Lesson 3: How Plants Move Water through Water Potential, Transport and Vapour Pressure Deficit (VPD)
How does water reach the top of a 30 m coconut palm with no pump and no heart? This lesson follows a single water molecule from the soil, through the plant, and out into the air. Along the way, we build the one idea that explains it all: water potential.
Part A: Why water is special
Small molecule, strong team. In a water molecule, oxygen pulls the shared electrons closer, so the molecule has a slightly negative end (oxygen) and two slightly positive ends (the hydrogens). Neighbouring molecules attract each other through hydrogen bonds. One hydrogen bond is weak, about 20 kJ/mol, compared with about 463 kJ/mol for the covalent O–H bond inside the molecule. Dividing 463 by 20 gives about 23, so a hydrogen bond has about one twenty-third of the strength. But billions of them together give water its unusual properties. That’s why water boils at 100 °C, while hydrogen sulphide, a similar small molecule, boils at −60 °C.
One molecule, many jobs. Growing plant tissue is typically 80–95% water. Because of hydrogen bonding, water:
dissolves mineral ions and sugars, making it the transport medium in xylem and phloem;
holds together under tension (cohesion), keeping the xylem water column continuous;
clings to cellulose (adhesion), keeping cell walls wet;
resists temperature change (high specific heat);
takes a lot of energy to evaporate (high latent heat), which cools leaves;
resists compression, so turgor pressure can support soft tissues.
Specific heat. Warming 1 kg of liquid water by 1 °C takes 1,000 g × 4.18 J/(g·K) × 1 K = 4,180 J. The same 4,180 J given to 1 kg of dry mineral soil warms it by 4,180 ÷ (1,000 × 0.8) ≈ 5.2 °C, about five times more. Water-rich fruits, stems, tubers and flooded paddies therefore change temperature slowly. Thin leaves hold little water per unit area, so they rely mainly on evaporation to stay cool.
Latent heat. At a leaf temperature of 25 °C, evaporating water takes about 44 kJ/mol (about 2,442 kJ/kg). Evaporating 1 g removes 2,442 J, while warming 1 g by 1 °C needs only 4.18 J. Dividing 2,442 by 4.18 gives about 584, so evaporating one gram removes as much heat as warming 584 g of water by 1 °C. A well-watered leaf transpiring 5 mmol of water per square metre per second loses 0.005 × 44,000 = 220 W/m² as latent heat. When stomata close in strong sun, that cooling is lost and the leaf can become several degrees warmer than the air.
The water cost of carbon. An open stoma lets carbon dioxide (CO₂) in, but water vapour escapes at the same time. Textbook benchmarks (Taiz et al., 2015) are about 400 mol of water lost per mol of CO₂ fixed for C₃ crops such as rice, oil palm and soybean; about 150 for C₄ crops such as maize and sugarcane; and about 50 for crassulacean acid metabolism (CAM) plants such as pineapple. These are benchmarks, not fixed constants, because the ratio changes with air dryness, species and conditions.
Cohesion, adhesion and surface tension: the pull starts in the leaf. Capillary rise alone lifts water only about 0.73 m in a xylem vessel 20 µm in radius. The real lifting force comes from the leaf. Water evaporates from tiny pores in the mesophyll cell walls, the meniscus curves inward, and the resulting tension is passed down through the connected water to the roots. A 10 nm pore can in principle sustain about 14 MPa of tension, far more than the roughly 0.5–2 MPa a leaf usually needs.
Cohesion–tension: soil to air. Example values for a well-watered crop at midday, 25 °C:
Moist soil: −0.03 MPa
Root: −0.2 MPa
Stem xylem: −0.6 MPa
Leaf xylem: −1.5 MPa
Air at 25 °C and 50% relative humidity: about −95 MPa
Water moves toward lower water potential, and the largest drop is from leaf to air. Gravity also has a cost. Lifting water takes 1,000 × 9.81 = 9,810 Pa per metre, which is about 0.01 MPa per metre, so a 30 m palm needs about 0.29 MPa just to hold its water column up.
