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Very Difficult : Finding final Volume gasses (for me)

  1. Dec 2, 2004 #1
    (I Solved It Thanks)

    The problem:

    a cylinder contains 3g of liquid octane and an initial o2 pressure of 1atm. the cylinder is fitted with a piston and initially occupies a volume of 10L at 298 K. The octane is completely consumed by the reaction. Assume The piston is massless and frictionless and that the cylinder neither absorbs nor releases heat. Calculate the final volume occupied by the cylinder.

    now what i have so far

    C8H18 + 25O2 ----> 8 CO2 + 9H2O

    this is the reaction after converting 3 g of octane to moles new reaction

    .0263 <octane> + .409 O2 (.329 mols react to consume all octane) --> .2104 C02 + .2367 H2O <theres .0799 leftover mols of O2 from the reaction .409 - .329>

    ok i assume that pressure is constant. because after the reaction the piston is going to move to equalize the pressure.

    so i need to find the final temperture or maybe theres a way to find volume without doing temperture. i calculated enerergy released which is delta H = -5470kj if this helps any.

    if you have any idea how to finish this problem it would be helpful you dont have to solve it i guess i just need to know what to use to figure it out ive played with q = mc delta T <q = delta H @ constant P> but i dnt knw if it will work and im not sure exactly how to use it. ive worked a long time on this and i have had no luck

    anything will help maybe. Thanks

    i turn it in tomorrow so if i dont get any ideas by then he will explain it for us but i wont get the credit oh well its extra credit.

    Thanks again.
     
    Last edited: Dec 3, 2004
  2. jcsd
  3. Dec 2, 2004 #2

    chem_tr

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    Hello, your reaction seems a bit wrong, as 25/2 moles of oxygen gas is needed. So divide the moles of oxygen you found in your post by 2.
     
  4. Dec 2, 2004 #3

    GCT

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    To get you started,

    You know that the moles of gas changes, since it is given that the reaction goes to completion at the current temperature, you can find the moles of carbon dixodie formed from this information, also you can find the number of moles of oxygen gas consumed simply by deducing-use pv=nrt-and solve for moles.......subtract from this the number of moles 02 consumed in the reaction. Add these two values (moles of C02 and 02) and you'll have the new total number of moles.

    Also, the temperature changes. There is a net change in enthalpy and this energy is used to do what? Change the internal energy (temperature). Find the equation.

    These two factors now contribute to changing the pressure of the system, and at this pressure a certain volume will be occupied. It is important to remember that PV=nRT pertains to a state function. You should read up on this. Don't worry about equilibrium.

    I'll check up on this post tommorrow.
     
  5. Dec 2, 2004 #4
    I turned it in and i was the closest one to solving it in the class.

    First srry about the 25/2 i had that on my paper. second

    If you look after my equation i have all that. next

    using heat releassed you can plug it in q = mc delta T

    What i did whcih was wrong for mc i used the average of the first two substances octane and oxygen. The correct way of doing this was using carbon dioxide and water as the two substances for finding the new temperture.

    Pressure is constant.

    .......m1*c1*T1 + m2*c2*T2
    Tf = ------------------------
    ............m2*c2 + m1*c1

    thats the equation after solving for two substances and the new temperature,

    m1 mass of substance 1
    c1 = specific heat of that substance

    and m2,c2 same 2nd substance

    find new temperature and plug it in

    solve PV = nRT removing constant var and setting the non constant side equal to the final side

    (Pi Vi) / (ni Ti)
    = (Pf Vf) / (nf Tf)

    solve for Vf and plug in the values.
     
  6. Dec 3, 2004 #5

    GCT

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    Is this is the solution your teacher gave you? I would have simply solved for the new PV value and deduce from there; [square root of (PfVf/PiVi) x Vi ]= Vf .Solving for the temperature, I admit is quite a bit tedious.

    Pressure is not constant. The volume changes because of the new pressure value. And what factors into pressure.......the internal energy of the molecules as well as the number of them. In the end it's a direct pressure/volume situation. That's how I found the solution without all of the perfunctory math, since pressure changes in direct proportion to volume.

