<u>Answer:</u> The molar volume of the Argon is 0.321 L/mol and compression factor is 0.658
<u>Explanation:</u>
To calculate the compression factor, we use the Virial equation to the second order, which is:
![Z=1+B'P+C'P^2](https://tex.z-dn.net/?f=Z%3D1%2BB%27P%2BC%27P%5E2)
where,
Z = compression factor
B' = second virial constant =
(From standard values)
C' = third virial constant =
(From standard values)
P = pressure of the gas = 53 atm
Putting values in above equation, we get:
![Z=1+(-0.00616atm^{-1}\times 53tm)+(1.52\times 10^{-7}atm^{-2}\times (53)^2)\\\\Z=0.658](https://tex.z-dn.net/?f=Z%3D1%2B%28-0.00616atm%5E%7B-1%7D%5Ctimes%2053tm%29%2B%281.52%5Ctimes%2010%5E%7B-7%7Datm%5E%7B-2%7D%5Ctimes%20%2853%29%5E2%29%5C%5C%5C%5CZ%3D0.658)
Now, calculating the molar volume, by using the equation of compression factor:
![Z=\frac{PV_m}{RT}](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BPV_m%7D%7BRT%7D)
where,
Z = compression factor = 0.658
P = pressure of the gas = 53 atm
= molar volume of gas = ?
R = Gas constant = ![0.0821\text{ L atm }mol^{-1}K^{-1}](https://tex.z-dn.net/?f=0.0821%5Ctext%7B%20L%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D)
T = Temperature of the gas = 315.2 K
Putting values in above equation, we get:
![0.658=\frac{53atm\times V_m}{0.0821\text{ L atm }mol^{-1}K^{-1}\times 315.2K}\\\\V_m=\frac{0.658\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 315.2K}{53atm}=0.321L/mol](https://tex.z-dn.net/?f=0.658%3D%5Cfrac%7B53atm%5Ctimes%20V_m%7D%7B0.0821%5Ctext%7B%20L%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20315.2K%7D%5C%5C%5C%5CV_m%3D%5Cfrac%7B0.658%5Ctimes%200.0821%5Ctext%7B%20L%20atm%20%7Dmol%5E%7B-1%7DK%5E%7B-1%7D%5Ctimes%20315.2K%7D%7B53atm%7D%3D0.321L%2Fmol)
Hence, the molar volume of the Argon is 0.321 L/mol and compression factor is 0.658