Respuesta :
Answer: The molar volume of the Argon is 0.321 L/mol and compression factor is 0.658
Explanation:
To calculate the compression factor, we use the Virial equation to the second order, which is:
[tex]Z=1+B'P+C'P^2[/tex]
where,
Z = compression factor
B' = second virial constant = [tex]-0.00646atm^{-1}[/tex] (From standard values)
C' = third virial constant = [tex]0.000000152atm^{-2}=1.52\times 10^{-7}atm^{-2}[/tex] (From standard values)
P = pressure of the gas = 53 atm
Putting values in above equation, we get:
[tex]Z=1+(-0.00616atm^{-1}\times 53tm)+(1.52\times 10^{-7}atm^{-2}\times (53)^2)\\\\Z=0.658[/tex]
Now, calculating the molar volume, by using the equation of compression factor:
[tex]Z=\frac{PV_m}{RT}[/tex]
where,
Z = compression factor = 0.658
P = pressure of the gas = 53 atm
[tex]V_m[/tex] = molar volume of gas = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = Temperature of the gas = 315.2 K
Putting values in above equation, we get:
[tex]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[/tex]
Hence, the molar volume of the Argon is 0.321 L/mol and compression factor is 0.658
