Download Week 11 - Guelph Physics

11/1
Week 11
Topics:
• More on diffusion (Study guide 17. Section 13.8.)
• Osmosis (Study guide 17. Sections 13.9 and 13.10.)
• Heat capacity (Study guide 18. Section 14.2.)
• Phase transformations (Study guide 18. Section 14.3)
Next deadlines:
Friday Nov. 14 — Quiz #9 (14–15); requires lab 14
Thursday Nov. 27 — Quiz #10 (16–17); requires lab 17 (on pretest
page in Blackboard)
Final exam:
Thursday December 04, 11:30am to 1:30pm, TBA
11/2
More on diffusion (Sec. 13.8)
We have seen that thermal agitation causes a typical molecule to
undergo a random walk, so that after a time t it has diffused by a
typical distance
p
√
2
Rrms = R = 6Dt
For a spherical molecule, the diffusion constant is given by
D=
kB T
6πηr
where r is the molecule’s radius, and η is the viscosity of the fluid
in which the molecule is moving.
11/3
How can D be determined?
One way is to let the molecules diffuse through a porous membrane,
and to measure the concentration of molecules as time goes by.
(The concentration C is the number of moles per unit volume.)
Let a container be separated in half by a porous membrane of
thickness ∆x and area A.
C2(0) = 0
V
V
C1(0)
∆x
Initially, side 1 of the container contains all the molecules, and side
2 is empty; the initial concentration C1 (0) of side 1 is high, and
C2 (0) = 0.
11/4
As time goes by, C1 (t) decreases and C2 (t) increases, until
equilibrium is reached and the concentrations becomes equal.
C1
C2
t
The number of moles of molecules going through the membrane per
unit time is given by
Flux (J) across membrane =
C1 − C 2
1 ∆n
=D
A ∆t
∆x
(Fick’s law)
11/5
As a consequence of this equation, the time evolution of C1 is given
by
¸¾
½
·
1
2DA
t
C1 (t) = C1 (0) 1 + exp −
2
V ∆x
As t → ∞, C1 → 12 C1 (0) and C2 → 12 C1 (0); we have equal
concentrations.
Because A, ∆x, and V are all known, D can be determined by a
measurement of C1 as a function of time.
But molecules generally don’t actually diffuse across the whole
membrane area A, but only through the pores in the membrane,
which have an “effective” area Aef f < A, e.g., Aef f = 0.04A in Lab
17. Then you have to multiply your measured D by the ratio
A/Aef f (= 25 in Lab 17). See also Text p. 13-29 and exercise
13-13.
11/6
What can a measurement of D tell us?
A measurement of D can be converted into a determination of the
molecule’s molar mass M .
The molar mass is the mass of NA = 6.02 × 1023 molecules:
4π
NA ρr3
M = NA (ρV ) =
3
where ρ is the molecule’s density.
But the molecule’s radius r is determined by the diffusion constant:
kB T
D=
6πηr
kB T
r=
6πηD
⇒
So we find
µ
kB T
4π
NA ρ
M=
3
6πηD
¶3
11/7
Molar masses calculated using measured values of D from the last
equation are sometimes too high, due to the presence of so-called
“bound” water or due to non-spherical shape. Both of these effects
reduce the measured value of D.
Example 1:
In a given time, protein A diffuses 3 times as far as protein B.
What is the ratio MA /MB of their molar masses? (Assume that
the proteins are spherical and have the same density.)
11/8
Osmosis (Secs. 13.9 and 13.10)
The term osmosis refers to the diffusion of solvent molecules
across a semi-permeable membrane that allows the solvent to pass
through, but prevents the passage of the larger solute molecules.
Osmotic pressure
Consider the flow of water — the solvent — in the following
experiment:
A
B
pure water
water + solute
11/9
Water flows from A to B because its concentration is initially
smaller in B than in A (a number of water molecules were removed
by the larger solute molecules).
A
B
At equilibrium, the concentrations of water have equalized, but a
pressure difference (measured by the difference in levels) has
developed.
