Question

A closed system consists of 0.5 kmol of ammonia occupying a volume of 6 m3. Determine (a) the weight of the system, in N, and (b) the specific volume, in m3/kmol and m3/kg. Let g 5 9.81 m/s2.

Answer 1

a)
`w=83.385\ N`

b)
`\bar V=12\ m^3.kmol^(-1)`

`\b V=0.7059\ m^3.kg^(-1)`

Step-by-step explanation:

Given:

no. of moles of ammonia in a closed system,
`n=0.5\ kmol=500\ mol`

volume of ammonia,
`V=6\ m^3`

We know the molecular formula of ammonia:
`NH_3`

The molecular mass of ammonia:

`M=14+3* 1=18\ g.mol^(-1)`

Now the mass of given ammonia:

`m=n.M`

`m=500* 17`

`m=8500\ g=8.5\ kg`

a)

Now weight:

`w=m.g`

`w=8.5* 9.81`

`w=83.385\ N`

b)

Specific volume:

`\bar V=(6)/(0.5)`

`\bar V=12\ m^3.kmol^(-1)`

also

`\b V=(V)/(m)`

`\b V=(6)/(8.5)`

`\b V=0.7059\ m^3.kg^(-1)`

Answer 2

Step-by-step explanation:

It is known that the molecular weight of ammonia (
`NH_(3)`) is as follows.

Molecular weight (
`NH_(3)`) =
`14 + 3 * 1` = 17

(a) Therefore, we will calculate the mass as follows.

`0.5 kmol * ((1000 mol)/(1 kmol)) * ((17 g)/(1 mol))`

= 8500 g

Now, formula to calculate weight of the system in N is as follows.

Weight = mass × g

=
`8500 g * ((1 kg)/(1000 g)) * (9.8 m/s^(2))`

= 83.3 N (1
`kg m/s^(2)` = 1 N)

Hence, the weight of the system is 83.3 N.

(b) Relation between specific volume and number of moles is as follows.

`v (m^(3)/kmol) = (V)/(n)`

Therefore, calculate the specific volume as follows.

`V_m = (6 m^(3))/(0.5 k mol)`

= 12
`m^(3)/k mol`

Also,

`v (m^(3)/kmol) = (V)/(m)`

v =
`(6 m^(3))/(8.5 kg)`

= 0.705882
`m^(3)/kg`

Therefore, we can conclude that the value of specific volume is 12
`m^(3)/k mol` and 0.705882
`m^(3)/kg`.