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ajprincess

Plzz help. Typical bedroom contains 2500 moles of air. Find the change in internal energy of the air, when it is cooled from 23.9 degree C to 11.6 degree C at a constant pressure of 1atm. Treat this air as an ideal gas with gamma = 1.4.

  • one year ago
  • one year ago

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  1. UnkleRhaukus
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    do you have a formula yet/

    • one year ago
  2. ajprincess
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    |dw:1338339090808:dw|

    • one year ago
  3. ajprincess
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    |dw:1338339348203:dw|

    • one year ago
  4. UnkleRhaukus
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    and the question asks for \(\frac{\text d u}{\text dt}\)

    • one year ago
  5. ajprincess
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    jst du.

    • one year ago
  6. UnkleRhaukus
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    oh ok.

    • one year ago
  7. amistre64
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    cant say id even know how to begin with this one :/ srry

    • one year ago
  8. ajprincess
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    it's ok. Thanxxx a lot anyways.

    • one year ago
  9. UnkleRhaukus
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    it is really annoying because it seams like a simple question, and i know i have done an almost identical question about a year ago, im gonna look through my old books

    • one year ago
  10. ajprincess
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    ha k.

    • one year ago
  11. ajprincess
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    thanxxx

    • one year ago
  12. UnkleRhaukus
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    The pressure is constant \(p\) is pressure; \(n=\frac NV\) is the moles per unit volume ; \(k_b\)the Boltzmann constant; \(T\) is the Temperature in Kelvin; \(u=\frac UV\) is the kinetic energy per unit volume \[p=nk_bT\] \[p=\frac23u\] \(U=\frac 32nk_BT\) \(\Delta U=\frac 32nk_B\Delta T\)

    • one year ago
  13. ajprincess
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    Thanxxxxx a lottt

    • one year ago
  14. UnkleRhaukus
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    there are too many formulas in thermodynamics i found this subject very difficult

    • one year ago
  15. ajprincess
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    :)

    • one year ago
  16. ajprincess
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    Bt y hav they given gamma @unklerhaukus.

    • one year ago
  17. UnkleRhaukus
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    i cant see how to use gamma for this question, the method of cooling has not be specified

    • one year ago
  18. ajprincess
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    |dw:1338353654846:dw| When I solve using this and the above formula i posted I end up with a different answer.

    • one year ago
  19. UnkleRhaukus
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    what were the two answers you got?

    • one year ago
  20. ajprincess
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    one is 635,215.625 and the other is 2.30*10^-20

    • one year ago
  21. UnkleRhaukus
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    and the units that belong to those numbers are ?

    • one year ago
  22. ajprincess
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    first one is Joules bt the second is the answer I found using ur formula. I am confused with its units.

    • one year ago
  23. ajprincess
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    second answer has to be 2.55*10^-22

    • one year ago
  24. ajprincess
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    @unklerhaukus

    • one year ago
  25. UnkleRhaukus
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    in physics , units are most important , did you stick to SI units , and remember to convert temperatures to Kelvin

    • one year ago
  26. ajprincess
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    ya I did.

    • one year ago
  27. eashmore
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    Let's first make a couple of assumptions to better define the problem. 1) Closed system. 2) Constant specific heats 3) Ideal gas 4) Reversible We also know that the process happens at constant pressure. With Assumptions 1, 2, and 3, we can say that\[dU = m C_v dT\]However, we can look up the specific heat on a molar basis. The expression now becomes\[dU [ kJ] = n [mole] \cdot \bar C_v \left [ kJ \over {mol \cdot K } \right ] \cdot dT [K ~ or~ C]\] (Note that bar over the \(C_v\) indicates molar units.) But, because the process happens at constant pressure and not constant volume, we cannot use the constant volume specific heat value (\(C_v\)). Let's recall that change in enthalpy can be expressed as (under conditions specified by Assumptions 1, 2, and 3) \[dH = m C_p dT = n \bar C_p dT\]Let's also recall that enthalpy can be expressed in terms of internal energy as\[H = U + pV\]Differentiating, we obtain \[dH = dU + dp \cdot V + p \cdot dV\]Since pressure is constant, \(dp \cdot V = 0\)\[dH = dU + p \cdot dV\]Therefore, \[dU = n \bar C_p dT - p \cdot dV\]From Assumption 3, we can express the change in volume as\[dV = {nR(dT) \over p}\]\[\therefore dU = n \bar C_p (dT) - {nR(dT) \over p}\]With Assumption 2 and 3 and the fact that pressure and temperature are state quantities, we can expression the differential as\[\Delta U = n \bar C_p (\Delta T) - {n R (\Delta T) \over p}\] We don't know \(C_p\) however, but we know gamma and the gas constant. Let's work a couple of relations to get \(C_p\) in terms of gamma and R. It can be shown that\[C_p = {\gamma n R \over \gamma - 1}\] We obtain the following final expression\[\Delta U = n \left [ \gamma n R \over \gamma -1 \right] \Delta T - {n R \Delta T \over p}\] Be sure the units on R matches the units on n and p.

    • one year ago
  28. UnkleRhaukus
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    are you sure the pressure is constant, not the volume @eashmore?

    • one year ago
  29. eashmore
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    The problem statement explicitly states "at a constant pressure of 1 atm." Counter intuitive? Sure, but I'll stick to the problem statement.

    • one year ago
  30. UnkleRhaukus
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    ok that isa good point, but i still dont see how come you can assume reverseable cooling

    • one year ago
  31. ajprincess
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    Thanxxxx a lottt.

    • one year ago
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