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Homework Help: Boyle's Law Question/Problem

  1. Apr 13, 2006 #1
    Hi Everyone, first semester Physics student here and we are working on a tough lab this week, having to do with Boyle's Law...argghh!!! :cry:

    Anyways, here is my problem, and trying to work through it to obtain the correct answer...but it is stumping me!!

    The question as follows:

    1. Suppose data was collected for a sample of 3.oo moles of the ideal gas represented by the Pressure versus Volume -1 graph below. What is the Kelvin temperature of the gas?

    the graph has P (pressure) for the Y axis, and 1/V for the x axis.

    The slope =45.6 atm L and is at an approximate 45 degree angle (straight line) on the graph.

    Any thoughts?

    I am trying to use the equation of P=nRT/V=constant/V to solve this, but am not sure where/how to start.
    P=total pressure of the gas
    V=Volume of the gas
    n=# of moles of gas
    T absolute temperature og gas in Kelvin

    It is given that R=8.314 J/mol and K=0.0821 atm liter/mol K

    And here is a real tough one...

    2. 2.00 moles of an ideal gas at -50 degrees Celsius has a volum of 1.00 liter. What is the pressure (in atm) in the container??

    I can't find the proper equation/setup for this one and am totaly lost here...these two problems are just evil!!!

    Please help, if you can!! Thanks!! :smile:
     
  2. jcsd
  3. Apr 13, 2006 #2

    Hootenanny

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    Well, you can use your equation;

    [tex]P = \frac{nRT}{V}[/tex]

    Now consider the general equation of a straight line;

    [tex]y =mx +c[/tex]

    Think about what the gradient of your line represents, it may be more useful if we re-write your equation;

    [tex]P = nRT \cdot \frac{1}{V}[/tex]

    Regards,
    -Hoot

    [Edit] I've just seen your other post now thunder, welcome to PF![/Edit]
     
    Last edited: Apr 13, 2006
  4. Apr 13, 2006 #3
    Hi Hoot, for some reason, when I click on this topic, I can't pull up your reply. just joined today so I may not be hitting the right buttons yet; Cool I see it now!! Hey thanks, hoot!! Great to be here!! :)
     
  5. Apr 13, 2006 #4

    Hootenanny

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    For your second question, you need only apply the equation

    [tex]pV = nRT[/tex]

    Ensure that the volume is in [itex]m^3[/itex] and the temperature is in kelvin. Also note that 1atm is approximately [itex]1.01\times 10^5[/itex] pascals.

    Reagards,
    -Hoot
     
  6. Apr 13, 2006 #5
    Ok so here is my calculation method for question #1:

    Suppose data was collected for a sample of 3.00 moles of the ideal gas represented by the Pressure versus Volume -1 graph below. What is the Kelvin temperature of the gas?

    the graph has P (pressure) for the Y axis, and 1/V for the x axis.

    The given slope = 45.6 atm L and is at an approximate 45 degree angle (straight line) on the graph.

    Using equation P=nRT x 1/V

    slope = 45.6 atm = nRT

    R=universal gas constant=8.3145 J/mol K

    n=# of moles=3.00 moles for this problem

    so, I solve for T as follows:

    nRT = 45.6

    (3.00)(8.3145)T=45.6

    DIVIDE both sides by (3.00)(8.3145) OR (24.9435)

    T= 45.6/(24.9345)

    So then solving for T=1.82813157736 degrees Celsius

    So my answer in Kelvin would be 1.82813157736 + 273

    Which is approximately 274.8 Kelvin

    Is this correct? The question is asking for me to obtain the correct Kelvin temperature of the gas.
     
  7. Apr 13, 2006 #6

    Hootenanny

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    Not quite. As the gradient is given in atm L / mol, you need to use that form of the universal gas constant (i.e 0.0821). Also the T in the equation represents absolute temperature, so there's no need to add 273 to your answer.

    Regards,
    -Hoot
     
  8. Apr 13, 2006 #7
    Thanks!!

    so for 1 liter, I get 1.0 x 10^-3 m^3

    "1.0 times ten to the minus three meters cubed"

    Using the equation PV=nRT

    and solve for P

    P={(2.00 moles)(8.314 J/mol)(-50 dgrees Celsius + 273K)/1.0 x 10^-3 m^3}

    P={(2.00)(8.314)(223K)/(.001 meters cubed)

    P=3708044 (what is the unit for this answer...is it millimeters or ?)

    So to obtain the correct pressure (in atm), I divided my answer of 3708044 by (1 x 10^5 pascals) to get approximately 37.08044 atm or 37.1 atm??

    Am I close?? LOL :smile:
     
  9. Apr 13, 2006 #8

    Hootenanny

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    Looks right to me :smile:

    Regards,
    -Hoot
     
  10. Apr 13, 2006 #9
    LOL, oops, ok thanks. Let's see then what I need to do is as follows:

    nRT=45.6 atm

    replace n with 2.00 moles
    repalce R with .0821 atm L/mole

    and solve for T

    (2.00)(.0821)T=45.6

    DIVIDE (2.00)(.0821) OR (.1642) from both sides

    and solve for T

    T=45.6/.1642

    T=277.710109622 Kelvin OR 277.7 Kelvin

    Correct?? :smile:
     
  11. Apr 13, 2006 #10

    Hootenanny

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    Almost, you've used 2 moles when it should be 3, I think you've confused it with your second question :smile:

    Regards,
    -Hoot
     
  12. Apr 13, 2006 #11
    Yippeee!!!!! Thanks!!! Ok so that takes care of problem #2.

    I redid problem # 1, where I am looking for the correct Kelvin temperature of the gas (see my last post above). Am I close? How'd I do? Really appreciate you helping me work thrugh this to find the problem.
     
  13. Apr 13, 2006 #12
    oops!! LOL

    Yep, I'll correct that!! Thanks!! :smile:
     
  14. Apr 13, 2006 #13

    Hootenanny

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    I'm talking about you re-do for question one. You've used 2 moles when it should be 3!!
     
  15. Apr 13, 2006 #14
    ok , gotcha...

    I reworked it as follows:

    nRT=45.6 atm L

    n=3.00 moles :smile:
    R=.0821
    and solve for T

    (3.00 moles)(.0821)(T)=45.6

    DIVIDE both sides by (3.00)(.0821) OR (.2463)

    and solve for T

    T=45.6/.2463

    T=185.140073082 Kelvin or approximately 185 Kelvin


    So, as Regis would say, my final answer for problem # 1 is approximately 185 Kelvin.

    And for problem #2, I got approximately 36.7 atm.

    Whewww!!! Correct?? o:)
     
  16. Apr 13, 2006 #15

    Hootenanny

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    They look correct to me, they're the same as what I got anyway :biggrin:

    Well done.

    Regards,
    -Hoot
     
  17. Apr 13, 2006 #16
    Awesome!! Thanks so much for you assistance! Really appreciate it! BTW, I love that Einstein quote!! :biggrin:
     
  18. Apr 13, 2006 #17

    Hootenanny

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    My pleasure
     
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