Question

Osmosis in plants. Consider the problem of how plants might lift water from ground level to their leaves. Assume that there is a semipermeable membrane at the roots, with pure water on the outside, and an ideal solution inside a small cylindrical capillary inside the plant. The solute mole fraction inside the capillary is \(x=0.001\). The radius of the capillary is \(10^{-2} \mathrm{~cm}\). The gravitational potential energy that must be overcome is \(m g h\), where \(m\) is the mass of the solution, and \(g\) is the gravitational acceleration constant, \(980 \mathrm{~cm} \mathrm{~s}^{-2}\). The density of the solution is \(1 \mathrm{~g} \mathrm{~cm}^{-3}\). What is the height of the solution at room temperature? Can osmotic pressure account for raising this water?

Short answer

The height is approximately 0.14 cm. Osmotic pressure alone cannot account for raising water to the leaves.

Step-by-step solution

- Identify given values

List the given quantities in the problem statement: - Mole fraction inside the capillary, \( x = 0.001 \)- Radius of cylindrical capillary, \( r = 10^{-2} \text{ cm} \)- Gravitational acceleration constant, \( g = 980 \text{ cm/s}^2 \)- Density of the solution, \( \rho = 1 \text{ g/cm}^3 \)

- Express osmotic pressure

Use the formula for osmotic pressure of an ideal solution, \[ \text{Osmotic Pressure} (\text{ } \Pi\text{ }) = i CRT \]where - \( \Pi \) is the osmotic pressure,- \( i \) is the van 't Hoff factor, approximately 1 for dilute solutions,- \( C \) is the molar concentration of solute particles, and- \( R \) is the gas constant, \( R = 8.314 \text{ J/mol⋅K} \), and - \( T \) is the temperature in Kelvin. Room temperature is approximately 298K.

- Calculate molar concentration

Since \( x = 0.001 \) represents the mole fraction: \( C = x \times \text{number of moles of solvent per liter} \). The density of the solution tells us that 1 liter of solution contains 1000 grams of water. Thus, \[ C = x \times 1000 \text{ g/L} \times 1 \text{ mol/18.015 \text{ g}} = 0.001 \times \frac{1000}{18.015} \text{ mol/L} \approx 0.0555 \text{ mol/L} \]

- Calculate osmotic pressure

Substitute the values of \( C \), \( R \), and \( T \):\[ \Pi = 1 \times 0.0555 \times 8.314 \text{ J/mol⋅K} \times 298 \text{ K} \approx 137.45 \text{ Pa} \]

- Calculate the mass of the solution

To find the mass:\[ m = \text{density of the solution} \times \text{volume of the solution} \]The volume of the cylindrical capillary is,\[ V = \pi r^2 h \], thus,\[ m = 1 \text{ g/cm}^3 \times \pi (10^{-2} \text{ cm})^2 h \]\[ m = \pi \times 10^{-4} \text{ g/cm} \times h \]

- Equate gravitational potential energy and osmotic pressure

Osmotic pressure provides an upward force equal to the weight of the column of the solution raised:\[ m g h = \Pi V \]. Substitute the expressions for \( m \), \( g \), and \( V \):\[ \pi 10^{-4} g h \times h = \Pi \pi 10^{-4} h \]This simplifies to,\[ 10^{-4} \times 980 h^2 = 137.45 \times 10^{-4} h \]\[ h = \frac{137.45}{980} \text{ cm} \approx 0.14 \text{ cm} \]

- Conclusion

The height calculated, \( h \approx 0.14 \text{ cm} \), is the height the osmotic pressure can raise the water. Given the height a plant needs to lift water to its leaves is far greater, we conclude that osmotic pressure alone cannot account for raising this water.

Key concepts

semipermeable membrane

A semipermeable membrane is crucial in osmosis. It allows only certain molecules, usually small ones like water, to pass while blocking larger molecules or particles.
This selective permeability helps plants absorb water from the soil through their roots.
In our exercise, the membrane at the roots ensures that pure water enters the plant while solutes remain outside.
This creates a concentration gradient, driving water into the plant by osmosis.

osmotic pressure

Osmotic pressure is the driving force behind water movement in osmosis.
It occurs when water moves from an area of lower solute concentration to one of higher solute concentration, creating pressure.
The formula for osmotic pressure is given by:i CRTwhere:

• i
- van 't Hoff factor (approximately 1 for dilute solutions)
• C
- molar concentration of solute particles
• R
- gas constant (8.314 J/mol⋅K)
• T
- temperature in Kelvin

In the case of plants, this pressure helps move water up through the roots and stems.

gravitational potential energy

Gravitational potential energy needs to be considered when lifting water within a plant.
It is the energy required to move an object against Earth's gravitational pull.
The formula to calculate it is:m g hwhere:

• m - mass of the water
• g - gravitational acceleration constant (980 cm/s²)
• h - height to which water is raised

In the exercise, the osmotic pressure must counteract this gravitational energy to lift the water up through the capillary.

mole fraction

The mole fraction in a solution is a way to express the concentration of a component as a ratio of the number of moles of that component to the total number of moles of all components.
The mole fraction formula is:Mole fraction, x = (number of moles of solute) / (total number of moles in solution)In the capillary of the plant, we have a mole fraction of 0.001 for the solute.
This low value indicates a very dilute solution, important for the efficient application of osmotic pressure.

ideal solution

An ideal solution assumes that the solute and solvent molecules interact similarly, making the solution's properties predictable based on the sum of their individual properties.
In an ideal solution, the behaviors like osmotic pressure closely follow theoretical models, simplifying calculations.
In our exercise, the plant solution is assumed to be ideal, allowing us to use simplified formulas for osmotic pressure and other physical properties.
These assumptions help in understanding and solving complex plant physiology problems effortlessly.