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Year 12: Cambridge Physics Problem (heat and gases)

  1. Jun 4, 2012 #1
    Guys I'm weak in heat and kinetic theory, so I'm gonna need extra guide and pointers from you guys to solve this and the coming questions. Thank you.

    A constant-volume gas thermometer of volume 0.001m3 contains 0.05mol of a gas and is used to deduce values of temperature on the assumption that the ideal gas law is obeyed. In fact, one mole of the gas satisfies the relation [tex](p + \frac{a}{V_m^2})(V_m -b) = RT[/tex],
    where a correct for intermolecular forces and b corrects for the effective size of the gas molecules.
    (i) What form of the equation should be used for x moles of gas? Explain your answer.
    (ii) Given that the thermometer was calibrated at the triple point of water, use the data below to calculat the error when the thermometer is used to measure a temperature near to that of boiling water: a = 0.08 Pa m6 mol -2; b = 3E-5 m3 mol-1.

    Attempt:
    I honestly don't know. Pointers anyone?
     
    Last edited: Jun 4, 2012
  2. jcsd
  3. Jun 5, 2012 #2
  4. Jun 5, 2012 #3
    No, I didn't know about Van der Waals' equation. I'll have a look at them and reattempt the questions soon. Thank you very much!

    P.s. I'm actually waiting for my degree to start in July. These exercises aren't homework; I'm doing them for sport. So please do consider giving me extra pointers/partial solutions via PM as I do not have any reference materials with me, and googling usually won't help much either (they're either too brief or too complicated). Thank you again!
     
  5. Jun 6, 2012 #4

    ehild

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    The constant-volume gas thermometer uses the change of pressure of a gas at constant volume to find the temperature, applying Gay-Lussac's Law p/p0=T/T0 valid for ideal gases. http://www.brighthub.com/engineering/mechanical/articles/26627.aspx
    For reference, the triple point of water is used with T0=273.16 K and p0=611.73 Pa.
    The ratio of the actual pressure to that of the pressure at the triple point is measured by a mercury manometer. The temperature reading is then 273.16 p(P0)

    Your first task is to rewrite the given Van der Waals equation of state in terms of volume V and number of moles n, using that Vm=V/n.

    Then determine p/p0for T=373 K, boiling point of water, and substitute that value into the formula Tmeasured=273.16 p/p0.
    Determine the percent derivation from the real temperature.

    ehild
     
  6. Jun 7, 2012 #5
    [itex](p+\frac{n^2a}{V^2})(V/n-b)[/itex]

    At T0=273.16K, and substituting values of a, b, n, and V into the equation above,

    p=113. 529.6531 Pa ?!
     
  7. Jun 7, 2012 #6

    ehild

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    There are two decimal points in your result... What do you mean really?


    ehild
     
  8. Jun 7, 2012 #7
    Sorry. I meant 133, 529.6531 Pa ?!
     
  9. Jun 7, 2012 #8
    Hmm what is that 133??

    You can consider the change in pressure-temperature relations this way too. The ideal behavior gives,

    [tex]\Delta P \cdot V=nR\Delta T[/tex]

    Similarly, from the Van der Waals' equation, you have will have a relation between change in P and T for the real gas. V of the thermometer is constant, so the volume of the gas inside it will also be constant.

    Compare the difference in temperatures from both equations, from which you can calculate the percent deviation.
     
  10. Jun 7, 2012 #9

    ehild

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    So the pressure of the gas at the triple point of water is p0=113530 Pa, calculated from the van der Waals equation. Calculate the pressure p also at 373 K, at the boiling point of water.

    The pressure change is evaluated by assuming that the gas is ideal. As the volume is constant, the measured temperature is T(measured)=273.16 p/p0. Determine the temperature T(measured), obtained by the thermometer, when the real temperature is 373 K.

    ehild
     
  11. Jun 8, 2012 #10
    Got it. Phew. Thank you so much!
     
  12. Jun 8, 2012 #11

    ehild

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    It was a nice problem. What difference have you got?


    ehild
     
  13. Jun 8, 2012 #12
    0.18K too high. The actual answer is 0.15K but I think it's just because I rounded off the values a bit too much.

    It is indeed a nice problem. Otherwise it wouldn't have come from Cambridge! I have tonnes of questions harder than this yet to be solved. (gasp!)
     
  14. Jun 8, 2012 #13

    ehild

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    The calculation would have been much easier and also more accurate if you calculated with the pressure and temperature differences between 273.16 and 373 K, as Infinitum suggested in Post #8.

    Isolating p from the Van der Waals equation, you get

    [tex]p=\frac{nRT}{V-nb}-\frac{a n^2}{V^2}[/tex]

    The same at T0, the triple point:

    [tex]p_0=\frac{nRT_0}{V-nb}-\frac{a n^2}{V^2}[/tex]

    Subtracting the pressures:

    [tex]p-p_0=\frac{nR(T-T_0)}{V-nb}[/tex]

    Assuming ideal gas: [tex]p-p_0=\frac{nR(T_m-T_0)}{V}[/tex]

    So the measured temperature difference is

    [tex]T_m-T_0=(T-T_0)\frac{V}{V-nb}[/tex]

    ehild
     
  15. Jun 8, 2012 #14
    Ehild nailed it perfectly!! :smile:

    I get approximately (0.15 ± 0.01) using my method. I think that should suffice :tongue2:
     
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