Water Potential and Water Content
where C is the concentration of solute (molkg), $is the osmotic coefficient, v is the number of ions per molecule (e.g., 2 for NaC1,3 for
CaC12, and 1 for sucrose), R is the gas constant (8.3143 Jmol-'KT'), and
T is the kelvin temperature. The osmotic coefficient has a value of one
for an ideal solute, and is generally within ten percent of that value for
solutions encountered in organisms and their environment. More accurate values are available in Robinson and Stokes (1965).
Two examples from nature illustrate the balance of potentials and the
way they sum. In plant cells, concentrations of solutes are quite high.
The cell membrane is permeable to water but not to the solutes, so water
tends to move into the cell. The cell wall prevents volume expansion,
so the pressure inside the cell increases. When the sum of the pressure
and osmotic potential is equal to the water potential in the xylem, water
ceases to move into the cell. If the cell walls of plants were not rigid and
able to withstand high pressures, water would continue to move into the
cell, diluting its contents until life processes would cease.
The other example has to do with blood in the circulatory system of
animals. Solutes are free to diffuse through the walls of the capillary
system, but proteins are too large and are kept in the blood stream. The
negative matric potential of the blood proteins just balances the positive
blood pressure potential. The blood matric potential (referred to as colloid
osmotic pressure in the medical literature), provides just enough "suction"
to keep the blood in the circulation system.
Example 4.1. If the reference for gravitational potential is the water table
at 2 m depth, what is the gravitational potential at the soil surface?
Solution. Using Eq. (4.3),
= 2 m x 9.8 m s - ~ = 19.6 m
2 s-'. From
the example in Ch. 1, we know that this is equivalent to 19.6 Jkg.
Example 4.2. If the osmotic potential of plant sap is equivalent to 0.3
molal KC1, and the total water potential of the tissue is -700 Jkg, what
is the turgor pressure?
Solution. Using Eq. (4.6) to get the osmotic potential, with C =
0.3 mollkg, 4 = 1, and v = 2 gives:
mol
J
I,+o = -0.3- x 1 x 2 x 8.31 -
J
x 293K = -1461 -.
kg
mol K
kg
Now use Eq. (4.2) to obtain the turgor pressure. Assume that all components except the osmotic and pressure components are negligible: +, =
I , + -
= -700 Jkg - (-1461 Jkg) = 761 Jkg. Using Eq. (4.5),
P = 761 Jlkg x 1000 kg/m
3 = 76 1 Wa. One atmosphere is 101 Wa, so
the pressure inside the cell is 7.5 atmospheres. Ifthe plant were fully turgid
where C is the concentration of solute (molkg), $is the osmotic coefficient, v is the number of ions per molecule (e.g., 2 for NaC1,3 for
CaC12, and 1 for sucrose), R is the gas constant (8.3143 Jmol-'KT'), and
T is the kelvin temperature. The osmotic coefficient has a value of one
for an ideal solute, and is generally within ten percent of that value for
solutions encountered in organisms and their environment. More accurate values are available in Robinson and Stokes (1965).
Two examples from nature illustrate the balance of potentials and the
way they sum. In plant cells, concentrations of solutes are quite high.
The cell membrane is permeable to water but not to the solutes, so water
tends to move into the cell. The cell wall prevents volume expansion,
so the pressure inside the cell increases. When the sum of the pressure
and osmotic potential is equal to the water potential in the xylem, water
ceases to move into the cell. If the cell walls of plants were not rigid and
able to withstand high pressures, water would continue to move into the
cell, diluting its contents until life processes would cease.
The other example has to do with blood in the circulatory system of
animals. Solutes are free to diffuse through the walls of the capillary
system, but proteins are too large and are kept in the blood stream. The
negative matric potential of the blood proteins just balances the positive
blood pressure potential. The blood matric potential (referred to as colloid
osmotic pressure in the medical literature), provides just enough "suction"
to keep the blood in the circulation system.
Example 4.1. If the reference for gravitational potential is the water table
at 2 m depth, what is the gravitational potential at the soil surface?
Solution. Using Eq. (4.3),
= 2 m x 9.8 m s - ~ = 19.6 m
2 s-'. From
the example in Ch. 1, we know that this is equivalent to 19.6 Jkg.
Example 4.2. If the osmotic potential of plant sap is equivalent to 0.3
molal KC1, and the total water potential of the tissue is -700 Jkg, what
is the turgor pressure?
Solution. Using Eq. (4.6) to get the osmotic potential, with C =
0.3 mollkg, 4 = 1, and v = 2 gives:
mol
J
I,+o = -0.3- x 1 x 2 x 8.31 -
J
x 293K = -1461 -.
kg
mol K
kg
Now use Eq. (4.2) to obtain the turgor pressure. Assume that all components except the osmotic and pressure components are negligible: +, =
I , + -
= -700 Jkg - (-1461 Jkg) = 761 Jkg. Using Eq. (4.5),
P = 761 Jlkg x 1000 kg/m
3 = 76 1 Wa. One atmosphere is 101 Wa, so
the pressure inside the cell is 7.5 atmospheres. Ifthe plant were fully turgid