Question

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 = 9.81 m/s2

Answer 1

(a). The weight of the system is 336.32 N.

(b). The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.

Step-by-step explanation:

Given that,

Weight of octane = 0.3 kmol

Volume = 5 m³

(a). Molecular mass of octane

`M=114.28\ g/mol`

We need to calculate the mass of octane

Mass of 0.3 k mol of octane is

`M=114.28*0.3*1000`

`M=34.284\ kg`

We need to calculate the weight of the system

Using formula of weight

`W=mg`

Put the value into the formula

`W=34.284*9.81`

`W=336.32\ N`

(b). We need to calculate the molar volume

Using formula of molar volume

`\text{molar volume}=(volume)/(volume of moles)`

Put the value into the formula

`\text{molar volume}=(5)/(0.3)`

`\text{molar volume}=16.6\ m^3/k mol`

We need to calculate the mass based volume

Using formula of mass based volume

`\text{mass based volume}=(volume)/(mass)`

Put the value into the formula

`\text{mass based volume}=(5)/(34.284)`

`\text{mass based volume}=0.145\ m^3/kg`

Hence, (a). The weight of the system is 336.32 N.

(b). The molar volume is 16.6 m³/k mol.

The mass based volume is 0.145 m³/kg.