1. The osmotic potential at plasmolysis can be a variety of values, depending on the amount of solutes in the vacuole and cytoplasm. There is no set value. However, the turgor potential is defined as being zero at plasmolysis.

2. Water potential is simply the sum of the osmotic and turgor potentials, so if the turgor potential
= 1.0 MPa and the osmotic potential is -2.5 MPa, then:

                                                              Y_w = 1.0 + (-2.5) = -1.5 MPa

3. If a cell loses water by transpiring, the water volume of the cell will decrease. Osmotic potential is determined by the number of solutes per Kg of water. Thus if the amount of water decreases, this in effect raises the ratio of solutes to water, which in turn, is equivalent to lowering (making more negative) the osmotic potential. As for turgor potential, this will decrease rapidly as the cell loses water, and the force exerted by intracellular water on the cell walls will be eliminated. Eventually, the turgor pressure will drop to zero (plasmolysis).  After turgor loss, water potential is determined solely by the osmotic potential.

You can use V'ant Hoff's relationship to model how osmotic potential declines with water loss, but no such fixed relationship exists for turgor pressure. Instead, you must empirically model it, since it differs from plant to plant, depending on the bulk modulus of elasticity (how stiff the cell walls are).

4. Not to our knowledge in plants.  The reason is that under normal circumstances, a cell will equilibrate with the water potential of the apoplast.  If a plant is rooted in a saturated soil, with a soil water potential of 0 MPa, then, water will move into the plant until the plant's water potential is also zero.  When that happens, the gradient in potential is reduced to zero, and no more water flow occurs.  At that point, the cellular water potential must also equal zero, or else water will continue to flow somewhere.  But if the apoplast has equilibrated to zero, then so must the cell.  If cellular water potential was above zero, water would flow into the apoplast from the cell.  But at equilibrium, the apoplast is already at zero, and there is no mechanism for overhydrating the cell, making such a condition impossible.

5. The formula for water potential is:

                                                                                      Y_w = [RTln(e/e^o)]/V_w

where:    R = gas constant = 0.0831 liter-bar/degree-mole
              T = temperature in Kelvin (20^oC)
              ln(e/e^o) = ln of relative humidity (RH in decimal form, not percent)
              V_w = partial molal volume of water (18 cm³/mole)

Substituting 0.99 for e/e^o one gets ln(0.99)     =     -0.0100503

Thus:
                                                                                    Y_w= (0.0831)(293)(-0.0100503)/18 = -0.0136

The units are:                                                            [liter*bar*K/K*mole]/cm³/mole

After rearranging, one finds that the units do not fully cancel, but instead, result in: liter*bar/cm³
However, there are 10³ cm³ per liter, and the cm³ cancel after converting liters to cm³ and multiplying by 1000. This gives -13.6 bars as the answer.

If the temperature changed by +0.5^oC, there would be a new relative humidity, since the saturation vapor pressure changes with temperature. Now, the new temperature is 20.5^oC, with a saturation vapor pressure of (you have to use extrapolation to get at this one!): 2.412 kPa. If the original relative humidity was 99%, it is now lower, since the temperature has risen. From our data, we can calculate that the actual vapor pressure of the air was (0.99) * 2.338 = 2.362 kPa. The new relative humidity is 2.362/2.412 = 97.9%. Now, substitute the new values (temperature and vapor pressure) into the equation above:
                                                       Y_w= (0.0831)(293.5)ln(0.979)*1000/18 = -28.76 bars

Note that a change of only 0.5^oC was enough to change the answer from -13.6 to -28.8 bars!! A 211% error!!!

   As for the graphing, I leave that to you to do on MS Excel.

