Answer :
Explanation:
(a) The given data is as follows.
T = [tex]55^{o}C[/tex] = (55 + 273) K
= 328 K
P = 35 bar = 3500 kPa
Let us assume that moles of ethylene present are 1 kmol and according to the ideal gas equation, PV = nRT.
Or, V = [tex]\frac{nRT}{P}[/tex]
= [tex]\frac{1 \times 8.314 \times 328}{3500 kPa}[/tex]
= 0.779 [tex]m^{3}[/tex]
Hence, the volume occupied by 18 kg of ethylene at [tex]55^{o}C[/tex] and 35 bar is 0.779 [tex]m^{3}[/tex].
(b) The given data is as follows.
V = 0.25 [tex]m^{3}[/tex]
T = [tex]50^{o}C[/tex] = (50 + 273) K
= 323 K
P = 115 bar = 11500 kPa
Using ideal gas equation first, we will calculate its moles as follows.
n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{11500 \times 0.25}{8.314 \times 323 K}[/tex]
= 1.07 mol
Since, moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
Hence, mass of ethylene will be calculated as follows.
moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
1.07 mol = [tex]\frac{mass}{28 g/mol}[/tex]
mass = 29.96 g
Therefore, mass of given ethylene is 29.96 g.