Respuesta :
Explanation:
It is known that the molecular weight of ammonia ([tex]NH_{3}[/tex]) is as follows.
Molecular weight ([tex]NH_{3}[/tex]) = [tex]14 + 3 \times 1[/tex] = 17
(a) Therefore, we will calculate the mass as follows.
[tex]0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})[/tex]
= 8500 g
Now, formula to calculate weight of the system in N is as follows.
Weight = mass × g
= [tex]8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})[/tex]
= 83.3 N (1 [tex]kg m/s^{2}[/tex] = 1 N)
Hence, the weight of the system is 83.3 N.
(b) Relation between specific volume and number of moles is as follows.
[tex]v (m^{3}/kmol) = \frac{V}{n}[/tex]
Therefore, calculate the specific volume as follows.
[tex]V_m = \frac{6 m^{3}}{0.5 k mol}[/tex]
= 12 [tex]m^{3}/k mol[/tex]
Also,
[tex]v (m^{3}/kmol) = \frac{V}{m}[/tex]
v = [tex]\frac{6 m^{3}}{8.5 kg}[/tex]
= 0.705882 [tex]m^{3}/kg[/tex]
Therefore, we can conclude that the value of specific volume is 12 [tex]m^{3}/k mol[/tex] and 0.705882 [tex]m^{3}/kg[/tex].
Answer:
a) [tex]w=83.385\ N[/tex]
b) [tex]\bar V=12\ m^3.kmol^{-1}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
Explanation:
Given:
no. of moles of ammonia in a closed system, [tex]n=0.5\ kmol=500\ mol[/tex]
volume of ammonia, [tex]V=6\ m^3[/tex]
We know the molecular formula of ammonia: [tex]NH_3[/tex]
The molecular mass of ammonia:
[tex]M=14+3\times 1=18\ g.mol^{-1}[/tex]
Now the mass of given ammonia:
[tex]m=n.M[/tex]
[tex]m=500\times 17[/tex]
[tex]m=8500\ g=8.5\ kg[/tex]
a)
Now weight:
[tex]w=m.g[/tex]
[tex]w=8.5\times 9.81[/tex]
[tex]w=83.385\ N[/tex]
b)
Specific volume:
[tex]\bar V=\frac{6}{0.5}[/tex]
[tex]\bar V=12\ m^3.kmol^{-1}[/tex]
also
[tex]\b V=\frac{V}{m}[/tex]
[tex]\b V=\frac{6}{8.5}[/tex]
[tex]\b V=0.7059\ m^3.kg^{-1}[/tex]