6. Osmotic potential, derived from the Van't Hoff relationship, predicts that Y_s is related to the number of moles of solute by the following equation:

                                                                                Y_s = nRT/V

where:              n = # moles of solute
                       R = gas constant (see above)
                       T = temperature in Kelvin
                       V = volume of solvent (water)

Assuming we have a solution with a volume of 1 liter, we want to find out how many solutes it would take to lower the osmotic potential from 0 (pure water) to -1.0 MPa (or -10 bars). Substituting into the above equation and assuming a temperature of 25^oC:

                                                            -10.0 = n(0.0831)(298)/1

Now solve for n:                                         n = -10.0/[(0.0831)(298)] = 0.404 moles

7. Relative humidity is the ratio of the vapor pressure (or density) of water in air to what the air could hold at that temperature, were it saturated with water. Using the table of vapor pressures provided in class, we find that at 25^oC, the saturation vapor pressure is 3.167 kPa. A relative humidity of 50% implies that there is only 1/2 (50%) of the water actually available in the air at that temperature, or a vapor pressure of 3.167*(0.5), for an actual vapor pressure of 1.584 kPa. At 30^oC, the saturation vapor pressure is 4.243. Taking the ratio of vapor pressures, we can calculate the relative humidity of the air were it at 30^oC with the same amount of water as at 25^oC:
 

                                                      %RH = 1.584/4.243 = 0.37*100 = 37% RH

If the actual vapor pressure is 1.584 kPa, the dew point would be 13.86^oC. Use linear extrapolation to find the exact temperature. Even though the relationship is nonlinear, the small error introduced is not important.

8. If you have 2.5 kPa of water vapore pressure at 25^oC, then the relative humidity is this value divided by the saturation vapor pressure at 25^oC:

                                                        RH% = 2.500*100/3.167  =  78.9%

    If the air temperature rises to 30^oC, then the new RH would be:

                                                        RH% = 2.500*100/4.243  =  58.8%

9. The substomatal cavity is thought to be near or at saturation, so if the leaf is at 25^oC, then the substomatal vapor pressure is the saturation vapor pressure at that temperature: 3.167 kPa. If the air is at a temperature of 22^oC, its saturation vapor pressure would be 2.643 kPa. But the air is at only 65% RH, so its actual vapor pressure is: (0.65)(2.643) = 1.718 kPa. The VPD, or vapor pressure deficit, is the difference in vapor pressure between the leaf and the air, or:

                                                      VPD = 3.167 - 1.718 = 1.449 kPa

If leaf temperature rises by 4^oC, then the VP_leaf = 4.006 kPa and the new VPD would be:
 
                                                     VPD = 4.006 - 1.718 = 2.288 kPa.

10. The saturation vapor pressure is 1.705 kPa at 15^oC, and 4.243 kPa at 30^oC.  The air at 15^oC and 80% RH would have a vapor pressure of (1.705)(0.80) = 1.364 kPa, while the warmer air at 50% RH has a vapor pressure of (4.243)(0.50) = 2.122 KPa. Thus, the warmer air has more water per cubic meter by a factor of almost 1.5:1.  To convert water pressure to density, use the relationship PV = nRT and solve for n.  You will need to convert moles to grams, and be sure to get your units correct.  I leave that up to you.

11. For a cube with the given diameter, the volume is l*w*d = (50)³ = 125,000 µm³. Converting to m³:

                              m³= [125,000 µm³/1]*[1 m³/10¹⁸ µm³] = 1.25 x 10^-13 m³

Now simply substitute into V'ant Hoff's relationship (see answer to next question).

12. First, find the volume of the cell. The volume of a sphere is:

                                                                  Vol = 4pir³/3

Therefore, the volume of the cell is: 65449.8 um³

In m³, this comes out to be 65449.8 * 10^-18 m³, or 6.5 * 10^-14 m³

Only 90% of the volume is water in the vacuole, so the volume of the vacuole is

                                                        Vol = (0.90)(6.5)*10^-14 = 5.85 * 10^-14 m³.

Now it is only a matter of using the previously given equation (question 6) to obtain the number of moles of solute required to obtain this osmotic potential:

                                                                          Y_s = nRT/V

In this case, V = 5.85 * 10^-14 m³, R = 0.0831, T = 298 K and Y_s = -2.5 MPa  (-25 bars)
Convert V to liters: 1000 liters/m^3.

Solving for n, one gets:                      n = Y_s*V/RT = [(-25.0)(5.85 *10^-11)]/(0.0831)(298)

                                                      n = 5.98 * 10^-11 moles = What is this in picomoles?