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This question for Chemistry in the EK Chemistry lecture 2 GAS law

Discussion in 'MCAT: Medical College Admissions Test' started by christian15213, Mar 10, 2007.

  1. christian15213

    christian15213 Membership Revoked
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    WTHeck...

    LOL, ok here is the quesion:

    A force is applied to a container of gas reducing its volume by half. The temperature of the gas:

    A. decreases
    B. increases
    C. remains constant
    D. The temperature change depends upon the amount of force used.

    Ok, I know what the answer is and what the answer says... But, I am not following it. What does K.E. (and I am asuming this is the equation
    KE = 3/2 RT) have to do wit volume? or the equation PV = nRT...


    yea, lol I dunno what even to say I am sure I am missing something small.
     
  2. Funky

    Funky This space is for sale
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    I believe this has to do with the work energy theroem. As you put work into a system, you are converting it into kinetic energy and thus PV work. This is how the cornet engine works.
     
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  3. Funky

    Funky This space is for sale
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    whoops, duplicate
     
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  4. OP
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    christian15213

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    What do you mean thus PV work???? are you saying PV is equivalent to work? I am still confused ... sorry but I need this one in lamens terms. And not your fault but this is supposed to be only dealing with Gas laws and what not...
     
  5. Funky

    Funky This space is for sale
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    In the scope of the ideal gas law, you have to remember that average kinetic energy is proportional to the temperature of the gas. When you put in force into a container, you are increasing the average kinetic energy because you are transferring energy from the surroundings into the gas. As stated, PV=nRT. When the volume decreases, the pressure is increased. Temperature on the other hand would remain the same if you aren't counting the energy transfer but in this case you have to.
     
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  6. bigserve99

    bigserve99 Member
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    Ok, so this question is more involved than it would seem at first. Bear with me and it will make sense. The problem here is the explanation.....we know that PV=nRT tell us that V can change such that T doesn't (change in pressure). However, we have input energy via work (and there was no heat sink or source assoicated with the system). So how did this work change the internal energy of the gas (namely the kinetic energy). EK's explanation is horrible in this respect. Also, I don't believe they give you enough information to truly understand the molecular machinery at work in this problem.

    To answer this, remember:

    1. This is an ideal gas law problem.

    2. It is also an energy problem.

    First thing to remember is that temperature is directly proportional to avg kinetic energy and that avg.KE is = 1/2 (m)(rms-velocty)^2. Mass does not change in this equation(in this problem and most others), hence temperature is ultimately dependent only on the rms-velocity of the gas molecules.

    Secondly, remember that PV=nRT is a very simplified formula and is indeed only empirical. The mathematical underpinnings to this equation are as follows (and I believe enormously helpful when dealing with these problems):

    Pressure is proportional to impulse imparted per collision of gas molecule times the rate of these collisions. The impulse is proportional to momentum of the gas molecule. The rate of collisions is proportional to number of molecules per unit volume and the rms-velocities of these molecules. Basically, the more molecules, the less volume, and the faster they move....the more collisions per unit time (higher collision rate).

    (BEAR WITH ME, IT WILL ALL MAKE SENSE IN A MOMENT)

    therefore, essentially: Pressure is proportional to momentum * rate of collision
    or
    P is proportional to (mass*rms-velocity) * ((number of molecules / volume) * rms-velocity))

    just to bring it full circle, if avg. KE (1/2 * mass * rms-velocity^2) is proportional to TEMPERATURE, then just (mass * rms-velocity^2) is proportional to TEMPERATURE as well.

    Hence,
    P is proportional to (mass*rms-velocity) * ((number of molecules / volume) * rms-velocity))

    becomes,
    P is proportional to (mass*rms-velocity^2) * (number of molecules / volume)

    or when TEMPERATURE is substituted
    P is proportional to (temperature) * (number of molecules / volume)

    and when the molar gas constant is included (# of molecules becomes moles)

    P = nRT/V




    I took the time to explain all this, because I also did not understand at first why the temperature must change, yet knew there was an internal energy change because we have done work on the system. If we were to use the simplified PV=nRT, this would make no sense. Volume could be halved, and pressure doubled without any change in T. However, this is not a pure State Function problem.

    WE have done work on the gas. Work is nothing but F*d. A force is nothing but m*a and m*a is nothing but m*(delta velocity / delta time) which is nothing but (delta momentum / delta time) which is nothing but impulse. Since I assumed this to be an ideal gas, we know that the change in momentum is conserved, and the gas has therefore gained linear momentum. Since mass does not change, the rms-velocities of the gas molecules must have increased, and hence an increase in temperature.



    Please hit me up with questions or concerns.
     
  7. bigserve99

    bigserve99 Member
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    Oh, forgot to add....

