1. Limited time only! Sign up for a free 30min personal 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!

Thermodynamics 1st law problem

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Air occupies [itex]0.140m^3[/itex] while under gauge pressure of [itex]1.03x10^5[/itex]

    It is then expanded isothermically to atmospheric pressure.

    Then it is cooled at constant preesure back to its initial volume.

    And then I should calculate the work done by the gas.

    2. Relevant equations

    [itex] W=(nRT) \ln \frac{v_{1}}{v_{2}} [/itex]
    [itex] PV=nRT [/itex]

    3. The attempt at a solution

    First I need to know how many mols of air I have.
    So I convert [itex] 0.140 m^3 [/itex] to centimeters and since 1 mol of anything occupies [itex]22.4 L [/itex] or [itex]22400 cm^3[/itex] I find that I have 6.25mols of air.


    Then I find the initial temperature of the system using [itex] T_{i} = \frac{PV}{Rn} = 277.6K [/itex]

    And then I'm stuck here

    I tried before calculating the final volume and with this the ln between initial and final volumes.
    but I made the wrong assumption that temperature was constant which is not true.

    Now I'm stuck because I'm not sure how to calculate the final volume.
     
  2. jcsd
  3. Nov 1, 2011 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Hint: Draw the P-V diagram.
    The work done is the area inside the loop.

    Note: 1mol of a gas can have different volumes for different pressures and temperatures.
    You don't need to know the number of moles or the temperature: you need to know (nRT)
     
    Last edited: Nov 1, 2011
  4. Nov 1, 2011 #3

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    System starts at state A, goes to state B, then state C.

    A: PAVA=nRT1 ...(1)
    B: PBVB=nRT1 ...(2)
    C: PBVA=nRT2 ...(3)

    know: PA, VA, PB
    don't know: VB, (nRT1), and (nRT2)

    Work done is the area under the P-V diagram.
    From the diagram:

    [itex]W=(nRT_1)\ln |\frac{V_B}{V_A}| - P_B(V_B - V_A)[/itex] ...(4)

    from (1) ... [itex](nRT_1)=P_AV_A[/itex] ...(5)

    divide (1) into (2)

    [itex]V_B =P_AV_A/P_B[/itex] ...(6)

    substitute (5) and (6) into (4) and simplify:

    [itex]W=(P_AV_A)\ln |\frac{P_A}{P_B}| - V_A(P_A - P_B)[/itex]

    [edit]<sigh> didn't mean to post it yet! oh well... make sure you understand what I did.
     
    Last edited: Nov 1, 2011
  5. Nov 1, 2011 #4
    I honestly haven't thought about that...


    Anyways: the answer in the book is 5 700 N m (for some reason it didnt use Joules)
    Unless I miscalculated somewhere, trying your approach gives me a result of 2.75J
    When I tried another method it gave me 282.7 J
     
    Last edited: Nov 1, 2011
  6. Nov 1, 2011 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    1N.m = 1J

    Did you remember to add atmospheric pressure to the gauge-pressure for PA?
    Check your units are consistent.

    Note: the method I show has to be correct: these are the three state equations, and there are three unknowns, so the system is known completely and unambiguously.
    You do need to check my derivation... I may have misplaced a minus sign or something.

    You left off the units for pressure in post #1 so I cannot check.
    [itex]1.03 \times 10^5[/itex] what?
     
  7. Nov 1, 2011 #6
    I know, it just kinda feels awkward

    Nope.
    That is probably what is messing things up.
    edit: tried adding, still not solving.
    I keep getting 1.3J as answer

    Pa
     
    Last edited: Nov 1, 2011
  8. Nov 1, 2011 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK: I get -

    PA = 204kPa
    VA = 140L
    nRT1 = 29172kPa.L
    PB = 101kPa
    VB = 29172/101 = 289L
    nRT2 = 29172x140/289 = 14132kPa.L [*]

    WAB = 29172xln|289/140| = 21143kPa.L (isothermal)
    WBC = -101x(289-140) = -15049kPa.L (isobaric)
    WCA = 0 (isochoric) [*]

    WTOT = 21143 - 15049 = 6094kPa.L

    note:
    1kPa.L = 1000(N/m2)L and 1000L=1m3 so 1kPa.L=1N.m

    It's in the right ballpark - maybe less rounding off will get the 5700Nm needed?
    I cannot tell how you got such low figures without seeing your calculations.

    [*] these entries included for completeness.
     
    Last edited: Nov 1, 2011
  9. Nov 1, 2011 #8
    probably because I was using cubic meters instead of liters.
    giving

    PA = 204kPa
    VA = 0.140 m^3
    and then
    nRT1 = 28560Pa.m^3

    PB = 101kPa
    VB = 28560Pa/101kPa =0. 283m^3
    nRT2 = 28560x0.140/0.283 = 14129Pa


    WAB = 28560Pa x ln|0.283/0.140| = 20100Pa.m^3 (isothermal)
    WBC = -101kPa x(0.283 -0.140) = -14443Pa.L (isobaric)
    giving WTOT = 20100 - 14443 = 5657 J


    The answer in the book is 5700 J
    and the difference is probably because of rounding figures.

    Well, this time I got it right.
    But I'm not sure what was I doing that gave me those small numbers...


    Thanks for the help
     
  10. Nov 1, 2011 #9

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    giving WTOT = 20100 - 14443 = 5657 J

    Well done!

    The difference is actually because you didn't do enough rounding:

    The states are all known to 3 sig fig, so you should give your answer to 3 sig fig: 5657J is 4 sig fig - rounding to three gives you 5700J.

    When you give a formal answer you want to have:

    "Wtot = 5657J = 5700J (3sig.fig.)"

    as the 2nd to last line (last line is the same thing in words).

    aside: I just miskeyed a number into the computer :)
    What I've just modeled for you is a disciplined way to carry out analysis and calculations ... see how just listing everything like that clears a lot of thinking.
    Also very important is to draw the graph - most of your thermodynamics is is encoded in that and realizing that the work in any stage is the area under the graph.
    Even if you remember the equations, looking at the graph clears things up.
     
    Last edited: Nov 1, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Thermodynamics 1st law problem
Loading...