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

Explanation:

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

   Molecular weight ($NH_{3}$) = $14 + 3 \times 1$ = 17

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

     $0.5 kmol \times (\frac{1000 mol}{1 kmol}) \times (\frac{17 g}{1 mol})$

                       = 8500 g

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

            Weight = mass × g

             = $8500 g \times (\frac{1 kg}{1000 g}) \times (9.8 m/s^{2})$

             = 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.

            $v (m^{3}/kmol) = \frac{V}{n}$

Therefore, calculate the specific volume as follows.

       $V_m = \frac{6 m^{3}}{0.5 k mol}$

                   = 12 $m^{3}/k mol$

Also,  

               $v (m^{3}/kmol) = \frac{V}{m}$  

            v = $\frac{6 m^{3}}{8.5 kg}$

               = 0.705882 $m^{3}/kg$

Therefore, we can conclude that the value of specific volume is 12 $m^{3}/k mol$  and 0.705882 $m^{3}/kg$.

Answer 2

Answer:

a) $w=83.385\ N$

b) $\bar V=12\ m^3.kmol^{-1}$

$\b V=0.7059\ m^3.kg^{-1}$

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:

$M=14+3\times 1=18\ g.mol^{-1}$

Now the mass of given ammonia:

$m=n.M$

$m=500\times 17$

$m=8500\ g=8.5\ kg$

a)

Now weight:

$w=m.g$

$w=8.5\times 9.81$

$w=83.385\ N$

b)

Specific volume:

$\bar V=\frac{6}{0.5}$

$\bar V=12\ m^3.kmol^{-1}$

also

$\b V=\frac{V}{m}$

$\b V=\frac{6}{8.5}$

$\b V=0.7059\ m^3.kg^{-1}$