Part B: Water potential
What is water potential (Ψ)? Water potential measures the free energy of water relative to pure water at atmospheric pressure, at the same temperature and at a stated reference height. That reference is set at zero, much as sea level is the zero for height. Ψ is expressed per unit volume, which is why its unit is pressure: J/m³ is the same as N/m², which is the pascal (Pa). The key rule is that net passive water movement goes from higher Ψ to lower Ψ.
Components of water potential. Ψw = Ψs + Ψp + Ψg
Ψs (solute potential): zero in pure water, negative when solutes are present.
Ψp (pressure potential): positive under turgor and negative under tension, as in the xylem.
Ψg (gravitational potential): about +0.01 MPa for every metre above the reference height.
When comparing water at the same height, gravity cancels, so Ψw = Ψs + Ψp. For example, a turgid leaf cell with Ψs = −0.8 MPa and Ψp = +0.5 MPa has Ψw = −0.8 + 0.5 = −0.3 MPa.
Units. 1 MPa = 1,000 kPa = 10 bar. To convert MPa to bar, multiply by 10, so −1.5 MPa = −15 bar.
Calculating solute potential: Ψs = −iCRT. Worked example for 0.10 M sucrose at 20 °C:
Temperature in kelvin: T = 20 + 273.15 = 293.15 K.
Sucrose doesn’t split into ions, so the van ‘t Hoff factor i = 1, and i × C = 1 × 0.10 = 0.10 mol/L.
Multiply by the gas constant R = 0.008314 L·MPa/(mol·K): 0.10 × 0.008314 = 0.0008314.
Multiply by T: 0.0008314 × 293.15 = 0.244.
Add the minus sign: Ψs ≈ −0.244 MPa.
The most common mistake is using °C instead of kelvin, which gives an answer about 15 times too small.
The U-tube. Put pure water and a sucrose solution on either side of a membrane that passes water only. Water moves into the solution. If a piston resists it, pressure builds until it exactly cancels the solute effect, for example −0.219 + 0.219 = 0 MPa. At that point both sides have equal Ψ and there is no net movement. In a plant cell, the cell wall plays the part of the piston, which is how turgor builds.
Plant cells in different solutions. Start with the same leaf cell (Ψw = −0.3 MPa):
In pure water (0 MPa): water enters until the cell is fully turgid.
In about 0.33 M sucrose (−0.8 MPa): water leaves until the cell is flaccid.
In about 0.62 M sucrose (−1.5 MPa): water leaves until the protoplast pulls away from the wall. This is plasmolysis.
The surprise is that a solution “isotonic” to the cell sap still draws water out of a turgid cell, because turgor raises the cell’s total Ψ. Always compare total water potential, not solutes alone.
Part C: Water transport
Soil water and matric potential (Ψm). Soil holds water in small pores and as films on particle surfaces, which lowers its potential. Field capacity is commonly around −0.01 to −0.033 MPa, and −1.5 MPa is the conventional permanent wilting point. Water content is not water potential: at 15% water by volume, sand can still supply water to a plant, while clay cannot. Plants respond to Ψ, not to how much water is present.
How water crosses the root. Water moves through three routes: the apoplast (cell walls and spaces between cells), the symplast (cell to cell through plasmodesmata) and the transmembrane route. At the endodermis, the Casparian strip blocks the apoplast, so water must cross a membrane, which lets the plant control what enters the xylem. Aquaporins, the water channel proteins, make membranes more permeable to water. This also explains why crops can wilt in waterlogged soil: low oxygen acidifies root cells, aquaporins close, and water uptake falls even though the roots are surrounded by water.
Measuring plant water potential. In the Scholander pressure chamber, a cut leaf is sealed inside and pressurised until sap just appears at the cut surface. A gauge reading of +15 bar is 15 ÷ 10 = 1.5 MPa, so the estimated leaf Ψ ≈ −1.5 MPa. A predawn reading approximates the soil water potential around the roots, while a midday reading shows daytime water stress.
