Question

Ammonia gas is compressed from 21°C and 200 kPa to 1000 kPa in an adiabatic compressor with an efficiency of 0.82. Estimate the final temperature, the work required, and the entropy change of the ammonia.

Answer 1

Explanation:

It is known that efficiency is denoted by $\eta$.

The given data is as follows.

     $\eta$ = 0.82,       $T_{1}$ = (21 + 273) K = 294 K

     $P_{1}$ = 200 kPa,     $P_{2}$ = 1000 kPa

Therefore, calculate the final temperature as follows.

         $\eta = \frac{T_{2} - T_{1}}{T_{2}}$    

         0.82 = $\frac{T_{2} - 294 K}{T_{2}}$    

          $T_{2}$ = 1633 K

Final temperature in degree celsius = $(1633 - 273)^{o}C$

                                                            = $1360^{o}C$

Now, we will calculate the entropy as follows.

       $\Delta S = nC_{v} ln \frac{T_{2}}{T_{1}} + nR ln \frac{P_{1}}{P_{2}}$

For 1 mole,  $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

     $\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

                = $0.028 kJ/mol \times ln \frac{1633}{294} + 8.314 \times 10^{-3} kJ \times ln \frac{200}{1000}$

                = 0.0346 kJ/mol

or,             = 34.6 J/mol             (as 1 kJ = 1000 J)

Therefore, entropy change of ammonia is 34.6 J/mol.