).}$

Here, the difference in pressure of the two compartments ${\displaystyle \Pi \equiv p'-p}$ is defined as the osmotic pressure exerted by the solutes. Holding the pressure, the addition of solute decreases the chemical potential (an entropic effect). Thus, the pressure of the solution has to be increased in an effort to compensate the loss of the chemical potential.

In order to find ${\displaystyle \Pi }$, the osmotic pressure, we consider equilibrium between a solution containing solute and pure water.

${\displaystyle \mu _{v}(x_{v},p+\Pi )=\mu _{v}^{0}(p).}$

We can write the left hand side as:

${\displaystyle \mu _{v}(x_{v},p+\Pi )=\mu _{v}^{0}(p+\Pi )+RT\ln(\gamma _{v}x_{v})}$,

where ${\displaystyle \gamma _{v}}$ is the activity coefficient of the solvent. The product ${\displaystyle \gamma _{v}x_{v}}$ is also known as the activity of the solvent, which for water is the water activity ${\displaystyle a_{w}}$. The addition to the pressure is expressed through the expression for the energy of expansion:

${\displaystyle \mu _{v}^{o}(p+\Pi )=\mu _{v}^{0}(p)+\int _{p}^{p+\Pi }\!V_{m}(p')\,dp',}$

where ${\displaystyle V_{m}}$ is the molar volume (m³/mol). Inserting the expression presented above into the chemical potential equation for the entire system and rearranging will arrive at:

${\displaystyle -RT\ln(\gamma _{v}x_{v})=\int _{p}^{p+\Pi }\!V_{m}(p')\,dp'.}$

If the liquid is incompressible the molar volume is constant, ${\displaystyle V_{m}(p')\equiv V_{m}}$, and the integral becomes ${\displaystyle \Pi V_{m}}$. Thus, we get

${\displaystyle \Pi =-(RT/V_{m})\ln(\gamma _{v}x_{v}).}$

The activity coefficient is a function of concentration and temperature, but in the case of dilute mixtures, it is often very close to 1.0, so

${\displaystyle \Pi =-(RT/V_{m})\ln(x_{v}).}$

The mole fraction of solute, ${\displaystyle x_{s}}$, is ${\displaystyle 1-x_{v}}$, so ${\displaystyle \ln(x_{v})}$ can be replaced with ${\displaystyle \ln(1-x_{s})}$, which, when ${\displaystyle x_{s}}$ is small, can be approximated by ${\displaystyle -x_{s}}$.

${\displaystyle \Pi =(RT/V_{m})x_{s}.}$

The mole fraction ${\displaystyle x_{s}}$is ${\displaystyle n_{s}/(n_{s}+n_{v})}$. When ${\displaystyle x_{s}}$is small, it may be approximated by ${\displaystyle x_{s}=n_{s}/n_{v}}$. Also, the molar volume ${\displaystyle V_{m}}$ may be written as volume per mole, ${\displaystyle V_{m}=V/n_{v}}$. Combining these gives the following.

${\displaystyle \Pi =cRT.}$

For aqueous solutions of salts, ionisation must be taken into account. For example, 1 mole of NaCl ionises to 2 moles of ions.

See also

• Gibbs–Donnan effect

References