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anonymous

  • one year ago

Hi. My lecturer gave that "at 273K measurements on Argon gave..." She didn't give us a P. Is it possible to compute Z, compressibility factor without pressure?? Oor rather, how do i find P?

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  1. taramgrant0543664
    • one year ago
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    Z= p/nT where p is the pressure, n is the number density, T is the temperature in energy units (i.e. joules, by multiplication for the Boltzmann constant)

  2. Rushwr
    • one year ago
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    \[\frac{ PV }{ nRT }\] isn't it this @taramgrant0543664

  3. taramgrant0543664
    • one year ago
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    As for the compressibility of gases, the principle of corresponding states indicates that any pure gas at the same reduced temperature, Tr, and reduced pressure, Pr, should have the same compressibility factor. The reduced temperature and pressure are defined by Tr= T/Tc and Pr = P/Pc Here Tc and Pc are known as the critical temperature and critical pressure of a gas. They are characteristics of each specific gas with Tc being the temperature above which it is not possible to liquify a given gas and Pc is the minimum pressure required to liquify a given gas at its critical temperature. Together they define the critical point of a fluid above which distinct liquid and gas phases of a given fluid do not exist. I don't know if that makes sense

  4. taramgrant0543664
    • one year ago
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    That's just the ideal gas law in this we have the compressibility factor. You could attempt to solve for p if you have all the givens I assume

  5. anonymous
    • one year ago
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    Thanks Guys but hold up..I know the equations, although we haven't done Tr and Pr cos we doing critical points this week. The assignment she gave didn't have any Pressure stated. I wanna know if she was wrong not to give us that or if i am supposes to find it myself.. All we have is T- 273k, then the coefficients B & C

  6. Rushwr
    • one year ago
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    Since argon behaves ideally can't we just take Z as 1 cuz for ideal gases at stp z is 1

  7. taramgrant0543664
    • one year ago
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    What formula do you have?

  8. anonymous
    • one year ago
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    I can use any i guess. Isnt Rushwr right tho? I completely forgot argon was inert. So...Z will always be one for Argon yeah? Esp if its at 273k. Yay nay ??

  9. Rushwr
    • one year ago
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    According to theory the compresibility factor of an ideal gas is 1 at STP !! So what I'm asking is since Ar is some wht ideal can't we take the compress ability factor as 1>

  10. taramgrant0543664
    • one year ago
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    Yes argon acts as ideal at 273K so yes it should be 1 at STP

  11. Rushwr
    • one year ago
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    Awesome then !!!!!!!!! Thank you for confirming !!!! @taramgrant0543664

  12. taramgrant0543664
    • one year ago
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    STP would be 1atm for pressure then

  13. anonymous
    • one year ago
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    Thanks yall. This helped alot. So, if i understand correctly, there will be no interactions(attration/repulsion) at 273K b/w Ar atoms cos Z is one?

  14. Rushwr
    • one year ago
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    if a gas is believed to be ideal they don't interact right?

  15. anonymous
    • one year ago
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    ..true, yes.

  16. taramgrant0543664
    • one year ago
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    If a gas is ideal their speed overcomes any interactions so there is no attraction or repulsion interactions

  17. anonymous
    • one year ago
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    You guys are awesome. Thanks again

  18. Rushwr
    • one year ago
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    anytym !!!!! @sibaholo

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