UNITS OF PRESSURE: bars, pascals, psi, and even

UNITS OF PRESSURE: bars, pascals, psi, and even

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 Earths 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. p = 0p = 0 In which direction will the water flow? 1.

p = 0p = 0 Water moves from beaker to bag. Beaker has greater water potential: -3.25 bars Bag has lower water potential: -6.25 bars 2. p = 0p = 0

HIGHER p = 0 LOWER p = 0 The cell is hypotonic or hypertonic to the environment. The cell will gain water or lose water. Explain.

HIGHER p = 0 2. p = 0p = 0 LOWER p = 0 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 p = 0s Calculate p = 0 i=1 T = 22oC Solution 0.5 M 3. Cell at equilibrium.

Solution 0.5 M p = 0s = -iCRT = -1 x 0.5 moles/liter x 0.0831 liter bars/mole K x (22 oC + 273)K = -1 x 0.5 x 0.0831 bar x 295 = -12.28 bar p = 0= p = 0s + p = 0p p = 0= -12.28 bar + 0 = -12.28 bar

4. Calculate p = 0s for a 2.4 M solution at 24oC. p = 0s = -iCRT 4. Calculate p = 0s for a 2.4 M solution at 24oC. p = 0s = -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. p = 0s= -iCRT p = 0s 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 p = 0s= -2 x 0.5 moles/liter x 0.0831 liter bars/mole oK x 268oK = - 22.27 bars 5. p = 0s= -iCRT p = 0s 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 p = 0s= -2 x 0.15 moles/liter x 0.0831 liter bars/mole oK x 310oK = - 7.73 bars 5.

p = 0s= -iCRT p = 0s for you = - 7.73 bars p = 0s 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?

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