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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?
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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taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Z= 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)

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ PV }{ nRT }\] isn't it this @taramgrant0543664

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1As 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

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1That'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks 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

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Since argon behaves ideally can't we just take Z as 1 cuz for ideal gases at stp z is 1

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1What formula do you have?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 ??

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1According 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>

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1Yes argon acts as ideal at 273K so yes it should be 1 at STP

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1Awesome then !!!!!!!!! Thank you for confirming !!!! @taramgrant0543664

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1STP would be 1atm for pressure then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks 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?

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.1if a gas is believed to be ideal they don't interact right?

taramgrant0543664
 one year ago
Best ResponseYou've already chosen the best response.1If a gas is ideal their speed overcomes any interactions so there is no attraction or repulsion interactions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You guys are awesome. Thanks again
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