UNITS OF PRESSURE: bars, pascals, psi, and even more OH MY!
THIS IS NOT SOMETHING YOU WILL BE TESTED ON. BUT IT WILL MAKE THE
NUMBERS YOU CALCULATE MORE MEANINGFUL.
BAR - force of the Earth’s atmosphere against the surface at sea level; 1
kilogram per square centimeter; about 1 atmosphere of pressure; 100 kPa
PASCAL – SI unit of pressure, 1 newton per square meter (N/m2); 1 pascal =
0.00001 bar; megapascal – MPa – 1,000,000 Pa
POUNDS PER SQUARE INCH (psi) – 1 bar = 14.5 psi
For example an average pressure in a basketball is 8 psi = 0.55 bar
an average pressure in a car tire is 35 psi = 2.4 bar
1.
Ψp = 0
In which direction will the water flow?
1.
Ψp = 0
Water moves from beaker to bag.
Beaker has greater water potential: -3.25 bars
Bag has lower water potential: -6.25 bars
2.
Ψp = 0
HIGHER Ψ
LOWER Ψ
The cell is hypotonic or hypertonic to the environment.
The cell will gain water or lose water.
Explain.
2.
Ψp = 0
HIGHER Ψ
LOWER Ψ
The cell is hypertonic to the environment.
The cell will gain water.
The solution within the cell is higher in solutes. Bound water
molecules are not free to move into the surroundings. Water
moves from high to low water potential – from hypotonic to
hypertonic solutions – from high potential energy to low
potential energy
3. Cell at equilibrium.
For the solution:
Calculate Ψs
Calculate Ψ
i=1
T = 22oC
Solution
0.5 M
3. Cell at equilibrium.
Solution
0.5 M
Ψs = -iCRT
= -1 x 0.5 moles/liter x 0.0831 liter bars/mole K x (22oC + 273)K
= -1 x 0.5 x 0.0831 bar x 295
= -12.28 bar
Ψ= Ψs + Ψp
Ψ= -12.28 bar + 0 = -12.28 bar
4. Calculate Ψs for a 2.4 M solution at 24oC.
Ψs = -iCRT
4. Calculate Ψs for a 2.4 M solution at 24oC.
Ψs = -iCRT
= -1 x 2.4 moles/liter x 0.0831 liter bars/mole K x (24oC + 273)K
= -1 x 2.4 x 0.0831 bar x 297
= -59.23 bar
5. Ψs= -iCRT
Ψs for seawater
i = 2 (NaCl forms 2 ions in water)
C = 0.5 M (moles/liter)
R = 0.0831 liter bars/mole oK
T = 273 + -5oC = 268oK
Ψs= -2 x 0.5 moles/liter x 0.0831 liter bars/mole oK x 268oK
= - 22.27 bars
5. Ψs= -iCRT
Ψs for you
i = 2 (NaCl forms 2 ions in water)
C = 0.15 M (moles/liter)
R = 0.0831 liter bars/mole oK
T = 273 + 37oC = 310oK
Ψs= -2 x 0.15 moles/liter x 0.0831 liter bars/mole oK x 310oK
= - 7.73 bars
5. Ψs= -iCRT
Ψs for you = - 7.73 bars
Ψs for seawater = - 22.27 bars
Water moves from you to seawater.
You have greater water potential: -7.73 bars
Sea water has lower water potential: -22.27 bars
Can you discuss this problem using the concept of tonicity?
Do you know which solution has the greatest potential energy?
6. FIRST WE REVIEW TRANSPORT OF
WATER THROUGH THE XYLEM OF THE
PLANT: Ex. a tree
Good Review of properties of water!
a. Use water potential values – explain how
water moves from root xylem to trunk xylem
to leaf air spaces to outside air.
How does change in water potential explain
the upward movement?
Why do the solutes increase as the water
rises?
b. Which areas in the gradient have the highest
and lowest water potential?
c. Why is the water potential of the outside air
low? Would there be a difference between air
in Arizona versus Georgia?
7.
GUARD CELLS
Size of Stomata – holes – is
regulated by Guard Cells
Passage for CO2 O2 and H2O
Need CO2 for photosynthesis but if
Too much H2O escapes the plant is
Doomed!
Potassium ions K+ are used to control turgor of the guard cell –
open and close the stomata
7.
GUARD CELLS
a. Loss of water closes guard
cells. How is K+ related
to closing of stomata?
b. Explain using tonicity terms.
Explain again using water
potential differences.
c. When K+ channels open and K+ diffuses into the cell, why
does water follow? Explain using water potential terms.
7.
Selective Permeability
How does the cell membrane
of the guard cell move
K+?
How does water move
through the membrane?
How does do guard cells
“know” when to open and close?