# CHEMISTRY HELP

Discussion in 'Pre-Medical - MD' started by undergrad06, Jun 22, 2002.

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Please I need help with the following questions for my chemistry course. I'll appreciate it if you could respond before Monday (June 24). Thanks a lot.
1. A gas is diffused through a viscuous liquid to prepare the liquid for deep sea flujid breathing studies. The temperature of the system is lowered to simulate the temp. of deep sea dives. Describe the effect the decreased temp will have on the diffusion rate of the gas. If this liquid were exposed to temp approaching absolute zero, how would the diffusion rate be affected?

2. In the above study, the likquid is a mixture designed to simulate aminotic fluid, the gas is a mixture of Nitrogen,Helium, and Oxygen gases, Without increasing the temp, how would you increase the diffusion rate? What is the total pressure of the gas mixture (use an expression to demonstrate total pressure)? CHEMISTRY HELP

3. ### Mansoor Member

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If the temperature were lowered, the rate of diffusion of the gas would decrease (temperature = average kinetic energy, which is directly proportional to to velocity [KE=(1/2)mv^2]). So the atoms of gas travel slower, and diffuse slower. As the temperature nears zero, so does the speed, and rate of diffusion. It can never reach zero though.

Without increasing the temperature, you could increase the rate by lowering the pressure (PV=nRT). So i guess the volume would try to increase, and the molecules would be forced to diffuse faster (i think). The total pressure is equal to the partial pressures, eg: P(oxygen) + P(nitrogen) + P(helium).

4. ### jot

haha - i wish i knew about this place when i had problem sets to do - i would just throw them up here, there are bright pre-meds abound.
-jot

5. ### CD Senior Member

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Diffusion is dependent upon the molecular speed of the molecule. Therefore, anything which speeds up the speed of the molecule speeds up the diffusion rate. Things such as molecular weight, temperature, pressure, etc. all have an impact on the diffusion rate for this reason. Now the question is, HOW do each of these affect the rate? Look at molecular weight...a higher molecular weight decreases the diffusion rate because the molecule has a lower speed at the same kinetic energy (same temp) as a lighter molecule would have. (since K.E.= (one half of)mv^2). Temperature affects it the same way! If you heat molecules their kinetic energy goes up and the mass of the molecule stays the same so the speed (v) goes up! If the speed goes up then the diffusion rate goes up!

I really don't want to answer the whole question for you, so see if you can work it out from here. If you really get stuck I'll check back tonight or tomorrow and give another hint!

6. ### Mr.D insipidus maximus

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1.Assuming ideal conditions, you have two things in your system to consider: the viscous solution and the gas particles. Viscous means that the solution has a lot of intermolecular interactions (van der waals forces and H-bonds in particular)that prevent it from sliding past each other easily as an ideal liquid. Then you have your gas particles which will behave ideally (assuming PV=nRT). Diffusion is the process of random movement between atoms or molecules that allows them to intermix and achieve an overall more random state. The rate of diffusion is primarily governed by the size and temperature of the system. If you allow these two states of matter to mix, by decreasing the temp. you decrease the rate of diffusion in two ways:

1. By decreasing the temp ( Temp = ave. KE = 1/2 mass*velocity squared. ), you decrease the rate of the ave. velocity of the gas particles, thus diffusion is decreased.

2. At the same time, by decreasing the temp you also reduced the velocity of the molecules in the viscous fluid, thus further enhancing the decreased rate of diffusion (since by dec. the rate of the liquid molecules they have a higher tendency to build stronger intermolecular interactions and thus pass each less other easily and frequently). In effect the viscous liquid gets closer and closer to the physical state of a solid or gel- like matrix (assuming the viscous fluid has a lower melting point than the gas, but for all intents and purposes as this limit is approached everything in your system will be a solid anyway.)

7. ### Mr.D insipidus maximus

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Forgot to answer your second question. Since you want to increase the rate of diffusion without increasing the temp, then as a previous poster mentioned, the best way to attain this would be to increase the volume of the solution. Conceptually this makes sense, since the temp remains the same, but by allowing a greater surface area for the gas and liquid particles to interact, we allow the diffusion process to proceed faster until equilibrium is reached. Mathematically it holds since PV=nRT and n,R, and T are constant, then the only way to increase P (since we want to increase the rate of diffusion and recall the net effect of increased temp is to increase the KE of the system, thus increasing the velocity of the gas particles and in effect increasing the pressure of the gas), is by increasing V.

