Respuesta :
Explanation:
It is known that efficiency is denoted by [tex]\eta[/tex].
The given data is as follows.
   [tex]\eta[/tex] = 0.82,    [tex]T_{1}[/tex] = (21 + 273) K = 294 K
   [tex]P_{1}[/tex] = 200 kPa,   [tex]P_{2}[/tex] = 1000 kPa
Therefore, calculate the final temperature as follows.
     [tex]\eta = \frac{T_{2} - T_{1}}{T_{2}}[/tex]  Â
     0.82 = [tex]\frac{T_{2} - 294 K}{T_{2}}[/tex]  Â
     [tex]T_{2}[/tex] = 1633 K
Final temperature in degree celsius = [tex](1633 - 273)^{o}C[/tex]
                              = [tex]1360^{o}C[/tex]
Now, we will calculate the entropy as follows.
    [tex]\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}[/tex]
For 1 mole, Â [tex]\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}[/tex]
It is known that for [tex]NH_{3}[/tex] the value of [tex]C_{v}[/tex] = 0.028 kJ/mol.
Therefore, putting the given values into the above formula as follows.
   [tex]\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}[/tex]
        = [tex]0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}[/tex]
        = 0.0346 kJ/mol
or,       = 34.6 J/mol       (as 1 kJ = 1000 J)
Therefore, entropy change of ammonia is 34.6 J/mol.
