物理化学渗透压双语版

Van’t Hoff equation:
P cB RT
Osmometry - determination of molar mass by measurement of osmotic pressure – macromolecules (proteins and polymers)
Chapter 7 : Slide 1
-1
mB M BbB 0.3011180 103 kg dm-1
0.054kg dm-1 54g dm-1
(about5.4％)
•
Question：according to the result of on the topic, could you tell me 1 dm3 in physiological saline injection contains how many grams of NaCl is needed to have the same as the blood of the osmotic pressure.
Semipermeable Membrane
Chapter 7 : Slide 4
There are the application of plasma osmotic pressure in medicine. Plasma osmotic pressure for normal physiological function and physical health has a close relationship. On patient clinical transfusion, the use of certain drugs and preparation involves solution of osmotic pressure problem.Then we will discuss how to calculate the concentration of isotonic solution,likeNaCl injection and glucose injection
RTxB Vm ,A P
xB nB / nA , V nAVm, A As ，
RTnB V P
P cB RT
Osmosis
Height Proportional to Osmotic Pressure P
Solution
Solvent A with chemical potential A*(p)
• 血浆渗透压在医学上有很的应用，血浆 的渗透压对保持人体正常生理功能和身 体健康有着密切关系。临床上对病员进 行输液,某些药物的使用及配制等,都涉及 溶液渗透压问题。接下来我们将讨论如 何计算等渗溶液的浓度，像NaCபைடு நூலகம்注射液 ，葡萄糖注射液。
Blood is macromolecular aqueous solution, the human body blood freezing point of 272.59 K. (1) the body temperature of 37 ℃ when blood osmotic pressure; (2) at the same temperature, 1 dm3 glucose (C6H12O6) in aqueous solution containing how many grams of glucose is needed to have the same as the blood of the osmotic pressure. The freezing point of water is known to reduce coefficient of K kg mol 1 1.86 Tf K f ,bbB As
bB T f Kf 273.15 272.59 mol kg -1 0.30mol kg -1 1.86
Known as c =b，because blood is very dilute aqueous solution, think its density with water Same degree of approximation， c =b，So
Osmosis – for Greek word “push”
Spontaneous passage of a pure solvent into a solution separated from it by a semi-permeable membrane (membrane permeable to the solvent, but not to the solute) Osmotic pressure – P – the pressure that must be applied to the solution to stop the influx of the solvent
P cB RT
nB 0.3 RT [ 8.314 (273.15 37)]Pa 7.73 105 Pa V 1/1000
（2）As bB nB mA 0.30mol kg，when mA=1.0kg（think the density of glucose is ρ＝1kg· 3）， dm bB nB dm1 (mB M B )dm1
Vm ,A dp
RT ln x A
p P p
Vm ,A dp
Besides: -lnXa=-ln(1-Xb)=Xb change p to p+P， when change is very small，the integrator can be treated as constants。 so
Following we will show how to prove :
* A ( p) A ( xA , p P) *
* A ( xA , p P) A ( p P) RT ln xA
** ***
( p P ) ( p)
* A * A
p P p