8. ### CD Senior Member

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HOLD UP GUYS (OR GALS)!!!! When you mention PV=nRT you are ONLY talking about a gas. The question deals with a LIQUID AND A GAS. The two are NOT the same. This deals with chemical potential. The potential in a gas is lower than the potential in the liquid, therefore the gas would prefer to remain in the gas (that's why solubility of gases is relatively low in liquids). AS you increase pressure in the system, the gas molecules become closer together and more closely match the potential in the liquid, this allows the gas to diffuse into the liquid at a higher rate (until equilibrium is reached). The solubility of a gas will increase in the liquid at a higher pressure because of this. INCREASEING THE PRESSURE, INCREASES THE DIFFUSION OF THE GAS INTO THE LIQUID PHASE.

9. ### Spiderman [RNA Ladder 2003] Platinum Member

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When you decrease the temperature of the liquid, the diffusion rate actually increases. Think of global warming. With an increase of global temperature, there is less CO2 that is absorbed by water and there is a lot of CO2 that is being released from water.

10. ### CD Senior Member

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Okay, I looked it up in my Physical Chemistry book (not to be confused with general Chemistry) and for an ideal situation, Henry's law holds for the solute and Raoult's law holds for the solvent. Henry's law states that P=Kx where P is the partial pressure, K is henry's law constant, and x is the mole fraction. From x you can get the solubility of the gas. If the solubility increases then the NET diffusion INTO the liquid will increase while if solubility decreases then the NET diffusion OUT of the liquid will increase. The question is if the total pressure is increased what happens to the solubility? IT INCREASES because P=Kx (I know P is partial pressure and the P given is total pressure but the two are proportional). If the solubility increases then the NET diffusion into the liquid increases until equilibrium is reached.

TO CLARIFY:
If the temperature is increased the solubility is decreased and the gas diffuses OUT of the solution (net diffusion). If the temperature is decreased, the solubility is increased and the gas diffuses INTO the solution.

HOWEVER, at low temperatures the RATE of diffusion (in each direction) is lower and at high temperatures the rate of diffusion in each direction is higher. This sounds like a contradiction with what I wrote above, but it's not because here we are talking about the rate that the diffusion takes place and not the net direction of diffusion. In all cases diffusion takes place both into and out of solution (even at equilibrium it's just that diffusion into = diffusion out of solution at equalibrium)

IN SUMMARY For low temperatures: rate of diffusion is lower but NET diffusion into solution is higher. For high temperature: rate of diffusion is higher but NET diffusion out of solution is higher (into solution is lower).

Hope this helps! (p.s. I'm working on a M.S. in Chemistry!)

11. ### Spiderman [RNA Ladder 2003] Platinum Member

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Al right! I was talking about solubility but not about the rates. Now I remeber where it comes from. However, it was just my intuition.

12. ### Trek Grand Uranium Member

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Is this FLY returned? I"ll repeat what i said to him if it's not: DO YOUR OWN DAMN HOMEWORK, LAZY!!!!! --Trek

13. ### CD Senior Member

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To Trek; I understand what you are saying, however, there are times when you just get...."stuck" and need a little help getting a new concept. I worked as a Teaching Assistant for general Chemistry this last year and beleive me I saw plenty of lazy people, and plenty who truly worked hard but just "hit the wall" so to speak. I see nothing wrong with helping explain a concept as long as learning IS taking place, and it isn't a habit.

14. ### Trek Grand Uranium Member

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Perhaps- but it's more likely he wants someone to do his assignment for him (note the deadline) and a seeming lack of desire to understand the concept (note the lack of the question "why?"). All in all, there's a chance you maybe right, CD, but i'm betting it's just another lazy ass. --Trek

15. ### CD Senior Member

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You could be right (or maybe he just didn't want to type anymore than absolutely needed) and if you are it will show up on his tests! It all comes out in the wash!

17. ### Bikini Princess

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make some real friends to do your chem homework for you, don't ask us.

18. ### Jedi In Training Senior Member

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What does this thread have to do with getting into medical school? See Bikini Princess' advice above.

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Trek:
<strong>Is this FLY returned? I"ll repeat what i said to him if it's not: DO YOUR OWN DAMN HOMEWORK, LAZY!!!!! --Trek</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I remember those annoying FLY posts from earlier this year.

19. ### wgu Senior Member

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no no this is actually kinda fun! Obviously we are not giving up the answer b/c none of us may be right!
Here's what I think:
1) If the pressure is decreased as volume is held constant for the gas, I believe you can "push" the gas into the liquid faster and towards a different equilibrium. Think Soda pop. Excess gas is trapped and escape when pressure is lowered in the gas phase of the can. Meanwhile the can volume is fairly constant. In relation to the solubility statement before, hot pop means bad fizz!

2) Increasing surface area contact would increase the rate but not the equilibrium.

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