This pressure difference is called the osmotic pressure Π (upper
case pi).
11/10
If the solution is dilute, the osmotic pressure is given by
PB − PA = Π = RT C
where
n(solute)
C=
V
is the solute’s concentration (the number of moles of solute
molecules per m3 of solution), R = 8.315 J/(K · mol) is the ideal
gas constant, and T is the temperature (in K).
These relations give Π in units of pascals.
11/11
In the text, S.G. and quizzes, concentration is sometimes expressed
as m/V = ξ (grams of solute per unit volume). Conversion relation
is C = ξ/M where M is the molar mass (in grams/mole).
(But mole fraction is not used on quizzes.)
Example 2: (Similar to text, Example 13-12)
Suppose that 1.00 g of ovalbumin is dissolved in 1.00 L
(= 1.00 dm3 ) of water on one side of a semi-permeable membrane.
If the experiment is carried out at 20◦ C, by how much does the
level of water rise because of the osmotic pressure? (The molar
mass of ovalbumin is 44.6 kg/mol.)
11/12
Some terminology
The molarity C of a solution is the number of moles of solute
molecules per litre (1 L = 1 dm3 = 10−3 m3 ) of solution. (Notice
this definition of C differs from the previous, in terms of the
volume unit.)
The molality Cos of a solution is the number of moles of solute
molecules per kilogram of solution.
A solute molecule that cannot pass through the semi-permeable
membrane is said to be osmotically active; otherwise it is said to
be osmotically inactive.
(Only the osmotically active molecules contribute to the osmotic
pressure.)
11/13
At low concentrations, the relation between molarity and molality
is C = Cos ρ, where ρ is in units of kg/dm3 , but molality is not
needed on quizzes. For water at 20◦ C, ρ = 1.00 kg/dm3 , so that
molality and molarity are equal. Take this case in all following.
The osmolarity of a solution is the number of moles of osmotically
active solute species per litre of solution.
Depending on the chemistry of the solute molecules, the solution’s
osmolarity may not be equal to its molarity.
An important example is NaCl, which in a solution ionizes into
Na+ and Cl− . The osmolarity of the solution is therefore twice its
molarity.
11/14
Osmosis in biological cells
(For now, we assume the cells are placed in solution of osmotically
active solutes, which cannot diffuse through the cell membrane)
The membrane of a biological cell is semi-permeable, and the cell’s
interior (the lumen) contains osmotically active material.
The lumen is typically 0.30 osmolar.
If the cell is placed in an environment with the same osmolarity
(isosmotic or isotonic), then the concentration of water inside and
outside the cell is the same, and there is no flow of water — this is
good.
11/15
If the cell is placed in an environment with a smaller osmolarity
(hypotonic), then the concentration of water is larger outside the
cell, and water flows in. The cell swells and may burst (or lyse) —
bad.
If the cell is placed in an environment with a larger osmolarity
(hypertonic), then the concentration of water is larger inside the
cell, and water flows out. The cell shrinks — this is also bad.
11/16
Thus, the ideal cell environment has the same osmolarity as the
lumen.
If the cell is placed in a nonideal environment, then its volume will
increase or decrease according to
C1 (cell)V1 = C2 (cell)V2
where C1 (cell) is the initial osmolarity of the cell, C2 (cell) the final
osmolarity of the cell (which at equilibrium equals the osmolarity
Cout of the environment, which doesn’t change), V1 is the cell’s
initial volume, and V2 is its final volume.
The relation above follows because the number of moles n of the
lumen does not change, and Ci (cell) = n/Vi .
11/17
Example 2:
Red blood cells of osmolarity 0.30 are placed in a dilute solution of
0.12 molar NaCl in water. What will happen to the cells? If their
volume changes, what is the percentage change?
