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Homework Help: Thermodynamic - Work done in a system

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    The question is as following:

    One litre of an ideal gas, initially at atmospheric pressure, is held at a constant
    temperature of 300 K (by maintaining good thermal contact with a thermal reservoir
    held at this temperature). The volume is decreased until the pressure doubles.
    Calculate how much work is done on the gas. How much heat flows out of the gas?

    2. Relevant equations

    [tex]W = \int P(V) dV[/tex]

    3. The attempt at a solution

    Now I know that for work done in a varying pressure , I must use the following equation:

    [tex]W = \int P(V) dV[/tex]
    where I sub in ideal gas equation for pressure.

    [itex]W = nRT\ln{(V0/V)}[/itex] (since it's compression not expansion)

    Information that's given : 1 lite of water = 55.5 moles , T = 300k , Pressure 1 : 1x10^5 , pressure 2 : 2x10^5 , Volume one : 0.001 m3 , volume 2 : 5x10^-4 (since p1v1 = p2v2)

    After having plugged in the numbers.. I get work of 319 joules .
    Am I on the right track ? also since this is an isothermal cycle Internal energy stays constant i.e dE =0, So work done = heat out, right ?
  2. jcsd
  3. Oct 16, 2011 #2
    No ,I have just plugged int he same values I get 93 kJoules of work.. is this wrong ?
  4. Oct 16, 2011 #3
    Would you write a line of numbers showing the arithmetic you did in substituting into the formula for the polytropic process.
  5. Oct 16, 2011 #4
    Is the work done 69.3J?
  6. Oct 16, 2011 #5
    Sure .
    We know work = -p(V)dv which when we substitude int the ideal gas equation becomes : W = integral of nrt/v dv

    8.31 * 55.5 * 300 * ln(0.001/ 5x10^-4)

    Last night I got a value of 320 joules but now when I am plugging the same numbers into my calculator... an absurd quantity shows up.

    What am I doing wrong ?

    @ grzz: I don't know. hmm
  7. Oct 16, 2011 #6
    That is what I computed.
  8. Oct 16, 2011 #7
    Your equation shouldn't have n, R or T in it. All you need to compute P(V) is [itex] PV = P_0 V_0 [/itex] and you know [itex] P_0 [/itex] and [itex] V_0 [/itex]
  9. Oct 16, 2011 #8
    How did you get the value of the number of moles, n?
  10. Oct 16, 2011 #9
    I can't get that answer... hmm
    I am confused. The question does give t =300k.
    Well the given quantity of water is 1 litre = 1000g
    Moles = mass / mr = 1000 / 18 = 55.5 .
  11. Oct 16, 2011 #10
    PV = nRT
    PV/(RT) = n and i do not get 55.5 for n.
  12. Oct 16, 2011 #11
    What's wrong with my value of n ? I am totally lost on how to solve this question.
  13. Oct 16, 2011 #12
    I do not understand why you put 18.
  14. Oct 16, 2011 #13
    The substance is an ideal gas. It is not water.
  15. Oct 16, 2011 #14
    OMG. You're right! there's no mention of water!
    In this case would n = 0.040

    [itex]W = nRT\ln{(V0/V)}[/itex]

    Then let me try to compute in the answer.

    Last edited: Oct 16, 2011
  16. Oct 17, 2011 #15
    I used to tell my students to THINK and I used to remind them of that very often when correcting their homework.
    Let me share this with you. They heard that word so often that once a student turned up with a T-shirt printed 'THINK' on one side and 'E = mc[itex]^{2}[/itex] ' on the other.
  17. Oct 17, 2011 #16
    Thanks for sharing that. It made me chuckle. lol

    Such students should often think at a slower rate, there's no of urgency. (I ,myself am included into this category)
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