    For instance if pressure changes by a factor of 3, volume correspondingly increases by a factor of 3, the new PV value in relation to the initial will be by a factor of 9.

    [square root of (niRT/PiVi) x Vi] = Vf
     
    Last edited: Dec 4, 2004
  7. Dec 4, 2004 #6
    thats what he told me for finding the new temperature but he never told us the answer he wanted us to figure it out.

    as for pressure. you sure because its a piston in a cyalinder so the reaction releases heat -5470KJ which increases the temperature and this causes a increase of pressure but the cyliner therefore moves to equalize with the outside pressure of 1 atm (this is what makes the cylinder move) so the pressure is gona still be constant maybe?

    with that you can then solve for final volume. you cant say pressure is direct relation with volume unless you have constant temperature.
     
  8. Dec 5, 2004 #7

    Gokul43201

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    If the other side of the frictionless piston is exposed to atmosphere, the pressure in the container will remain constant. And I don't believe this is an unreasonable assumption to make, since most closed reactions are performed under isobaric conditions.
     
  9. Dec 5, 2004 #8

    GCT

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    The temperature increases, the number of moles increases.......a system to itself under these conditions will experience an increase in pressure. This was the whole point of figuring out the new temperature of the system, the heat increases the energy of the molecules which is directly related to temperature, temperature pertains to the change in energy of the molecules, their expansion. Isobaric conditions require an extra incentive, intervention I believe. The original problem did not indicate an isobaric condition and this requires more understanding of the problem and creativity, which probably qualified this problem for extra credit in the first place.

    Here's the basic logic of this problem:
    -The two independent variables are temperature and the number of moles, which pertains to the nRT side of the equation, of course nRT=PV, and since we have a new nRT value this will give a new PV value. There is no indication of isobaric conditions (pressure being constant, which would certainly make this problem much easier) unless you had not stated it in the original post.

    -So compare the old PV value with the new PV value. How do we find this relationship? For example, if the pressure increases by a factor of 3, then the volume also increases by same proportion. The new PV value thus will be 9 times the old PV value. Dividing the new PV value by the old one (PfVf/PiVi) should give you this factor, which you can square root to find the factor which you should multiply the old volume by.

    [square root(PfVf/PiVi) x Vi] = Vf

    The convenient aspect about PV=nRT is that it is a state function. Thus all that is required is simple algebra (PiVi to PfVf etc) and not calculus dealing with equilibrium problems. Know the beginning state and the end state. You don't need to visualize this in terms of equalizing with the outside pressure and such. It tells you the initial conditions and this is in a direct algebraic relation to the end condition and thus you don't need to worry about outside pressure since the two are proportional.

    It's too bad your teacher did not give you the answer for this problem.
     
    Last edited: Dec 5, 2004
  10. Dec 5, 2004 #9

    Gokul43201

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    I'll have to say I disagree with this.

    If you have 1 atmosphere on the outside of the piston, and a different pressure on the inside, there will be a net force on the piston that pushes it in or out (depending on the sign of the difference) till there is no net force. At this point, which is the final equilibrium state of the system, the inside pressure will have to match the outside pressure.

    In fact, I'd say that it takes special intervention to not allow the pressures to equalize. Typically, this is done by having a fixed volume (isochoric system) container, but that is not the case here. Any container with a frictionless piston is automatically an isobaric system unless you have some kind of pressure regulated manifold on the other side of the piston.

    Is it not because most chemical reactions are performed under isobaric conditions that the enthalpy is such an important quantity ?
     
  11. Dec 5, 2004 #10

    GCT

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    Yeah, you're right.......I misread this question (read it too quickly), confused it with another challenge question I was given during my undergrad year. My apologies (just count as a drunken post).
     
    Last edited: Dec 5, 2004
  12. Dec 5, 2004 #11

    Gokul43201

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    Not a big deal...I've made more mistakes than I'd care to recall.
     
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