I think that it is Covalent..
The correct answer to this question is 420 because it would be half of 741
Explanation:
Formula for compressibility factor is as follows.
z = ![\frac{P \times V_{m}}{R \times T}](https://tex.z-dn.net/?f=%5Cfrac%7BP%20%5Ctimes%20V_%7Bm%7D%7D%7BR%20%5Ctimes%20T%7D)
where, z = compressibility factor for helium = 1.0005
P = pressure
= molar volume
R = gas constant = 8.31 J/mol.K
T = temperature
So, calculate the molar volume as follows.
![V_{m} = \frac{z \times R \times T}{P}](https://tex.z-dn.net/?f=V_%7Bm%7D%20%3D%20%5Cfrac%7Bz%20%5Ctimes%20R%20%5Ctimes%20T%7D%7BP%7D)
= ![\frac{1.0005 \times 8.314 \times 10^{-3} m^{3}.kPa/mol K \times (60 + 273)K}{500 kPa}](https://tex.z-dn.net/?f=%5Cfrac%7B1.0005%20%5Ctimes%208.314%20%5Ctimes%2010%5E%7B-3%7D%20m%5E%7B3%7D.kPa%2Fmol%20K%20%5Ctimes%20%2860%20%2B%20273%29K%7D%7B500%20kPa%7D)
= 0.0056 ![m^{3}/mol](https://tex.z-dn.net/?f=m%5E%7B3%7D%2Fmol)
As molar mass of helium is 4 g/mol. Hence, calculate specific volume of helium as follows.
![V_{sp} = \frac{V_{m}}{M_{w}}](https://tex.z-dn.net/?f=V_%7Bsp%7D%20%3D%20%5Cfrac%7BV_%7Bm%7D%7D%7BM_%7Bw%7D%7D)
= ![\frac{0.0056 m^{3}/mol}{4 g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7B0.0056%20m%5E%7B3%7D%2Fmol%7D%7B4%20g%2Fmol%7D)
= 0.00139 ![m^{3}/g](https://tex.z-dn.net/?f=m%5E%7B3%7D%2Fg)
= 0.00139 ![m^{3}/g \times \frac{1 g}{10^{-3}kg}](https://tex.z-dn.net/?f=m%5E%7B3%7D%2Fg%20%5Ctimes%20%5Cfrac%7B1%20g%7D%7B10%5E%7B-3%7Dkg%7D)
= 1.39 ![m^{3}/kg](https://tex.z-dn.net/?f=m%5E%7B3%7D%2Fkg)
Thus, we can conclude that the specific volume of Helium in given conditions is 1.39
.