    Based on the increase in momentum (and therefore rms velocity and therefore kinetic energy), AND the change in volume (Vf = (1/2) * V0) the new pressure should be MORE than doubled.

    Pf > 2 * P0
     
  8. soonereng

    soonereng Double Trouble
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    What bigserve said but maybe a simpler explanation. You are inputting work into the system but think of it this way:

    1. Pressure is just evidence of the KE of the gas molecules bouncing off the walls of the container which is related to temp as described by bigserve.
    2. Halving the volume increases the pressure.
    3. This new pressure is higher, thus the KE of the gas molecules is higher, which is again related to temperature. Thus there is an increase in temperature.

    The equation PV=nRT only applies at the two end states of the system described, not in the interim because work is done on the system. The equation PV=nRT is not valid for the change in the system because we didn't just let it come to its natural state. If we had halved the volume by decreasing the temperature it would apply.

    I hope that makes it clear.
     
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  9. bigserve99

    bigserve99 Member
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    soonereng,

    I generally agree with what you have stated in your post except for the 3 points you have listed here......must be careful here, they do not hold true in all circumstances as this list would suggest, for example:


    EDITED: The process below is isolated and therefore adiabatic. NOT CLOSED AND ISOTHERMAL. (thanks soonereng!)
    Take a gas that is confined to one half of an otherwise evacuated (vacum) chamber via a removable partition. The gas is defined by the variables Pi, Vi, and Ti (initial). Upon removal of the partition, the gas will expand to fill the full volume of the chamber. Vf = 2*Vi. In this circumstance Pf = (1/2) Pi.

    As no work by the gas during the expansion (expanding against a vacum), the momentum of the gas particles will remain the same, hence the rms-velocities of the molecules will remain the same, and hence the avg. kinetic energy and temperature will remain the same. In this case, the temperature does not correlate with the pressure as you have suggested in your point number 2 - 3.. One cannot simply correlate temperature to pressure in such a circumstance.

    ~ BigServe99
     
  10. soonereng

    soonereng Double Trouble
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    I was describing the system in question, not any general system. Each system is different. This is why sometimes PV=nRT is applicable, and other times it is not. That was the point of my response.
     
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  11. bigserve99

    bigserve99 Member
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    point taken soonereng, my mistake. Just sounded like you were mentioning them like general axioms. Sorry.

    ~ BigServe99
     
  12. soonereng

    soonereng Double Trouble
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    No problem. I do have some comments about your example though.


    I am with you up to here.

    Here is where I disagree with you. For an isothermal process (T=Constant), W=nRTln(Vf/Vi). Since you say that there is no work done (W=0), there is no increase in mass, and the temp is constant, then the volumes must stay the same. However, in this situation, the volume does change so this must not be an isothermal process. The pressure decreases because the temperature of the gas decreases. Again rms is related to T which is related to P.

    P does correlate to T in your system.

    This is how an air conditioning system works. The gas expands against almost no pressure and is cooled.
     
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  13. OP
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    christian15213

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    Wow, thanks you both gave awesome explanations...

    However, I wanted to try and clean it up a little further.

    Tell me if I am thinking something wrong... First I like that you said know what we are working with

    1. Ideal gas problem

    2. It is also an energy problem from the the fact that it says there is work done... OK

    K.E. = 3/2 * RT

    and K.E. = 1/2 MV^2

    V can be equated to (Velocity ~= RMS)

    So, with this part of the equation here i.e. K.E. = 1/2 MV^2 if I increase rms then I will be increasing K.E. which is essentially like saying "work done" or work to the system or in this particular case a force is applied to the system.

    In fact it is a (tell me if I am wrong in associating it this way, you'll see in a minute why I make the argument this way) force such that the volume is decreased by half. <<<****now my point in here is what if it had said a force applied to the system where the volume is increased by 2. would that have lowered the temperature correct?*** and as if to say work alone / increased K.E. doesn't simply mean increased temperature... the more important point should have been that the volume was decreased by half. ****<<<< PLEASE COMMENT so that I have my HEAD STRAIGHT

    So, now that I can can say that there was a force applied and volume is decreased... I should know now that the RMS / velocity for my system is going to increase. And if V increases then K.E. increases which essentially increases temperature...


    WOW.... I think I have it... does that make sense????


    Ok, with all of that said I want to walk away with some things here to make sure I have my Gas bearings straight.

    PV=nRT would have been fine if per se we were not talking about force or anything to do with K.E. Since we are talking about K.E. then factors need to come into play like RMS / velocity, effusion and diffusion. which are all inversely proportional to MASS vs. RMS / velocity. and in the case of effusion rms is proportional to effusion which makes sense.


