Modern Engineering Thermodynamics

698 CHAPTER 17: Thermodynamics of Biological Systems
EXAMPLE 17.1
Using the concentration data provided in Table 17.2, determine the membrane potential in human cells of sodium, potassium,
and chlorine ions at 37.0°C.
Solution
Table 17.2 gives the concentration of sodium ions inside a human cell at 37.0°C as
while the concentration of sodium ions outside the cell is
c Na +
c
= 14:0 osmoles/cm 3
c Na +
o
= 144 osmoles/cm 3
The valence of a sodium ion is 1 kgmole electrons/kgmole Na + . Then, Eq. (17.9) gives the membrane potential of sodium as
E Na + = 26:7mVðkgmole electrons/kgmole Na+ Þ
z Na + kgmole electrons/kgmole Na + ln c !
Na o
+
c Na +
c
= 26:7mVðkgmole electrons/kgmole Na+ Þ
1 kgmole electrons/kgmole Na + ln 144
= 62:2mV
14:0
For potassium, Table 17.2 gives c K +
c
= 140 osmoles/cm 3 and c K þ
o
= 4:1 osmoles/cm 3 : The valence of a potassium ion is also
1 kgmole electrons/kgmole K + , and Eq. (17.9) gives the membrane potential of potassium in a human cell as
E K +
= 26:7mVðkgmole electrons=kgmole K+ Þ
1 kgmole electrons=kgmole K + ln 4:1
140 ¼ 94:3mV
Finally, Table 17.2 gives c Cl
− = 4:00 osmoles/cm 3 and c
c
Cl
− = 107 osmoles/cm 3 . The valence of a chlorine ion is –1 kgmole
o
electrons/kgmole Cl – , and Eq. (17.9) gives the membrane potential of chlorine in a human cell as
E Cl
−
=
26:7mVðkgmole electrons/kgmole Cl − Þ
−1 kgmole electrons/kgmole Cl − ln 107
4:00 = −87:8mV
Exercises
1. Determine the membrane potential in human cells of magnesium ions, Mg 2+ , at 37.0°C. Answer: E Mg 2 + = –40:4mV:
2. Determine the membrane potential in human cells of sulphate ions, SO 4 2– , at 37.0°C. Answer: E SO4
2 − = 9:3mV:
3. Determine the membrane potential in human cells of dihydrogen phosphate ions, H 2 PO 4 – , at 37.0°C.
Answer: E H2PO4 − = 45:5mV:
Actual measured potentials are generally in the range of –70 to − 90 mV and represent the cumulative effect of
all the ion species present. However, Na + ,K + ,andCl – are the primary high-transport ions in most mammal
membranes, and their cell potentials, listed earlier, average out to about the measured value.
Applying the open system energy rate balance equation to a living cell gives
_Q − _W +∑
in
_me −∑
out
_me =
dU
dt
Here, _Q is the irreversible metabolic heat transfer resulting from the life processes within the cell, ∑ in
_me is the food
energy intake, ∑ out
_me is the waste product output, and _W is the total work done on or by the cell. The food taken
into the cell can be generalized as glucose and molecular oxygen, and the waste products can be generalized as
carbon dioxide and water. The total work done on or by the cell is the electrochemical work done in maintaining
the chemical differences across the cell membrane, ðw EC Þ i = ðw EC Þ ic − ðw EC Þ io , and occasional p-V work done in
enlarging the cell plus γ-A surface tension work done in generating new membrane surface area. Then,
w _W = ∑ _m EC μ
i +∑ _m i + γ _A + p _V (17.10)
M i M i
where we have written all terms on a mass rather than a molar basis (using _n i = _m i /M i , where M i is the molecular
mass of species i) and the intensive properties have been assumed to be constant in time. Thus, the time rate of
change of the cell’s total internal energy is
dU
dt
cell
= _Q −∑ _m i
w EC
M
−∑ _m i
i
μ
M
i
cell
− γ _A − p _V +∑
in
_me −∑
out
_me