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Yahoo!

  • 2 years ago

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

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

  2. mahmit2012
    • 2 years ago
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    |dw:1351149292211:dw|

  3. mahmit2012
    • 2 years ago
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    |dw:1351149481464:dw|

  4. mahmit2012
    • 2 years ago
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    |dw:1351149608767:dw|

  5. mahmit2012
    • 2 years ago
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    |dw:1351149969466:dw|

  6. mahmit2012
    • 2 years ago
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    |dw:1351150077826:dw|

  7. mahmit2012
    • 2 years ago
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    I can't continue there is not Q ?

  8. mahmit2012
    • 2 years ago
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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.

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

  10. mahmit2012
    • 2 years ago
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    |dw:1351152257623:dw|

  11. mahmit2012
    • 2 years ago
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    I think you should have how percent was leaked ?

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

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

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

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

  16. experimentX
    • 2 years ago
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    i guess you can.

  17. Yahoo!
    • 2 years ago
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    But Not getting the Correcr Result:

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

  19. Yahoo!
    • 2 years ago
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    @Kryten

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

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