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?

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:

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:

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

where,

Z = compression factor = 0.658

P = pressure of the gas = 53 atm

= molar volume of gas = ?

R = Gas constant =

T = Temperature of the gas = 315.2 K

Putting values in above equation, we get:

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

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

b. examine and describe cells