Factors and Calculations of Transpiration Rate

Water vapour exits the leaf through tiny pores called stomata (singular: stoma), which are usually found on the underside of the leaf. Each stoma is surrounded by two guard cells. Guard cells have a uniquely uneven structure: a thick inner wall and a thin, flexible outer wall. When the plant has plenty of water, guard cells take in water by osmosis and become turgid (swollen). This causes them to curve outward, opening the stomatal pore. When the plant is short of water, the guard cells become flaccid (limp) and the stomata close to prevent further water loss.

Factors Affecting the Rate of Transpiration

You hang wet clothes outside to dry, knowing they dry fastest on a warm, dry, and windy day. Plants lose water in exactly the same way.

The rate of transpiration is affected by four main environmental factors:

• Temperature (Positive Correlation): Higher temperatures give water molecules more kinetic energy. This means they evaporate faster from the mesophyll cells and diffuse more rapidly out of the stomata. This rapid movement helps maintain a steep concentration gradient. However, at extreme temperatures (e.g., $40^\circ\text{C}+$ depending on the species), stomata will close to prevent wilting, causing the rate to drop sharply.
• Humidity (Inverse Relationship): High humidity means the air outside the leaf already has a high concentration of water vapour. This reduces the concentration gradient between the internal air spaces and the atmosphere, slowing down diffusion. Conversely, dry air creates a steeper concentration gradient, increasing transpiration.
• Air Movement / Wind Speed (Positive Correlation): In still air, water vapour builds up around the leaf, forming a humid layer that reduces the concentration gradient. Wind physically sweeps this humid layer away, maintaining a steep concentration gradient for faster diffusion.
• Light Intensity (Positive Correlation): Stomata open in the light and close in the dark. During the day, they open to allow carbon dioxide ($CO_2$) to diffuse in for photosynthesis. While open, water vapour can easily diffuse out. The rate will increase with light until another limiting factor (like temperature) restricts the process.

Measuring Transpiration with a Potometer

How do you measure an invisible gas leaving a plant? You estimate it by measuring the liquid water going in.

A potometer is a piece of apparatus used to measure the rate of water uptake by a cut plant shoot. We use uptake as an estimate for transpiration, though it is not perfectly identical. A small amount of the uptaken water is actually used by the plant for photosynthesis and to maintain cell turgidity. As water transpires, it pulls more water up the xylem, which draws an air bubble along a capillary tube.

To set up a potometer correctly and avoid breaking the continuous water column:

• Cut the plant stem at a diagonal angle to increase the surface area for water uptake.
• Assemble the apparatus underwater to prevent air bubbles from entering the xylem.
• Seal all joints with petroleum jelly to make the setup airtight.
• Dab the leaves dry, as wet leaves will artificially block stomata and reduce the rate.

Calculating Rates and Interpreting Graphs

To calculate the rate from a potometer experiment, you must measure the bubble movement over a set period.

$$\text{Rate of Transpiration} = \frac{\text{Distance moved by air bubble}}{\text{Time taken}}$$

Worked Example 1: Calculating Rate from Distance

• Step 1: Identify the known values. An air bubble moves $30\text{ mm}$ in $5\text{ minutes}$.

• Step 2: Substitute into the formula. $\text{Rate} = 30 \div 5$.

• Step 3: Calculate and state the units. $\text{Rate} = 6\text{ mm/min}$.

Worked Example 2: Calculating Rate from a Graph

Experimental data is usually plotted on a line graph because time and distance are continuous data. A straight line indicates a linear relationship. The rate of transpiration is equal to the gradient of the line.

$$\text{Gradient} = \frac{\text{Change in } Y (\text{distance})}{\text{Change in } X (\text{time})}$$

• Step 1: Pick two clear points on the line. For example, at $2\text{ mins}$ the bubble is at $5\text{ mm}$, and at $6\text{ mins}$ it is at $25\text{ mm}$.

• Step 2: Find the change in $Y$ and $X$. Change in $Y = 25 - 5 = 20\text{ mm}$. Change in $X = 6 - 2 = 4\text{ mins}$.

• Step 3: Divide to find the rate. $\text{Rate} = 20 \div 4 = 5\text{ mm/min}$.