    Wow, yea I think EK could have done a better job including some things and taking some things away from this section.... I guess it was partially my fault for not inputting more to the fact I darn well know K.E. = 1/2 MV^2... hahahah but this isn't physics yet. LOL JJ.

    lastly, I am sorry but BigServe... could you explain the K.E. = 3/2 RT I see it is proportional to temp but where did they get this equation? I can't even find it in my chem book... I have searched all through the gases but it escapes me. what is it good for or when asked a question like this in the future with a different form... is it smart to assume I need to take into account the original ke = 1/2 mv^2 and the fact v ~= RMS...

    And the whole point is that we are dealing with ideal gas which says the situation and enviorment for gas are very standard. however, when we deviate from that system we are more like real gases and exhibit features of real gas. i.e. lower temp = lower volume and higher pressure = lower volume.

    Ahhhhhh YES **** one more thing. VOLUME... volume in this sense isn't volume of the container? or am I wrong? I am thinking this means volume of the molecule? Again please correct me on this.

    Bigserve you say...
    Having no volume is a weird thing to me. Because for ideal gas it then says the SMV (standard molar volume) is 22.4 liters. ??? ok my point is here is that when it says gas molecules have zero volume ***should this really be saying that any ideal gas molecule whether it is N2 or O2 is not going to matter. The 0 volume in ideal gas theory says you will have the same molecular number irregardless of the gas molecule in question.***

    Now, going into grahms law and RMS this is still dealing with ideal gas but now just taking into consideration some actuall values and empircal data from the gases we are speaking of? I guess my point is here I was / am still a little bit perplexed that ideal gas is still relating to rms effusion diffusion and grahams law:

    But, I think I have a decent handle on it... What do you all think?

    Ok, so now when real gas molecules come into play they do have volume and the EK book says that this new volume must be added to ideal gas volume... which is funny because the ideal gas volume is zero.


    Thanks and I hope this helps others
     
  14. soonereng

    soonereng Double Trouble
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    Gas bearings...that's funny :D

    The equation for KE is actually Kbar = 3/2kT, where k=R/Na, and Na is Avagadro's number. Kbar is the average kinetic energy. It's derived from:

    Kbar=1/2mVrms^2

    Vrms=((3RT/M))^1/2 (M=molar mass)

    Substitute to give Kbar=3RT/2Na (M/m=Na)

    I had to dig my physics book out from 9 years ago to find that one.
     
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  15. bigserve99

    bigserve99 Member
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    The problem with your argument is that you assumed the hypothetical process I described was isothermal, when in fact it was adiabatic. I'm sorry if I didn't make it clear that it was isolated. The equation you gave, W=nRTln(Vf/Vi) is derived for an isothermal process (one in which heat can enter / leave a system). My example makes sense only when the system is isolated (as most PV=nRT problems on the MCAT usually are).

    In the isolated system I described, q = 0. W = 0 (because there is no pressure against which the gas must expand). The gas is essentially diffusing to fill the empty space of the isolated container. Hence, delta E (change in internal energy) is 0 and there is no temperature change.

    Furthermore, it can be seen simply by qualitative examination of the process at hand that the temperature could not change. There is no loss in momentum of gas particles. Could you explain how the momentum of the gas molecules is lost? This is the only way temperature can be changed!

    P is proportional to (mass*rms-velocity^2) * (number of molecules / volume)

    We've done nothing more than change the volume in this equation. No change in rms-velocity, mass, or number of molecules!


    Now, in this same example: if after following the adiabatic expansion of the gas to fill the volume, a piston were used compress the gas to the original volume, the temperature and pressure of the final state would be higher than the initial state (before original expansion). This is because a different PATH was used to compress the gas vs. expansion (work done on the system vs. no work done by the system)

    Finally, again, the refrigerator example you provide is again not an adiabatic system (it draws heat from the cold sink, in this case the interior of the refrigerator)
     
  16. OP
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    christian15213

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    When I FART the next time... boyyyy the things that will be running through my head... I tell ya. :laugh: :laugh: :laugh: :laugh:

    You guys are way to smart for this MCAT... LOL, don't over focus on one thing.

    But serioulsy thanks... It for sure answered my question.
     
  17. bigserve99

    bigserve99 Member
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    So you have this correct except a decrease in volume does not directly affect the rms-velocity and hence the kinetic-energy. A decrease in volume will affect the collision rate of the molecules on the wall of the container. The decrease in volume only directly affects pressure (boyle's law). What changes the rms-velocities is the F*D work provided by piston in compressing the gas.

    Keep in mind however: IF THE PRESSURE WAS MAINTAINED CONSTANT, as in Charle's law, then yes, the temperature (and hence rms-velocites) would drop as volume dropped to maintain the pressure.










