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

Answer:

(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.

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

$M=114.28\times0.3\times1000$

$M=34.284\ kg$

We need to calculate the weight of the system

Using formula of weight

$W=mg$

Put the value into the formula

$W=34.284\times9.81$

$W=336.32\ N$

(b). We need to calculate the molar volume

Using formula of molar volume

$\text{molar volume}=\dfrac{volume}{volume of moles}$

Put the value into the formula

$\text{molar volume}=\dfrac{5}{0.3}$

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

We need to calculate the mass based volume

Using formula of mass based volume

$\text{mass based volume}=\dfrac{volume}{mass}$

Put the value into the formula

$\text{mass based volume}=\dfrac{5}{34.284}$

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

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.