Why is the water potential ($\Psi_w$) of pure water taken as zero?

Water potential measures the kinetic energy or free energy of water molecules. Pure water at standard temperature and pressure has the maximum possible free energy and concentration of water molecules. By international thermodynamic convention, this reference state is defined as $0\text{ MPa}$. Adding solutes reduces this free energy, making $\Psi_w$ negative.

What is the key functional difference between DPD and Water Potential ($\Psi_w$)?

Diffusion Pressure Deficit ($DPD$) is a classical term defined as the suction power or thirst of a cell, which increases as water deficit increases ($DPD = OP - TP$). Water Potential ($\Psi_w$) is modern thermodynamic terminology. Conceptually, $DPD = -\Psi_w$. While $DPD$ is expressed in positive values, $\Psi_w$ is expressed in negative values.

How does potassium ($K^+$) ion movement control stomatal opening and closing?

During daylight, guard cells actively pump $K^+$ ions into their cytoplasm using ATP. This increases intracellular solute concentration, lowering solute potential ($\Psi_s$) and total water potential ($\Psi_w$). Water rushes into the guard cells via osmosis, making them turgid and bowing the thin outer walls outward, opening the pore.

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Solved High-Yield Sample Questions

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Question #1NEET 2024

Cell A has $\Psi_s = -12\text{ bar}$ and $\Psi_p = 8\text{ bar}$ . Cell B has $\Psi_s = -10\text{ bar}$ and $\Psi_p = 3\text{ bar}$ . What will be the direction of net water movement when Cell A and Cell B are placed in contact?

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate the net water potential $\Psi_w$ for Cell A using $\Psi_w = \Psi_s + \Psi_p$ . Here, $\Psi_{w,A} = -12\text{ bar} + 8\text{ bar} = -4\text{ bar}$ .

Step 2

Calculate net water potential $\Psi_w$ for Cell B: $\Psi_{w,B} = -10\text{ bar} + 3\text{ bar} = -7\text{ bar}$ .

Step 3

Compare values. Water moves from higher (less negative) to lower (more negative) water potential. Since $-4\text{ bar} > -7\text{ bar}$ , water flows from Cell A to Cell B.

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Question #2JEE Main 2023

A plant cell with an initial solute potential $\Psi_s = -0.8\text{ MPa}$ and turgor pressure $\Psi_p = 0.4\text{ MPa}$ is placed in a sucrose solution of water potential $\Psi_w = -0.6\text{ MPa}$ . At dynamic equilibrium, what is the final pressure potential $\Psi_p$ of the plant cell assuming its volume remains constant?

Solution Locked

Step-by-Step Solution & Concept

Step 1

At dynamic equilibrium, the water potential of the plant cell equals the water potential of the surrounding solution, so $\Psi_{w,\text{cell}} = -0.6\text{ MPa}$ .

Step 2

Use the formula $\Psi_{w,\text{cell}} = \Psi_s + \Psi_p$ . Assuming constant cell volume, solute potential remains $\Psi_s = -0.8\text{ MPa}$ .

Step 3

Substitute known values: $-0.6\text{ MPa} = -0.8\text{ MPa} + \Psi_p \implies \Psi_p = -0.6 + 0.8 = 0.2\text{ MPa}$ .

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Water Potential Equation

Solute Potential (Van 't Hoff Relation)

Plasmolysed Cell State

Fully Turgid Cell State

Diffusion Pressure Deficit (DPD) Equivalence

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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