Water links xylem and phloem. At the source leaf, sugar is loaded into the phloem: −1.7 + 0.8 = −0.9 MPa. The xylem at −0.8 MPa is higher, so water enters the phloem by osmosis. The pressure difference, 0.8 − 0.4 = 0.4 MPa, pushes sap toward the sink, such as a rice grain or a durian fruit. Notice that along the sieve tube, sap flows from Ψw −0.9 to −0.3 MPa. Within a tube, bulk flow follows pressure; the “high Ψ to low Ψ” rule applies when water crosses a membrane.
Part D: The atmosphere end
Humidity: three definitions.
Relative humidity (RH): actual vapour pressure ÷ saturation vapour pressure at the same temperature, × 100%.
Absolute humidity: mass of water vapour per volume of air.
Dew point: the temperature at which air becomes saturated if cooled without adding or removing water.
RH depends on temperature. With the same vapour pressure of 2.50 kPa all day, RH is 2.50 ÷ 2.81 ≈ 89% at 7 am (23 °C) but 2.50 ÷ 5.03 ≈ 50% at 1 pm (33 °C).
How negative is the water potential of air? Ψair = (RT ÷ V̄w) × ln(RH), where V̄w is the volume of one mole of liquid water. At 25 °C and 50% RH:
RT = 8.314 × 298.15 ≈ 2,479 J/mol.
2,479 ÷ 0.0000181 ≈ 137 MPa.
ln(0.50) ≈ −0.693.
137 × −0.693 ≈ −95 MPa.
Even air at 90% RH is about −14 MPa, far drier than a leaf at −1.5 MPa.
Vapour pressure deficit (VPD). Leaf-to-air VPD is the saturation vapour pressure at leaf temperature minus the actual vapour pressure of the air. RH alone doesn’t tell you the drying power of air. At 50% RH, VPD is 1.17 kPa at 20 °C but 2.81 kPa at 35 °C, which is 2.81 ÷ 1.17 ≈ 2.4 times the evaporative demand. Higher VPD increases evaporative demand; stomatal closure can limit the resulting water loss.
Calculating leaf-to-air VPD. Using es(T) = 0.611 × exp[17.502 × T ÷ (T + 240.97)] (Buck, 1981; Campbell and Norman, 1998), with air at 25 °C, a leaf at 23 °C and RH of 50%:
Air saturation vapour pressure = 3.166 kPa, so the air’s actual vapour pressure = 3.166 × 0.50 = 1.583 kPa.
Leaf saturation vapour pressure = 2.808 kPa.
VPD = 2.808 − 1.583 ≈ 1.23 kPa.
RH belongs to the air, so multiply 0.50 by the saturation pressure at air temperature, not leaf temperature. Assuming leaf temperature equals air temperature would overestimate VPD in this example.
Using VPD in practice. Standard VPD charts assume the leaf is at air temperature, so a cooler transpiring leaf has a lower real VPD and a warmer leaf a higher one. Recent research (Wong et al., 2022, Nature Plants) also shows that leaf air spaces near the stomata can become unsaturated. That qualifies the standard assumption behind VPD calculations and gas-exchange measurements.
The whole picture
Molecule: hydrogen bonds give water cohesion, adhesion and surface tension.
Gradient: water moves from higher to lower water potential, from soil to air.
Pull: evaporation at leaf cell walls creates tension that lifts the whole column.
Demand: VPD sets how hard the air pulls, and stomata control the loss.
The drop from soil to leaf is only about 1.5 MPa, while the drop from leaf to air is about 93.5 MPa. That is why stomata are the plant’s main control point.
What a farmer can change: irrigation and mulching keep soil water potential high; shading, windbreaks and greenhouse misting lower the evaporative demand; and choosing crops suited to the climate, including C₄ and CAM crops, reduces the water cost of each unit of carbon fixed.
References: Taiz et al. (2015), Plant Physiology and Development, 6th ed.; Nobel (2009), Physicochemical and Environmental Plant Physiology, 4th ed.; Buck (1981); Campbell and Norman (1998), An Introduction to Environmental Biophysics, 2nd ed.; Wong et al. (2022), Nature Plants 8: 971–978; Tournaire-Roux et al. (2003), Nature 425: 393–397.
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