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ParthKohli

  • one year ago

A gas undergoes a process such that the ratio of P and V is constant and equal to one. What's the molar heat capacity?

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  1. ParthKohli
    • one year ago
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    \[|dw:1444507155788:dw|\]

  2. ParthKohli
    • one year ago
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    \[C_{m} = \frac{\partial q }{\partial T}\]Is this relation right?

  3. hwyl
    • one year ago
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    MATH PROCESSING ERROR

  4. ParthKohli
    • one year ago
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    That drawing isn't important. Just a PV graph with slope one.

  5. ParthKohli
    • one year ago
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    \[q = \Delta U - w\]\[= \frac{3}2RT + pV = \frac{3}{2}RT + RT =\frac{5}{2}RT\]

  6. ParthKohli
    • one year ago
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    I don't think it's 5R/2 though.

  7. ParthKohli
    • one year ago
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    Looks like I nowhere used the fact that \(p=V\).

  8. thomas5267
    • one year ago
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    I suspect \(w=\int p(t)V(t)\,dt=\int p(t)^2\,dt\).

  9. ParthKohli
    • one year ago
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    Oh, what a stupid mistake.

  10. ParthKohli
    • one year ago
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    \[\frac{pV}{2}=\frac{p^2}{2}= \frac{V^2}{2}\]

  11. ParthKohli
    • one year ago
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    well duh

  12. ParthKohli
    • one year ago
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    then we're done here.

  13. thomas5267
    • one year ago
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    Why?

  14. hwyl
    • one year ago
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    WHY IS MATH NOT PROCESSING

  15. hwyl
    • one year ago
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    \(TEST\)

  16. hwyl
    • one year ago
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    WTF

  17. ParthKohli
    • one year ago
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    just refresh

  18. hwyl
    • one year ago
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    K NOW IT IS BACK

  19. thomas5267
    • one year ago
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    So the answer is 3/2R?

  20. ParthKohli
    • one year ago
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    No. 2R.

  21. thomas5267
    • one year ago
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    How does p^2=RT?

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