11/18
Things are different if cells are placed in a solution containing
osmotically inactive solutes, which can diffuse across the
membrane. In this case, an “isosmotic” solution is not necessarily
“isotonic”. (See text example 13-16 and Self-Test IV (SG 17), #4)
Example 3
A particular type of cell is impermeable to Na+ and Cl− ions, but
permeable to glycerol and water. A solution of 0.15 molar NaCL is
isotonic for this type of cell. Will a solution of 0.30 molar glycerol
be isotonic, hypotonic or hypertonic for this kind of cell?
11/19
Study guide 18
Heat
In this final study guide we examine the flow of heat and its
relation to changes in temperature.
We will look into:
heat capacity, phase transitions, heat conduction,
convection, heat radiation, and Newton’s law of cooling.
11/20
Heat capacity (Sec. 14.2)
Heat is a form of thermal energy that can be transferred from one
body to another (eg. from a burner to a coffee pot).
We denote a quantity of heat by the symbol Q; its unit is the Joule.
The transfer of heat typically comes with a change of temperature:
T increases if a body absorbs heat.
The heat capacity of a body is a measure of how much energy is
required to raise its temperature by 1 degree.
11/21
Since the amount of required heat increases with the mass of the
body, we have
heat
=
(heat capacity)(change in temperature)
=
(mass)(specific heat capacity)(change in temperature)
The specific heat capacity C is the body’s heat capacity per
unit mass.
In symbols,
Q = mC∆T
where m is the mass of the body.
The unit of the specific heat capacity is the J/(kg · K).
11/22
The specific heat capacity depends on the body’s composition, and
it may also depend on temperature (See Table 14-1, given below).
Substance
C (J/(kg · K))
Aluminum
886
Lead
127
Ice
2100
Water (20◦ C)
4186
Ethyl alcohol (20◦ C)
2500
A substance like lead has a low specific heat capacity, and raising
its temperature requires only a small quantity of heat.
A substance like water has a high specific heat capacity, and raising
its temperature requires a large quantity of heat (this is why water
takes a long time to boil).
11/23
Example:
A bucket containing 1.0 kg of water at 90◦ C is placed in contact
with another bucket that contains 2.0 kg of water at 20◦ C. Heat is
observed to flow from the hot bucket to the cold bucket until a
common equilibrium temperature is reached.
What is the equilibrium temperature? How much heat was lost by
the first bucket? How much heat was gained by the second bucket?
11/24
Phase transformations (Sec. 14.3)
Sometimes the absorbtion of heat by a substance does not translate
into an increase in temperature: the temperature might, in fact,
stay constant.
This occurs during a change of phase, for example, when ice turns
into liquid water.
heat added
−20 o
temperature
20o
0o
11/25
Instead of going into an increase of temperature, the absorbed heat
goes into the molecular reorganization that takes place during a
phase transformation.
(For example, ice’s crystalline structure is destroyed and the water
molecules become only loosely bound.)
The heat required to fuel a phase change for 1 kg of a given
substance is called the latent heat L (unit J/kg).
The heat required to fuel a phase change for a mass m of that
substance is therefore
Q(phase transformation) = mL
The latent heat depends on the substance, and on the type of
phase transformation (see Table 14-2 of the textbook).
11/26
Several types of phase transformations can occur:
fusion (or melting): from solid to liquid
vaporization: from liquid to vapour
condensation: from vapour to liquid
solidification: from liquid to solid
sublimation: directly from solid to vapour
Fusion and vaporization require heat, while condensation and
solidification give off heat.
The latent heats for fusion and solidification are equal; the latent
heats for vaporization and condensation are also equal.
11/27
Example:
An ice cube is dropped into a glass of scotch to cool the drink down
to 0◦ C from 20◦ C. Supposing that the glass contains 3.4 × 10−2 kg
of scotch (about 1.5 oz) and the ice cube has an initial mass of
7.5 × 10−3 kg, calculate the final mass of the ice cube after the
drink has cooled down.
(The specific heat capacity of scotch is 2.8 × 103 J/(kg · K) and the
latent heat of fusion of ice is 3.34 × 105 J/kg.)