    Do not worry about where they got the equation. Just know that 3/2(RT) is proportional to the avg. kinetic energy of the gas molecules.



    Volume, when dealing with ideal gases, is always the volume of the container the gas occupies (never the volume of the gas molecules itself).




    I said "halving" not "having".......I meant to divide the volume of gas by 2.



    ~ BigServe99
     
  18. soonereng

    soonereng Double Trouble
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    I think you need to look at this some more. Isothermal = same temperature. You are saying that the system is both adiabatic and isothermal at the same time.

    Check out this link: http://en.wikipedia.org/wiki/Adiabatic_process

    Hope this helps.
     
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  19. soonereng

    soonereng Double Trouble
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    Glad that I could help.

    I took the MCAT last August (13PS, 12VR, 9BS: I had no orgo or bio courses before I took it, hence the low BS score) and start med school this fall.

    I know so much about this stuff because I have been working as a mechanical engineer for 6 years and had to study this stuff like crazy to get my professional engineering license.
     
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  20. OP
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    christian15213

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    cool... any verbal tips? good job too and congradulations.
     
  21. soonereng

    soonereng Double Trouble
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    I have no verbal tips. Everyone complains that it is hard, but it just wasn't ever the case with me. I just read the passages and answered the questions.

    :luck:
     
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  22. bigserve99

    bigserve99 Member
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    Yah, so I know what adiabatic is and I know what isothermal is. See the problem is that you are just assigning names to the process without contemplating what they are intended for. An isothermal process maintains it's temperature using a heat source or sink while transfering q into usable work. An adiabatic process simply is a closed system with the added stiuplation that there is no heat exchange....thereby making it an isolated system.

    The problem with all these sources you are giving me is that all the work done by these process are AGAINST AN EXTERNAL PRESSURE. If it were not against an external pressure, NO WORK WOULD BE DONE. In this case, there is no change in internal energy because there is no loss in energy via WORK or via HEAT. You have already agreed to this fact in a previous post (W = 0 and Q = 0). SO HOW CAN TEMPERATURE DECREASE? Can you explain to me the molecular process by which this would occur? How do the molecules lose their rms-velocities if, considering they are ideal (not just approximations), their momentums are fully conserved and there is no net outside force being applied on or by the system!

    Engineering equations are not the end all be all of thermodynamics. Unfortunately, much of engineering thermodynamics are based on specific assumptions that usually understood by us engineers as a given. MCAT's make no assumptions. In my case, there is no outside force for the gas to expand against, and hence no work done by the gas. Just a heads up, I'm an engineer too so I understand where you're comming from.
     
  23. bigserve99

    bigserve99 Member
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    And yes, in this case, the process DOES end up being adiabatic AND isothermal, because the process does NOT do any WORK.
     
  24. emanrebmem

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    I realize this thread is quite old and has been thoroughly explained, however, I think I have a much simpler and quicker solution for anyone having issues with this problem.

    What are they asking for and what's given? V1, V2 and T1 are given, and they're asking for T2. Ask yourself this on every question!

    Some of you might recognize right away that this is Charles's Law: V1/T1=V2/T2. If not, that's okay because you only need to know PV=nRT, and you can eliminate the variables that are held constant (P, n, R) to find what you need (Charles's Law). If you chose A for your answer (decreases), it's probably because you tried to use PV=nRT straight up. However, this leaves out V2 and T2 and that's why it doesn't work. Recognize that you have to set the two conditions equal to one another: P1V1/nRT1=P2V2/nRT2

    Now just use proportions, which tell us that if you decrease V1, then you have to increase T2 because of their inverse relationship.

    Tip:
    Directly Proportional:
    opp/same: if the two variables in question are on the opposite side of the equals sign and the same side of the division sign, then those variables have a directly proportional relationship. (i.e. F and a in F=ma)​
    same/opp: just a manipulation of above--same side of equals sign with one variable in the numerator and the other in the denominator--as long as the quotient is held constant. (F and m in F/m=a)​

    Inversely Proportional:
    same/same: same side of equals sign and same side of divison sign. (i.e. m and a in F=ma)​
    opp/opp: just a manipulation of above--opposite sides of equals sign and opposite sides of division sign. (i.e. F and r^2 in F= GMm/r^2)​
    **Test it on the question above--even with Charles's Law manipulated to V1T2=V2T1: Since we're wanting to know what happens to T2 when V1 is manipulated we have are our variables in question. So, same/same = invers. prop. This means if you decrease V1, then there is an increase in T2.** ​

    ***DON'T memorize the above scenarios when all you need to know is that if the terms match it's Inversely and if they don't match it's Directly***
     
  25. kehlsh

    kehlsh Medic Commando
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    dood what's wrong with you bumping years old thread
    besides, all u need is PV = nRT
     

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