1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    My pleasure
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Boyle's Law Question/Problem
  1. Boyle's Law Question (Replies: 2)

  2. Boyle's Law Question (Replies: 7)

  3. Boyle's law problem. (Replies: 10)

Loading...