Question

Using the Virial equations to second order, what is the molar volume Vm of Ar at 315.2 K when p = 53 atm? What is the compression factor, Z, under the same conditions?

Answer 1

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:

$Z=1+B'P+C'P^2$

where,

Z = compression factor

B' = second virial constant = $-0.00646atm^{-1}$    (From standard values)

C' = third virial constant = $0.000000152atm^{-2}=1.52\times 10^{-7}atm^{-2}$    (From standard values)

P = pressure of the gas = 53 atm

Putting values in above equation, we get:

$Z=1+(-0.00616atm^{-1}\times 53tm)+(1.52\times 10^{-7}atm^{-2}\times (53)^2)\\\\Z=0.658$

Now, calculating the molar volume, by using the equation of compression factor:

$Z=\frac{PV_m}{RT}$

where,

Z = compression factor = 0.658

P = pressure of the gas = 53 atm

$V_m$ = molar volume of gas = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = 315.2 K

Putting values in above equation, we get:

$0.658=\frac{53atm\times V_m}{0.0821\text{ L atm }mol^{-1}K^{-1}\times 315.2K}\\\\V_m=\frac{0.658\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 315.2K}{53atm}=0.321L/mol$

Hence, the molar volume of the Argon is 0.321 L/mol and compression factor is 0.658