Question #208547

 Estimate the following: (a) The volume occupied by 18 kg of ethylene at 328.15 K (55°C) and 35 bar. (b) The mass of ethylene contained in a 0.25-m3 cylinder at 323.15 K (50°C) and 115 bar. 


1
Expert's answer
2021-06-22T05:08:36-0400

Part a

Tr=TTc=328.15282.3=1.162;Pr=PPc=3550.4=0.694T_r=\frac{T}{T_c}=\frac{328.15}{282.3}=1.162 ; P_r=\frac{P}{P_c}=\frac{35}{50.4}=0.694

From the table , Zo=0.838,Z1=0.033Z_o=0.838, Z_1=0.033

The correlation of the comprehensibility factor is Z=Z0+ωZ1=0.838+0.0870.033=0.841Z= Z_0+ωZ_1 = 0.838+0.087*0.033=0.841

Hence, PV=ZnRT    n=mM=1828=0.6428Kmol\implies n = \frac{m}{M}= \frac{18}{28}=0.6428 Kmol

V=ZnRTP=0.8410.64281038.3143281535105=0.421m3V= \frac{ZnRT}{P}= \frac{0.841*0.6428*10^3*8.314*32815}{35*10^5}=0.421 m^3

Part b

Tr=TTc=323.15282.3=1.145;Pr=PPc=11550.4=2.282T_r=\frac{T}{T_c}=\frac{323.15}{282.3}=1.145 ; P_r=\frac{P}{P_c}=\frac{115}{50.4}=2.282

From the table , Zo=0.482,Z1=0.126Z_o=0.482, Z_1=0.126

The correlation of the comprehensibility factor is Z=Z0+ωZ1=0.482+0.0870.126=0.493Z= Z_0+ωZ_1 = 0.482+0.087*0.126=0.493

Hence, PV=ZnRT    n=PVZRT=1151050.250.4938.314323.15=2.171Kmol\implies n = \frac{PV}{ZRT}= \frac{115*10^5*0.25}{0.493*8.314*323.15}=2.171 Kmol


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