A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b) the molar - and mass-based specific volumes, in m³/kmol and m³/kg respectively. Let g=9.81m/s².

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane (m):

Number of mole of octane (n) = 0.3kmol (given)

Molarmass of octane (M) = 114.23kg/kmol

m=n*M

m = (0.3kmol) * (114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane (W):

W=g*m

W = (9.81m/s^2) * (34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) = 5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be:

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg