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Yahoo!

A steel cylinder capacity 10L holds 20 Moles of O2 at 27 C an imperfection causes the gas to leak out slowly so that temperature remains Contant Ext Pressure is 1 atm The cylinder Finaaly Contain 1 atm Calculate Work Done in Calories

  • one year ago
  • one year ago

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  1. experimentX
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    this is isothermal process, work done is given by \[ n R T \log \left( V_b \over Va\right)\\ V_a = {10 \text{L}} \\ V_b = {300.4 \over 273.4}*20*22.4 \text{L}\]

    • one year ago
  2. mahmit2012
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    |dw:1351149292211:dw|

    • one year ago
  3. mahmit2012
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    |dw:1351149481464:dw|

    • one year ago
  4. mahmit2012
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    |dw:1351149608767:dw|

    • one year ago
  5. mahmit2012
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    |dw:1351149969466:dw|

    • one year ago
  6. mahmit2012
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    |dw:1351150077826:dw|

    • one year ago
  7. mahmit2012
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    I can't continue there is not Q ?

    • one year ago
  8. mahmit2012
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    Pay attention that in this case n is variable(leaking). so U is not constant(Adiabatic) So you can not find W(work) by int Pdv.

    • one year ago
  9. Vincent-Lyon.Fr
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    Is the answer 48.9 kJ = 11.7 Cal ?

    • one year ago
  10. mahmit2012
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    |dw:1351152257623:dw|

    • one year ago
  11. mahmit2012
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    I think you should have how percent was leaked ?

    • one year ago
  12. Yahoo!
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    it is ~11700 cal @Vincent-Lyon.Fr

    • one year ago
  13. Yahoo!
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    @experimentX : \[W=2.303*n*R*T* \log\frac{ P1 }{ P2 }\]

    • one year ago
  14. Yahoo!
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    \[P1 V = nRT\] \[P1=~48atm\]

    • one year ago
  15. Yahoo!
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    P2=1atm (given)......Can t we Do like this...?

    • one year ago
  16. experimentX
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    i guess you can.

    • one year ago
  17. Yahoo!
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    But Not getting the Correcr Result:

    • one year ago
  18. Yahoo!
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    but @Vincent-Lyon.Fr Got it correctly...)

    • one year ago
  19. Yahoo!
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    @Kryten

    • one year ago
  20. experimentX
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    http://www.physicsforums.com/archive/index.php/t-530754.html

    • one year ago
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