The following information is available for the unnamed fluid X: At T = 240 K the

Question

The following information is available for the unnamed fluid X: At T = 240 K the saturation pressure is Psat = 12:83 bar, the molar density of the saturated liquid is 24742:0 mol=m3, and the second virial coefficient is B =2:34*10-4 m3/mol

a) Calculate the compressibility factor of saturated vapor at 240 K

b) Calculate the volume of a tank that contains 1000 mol of a vapor-liquid mixture at 240 K that is 25% liquid by mol.

c) Calculate the volume of a tank that contains 1000 mol of a vapor-liquid mixture at 240 K that is 25% liquid by volume.

d) Calculate the volume of a tank that contains 1000 mol of this fluid at 240 K, 8 bar.

Explanation / Answer

Ans:- a) Here given

molar density= molar volume = 24742m3

Second virial coefficient B=2.34x10-4

Now Compressibility factor

Z=1+B/Vm

   = 1+2.34x10-4/24742

    =1

Ans b) Here P=12.83 bar

T=240 K

R=8.314 m3bar/K/mol

Number of moles n

In order to calculate the number of moles the statement given is 1000mol of a vapor - liquid mixture that is 25% liquid by mol

number of moles = 1000/0.25

                          =4000

Now

Z= PV/nRT

1=12.83 x V / 8.314 x240x4000

V= 8.314x240 x4000/12.83

= 622091m3

Ans d) Here P=8bar

T=240K

R=8.314 m3bar/K/mol

n=1000

V=nRT/P

=1000x8.314x240/12.83

   = 155522m3

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