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?

1. taramgrant0543664

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

$\frac{ PV }{ nRT }$ isn't it this @taramgrant0543664

3. taramgrant0543664

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

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

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

Since argon behaves ideally can't we just take Z as 1 cuz for ideal gases at stp z is 1

7. taramgrant0543664

What formula do you have?

8. anonymous

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

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

Yes argon acts as ideal at 273K so yes it should be 1 at STP

11. Rushwr

Awesome then !!!!!!!!! Thank you for confirming !!!! @taramgrant0543664

12. taramgrant0543664

STP would be 1atm for pressure then

13. anonymous

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

if a gas is believed to be ideal they don't interact right?

15. anonymous

..true, yes.

16. taramgrant0543664

If a gas is ideal their speed overcomes any interactions so there is no attraction or repulsion interactions

17. anonymous

You guys are awesome. Thanks again

18. Rushwr

anytym !!!!! @sibaholo