Question

Please consider 1 kg of NH3 that was compressed from 2.5 bar and 30°C to 5.0 bar in a well-insulated compressor. Please determine the (a) final temperature in C and (b) the work required in kJ.

Answer 1

Explanation:

(a)   The given data is as follows.

               mass = 1 kg = 1000 g        (as 1 kg = 1000 g)

          Molar mass of $NH_{3}$ = 17 g/mol

           $P_{1}$ = 2.5 bar = $2.5 \times 10^{5} Pa$    (as 1 bar = $10^{5} Pa$)

            $P_{2}$ = 5 bar = $5 \times 10^{5} Pa$

          $T_{1} = 30^{o}C$ = 30 + 273 = 303 K

For adiabatic process,   $PV^{\gamma}$ = constant = k

             $\gamma$ = 1.33 = $\frac{C_{p}}{C_{v}}$        

              $P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$

                $(\frac{V_{2}}{V_{1}})^{\gamma} = \frac{P_{1}}{P_{2}}$

        $\frac{V_{2}}{V_{1}} = (\frac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

    $V_{2} = (\frac{2.5 \times 10^{5}}{5 \times 10^{5}})^{\frac{1}{1.33}} \times \frac{nRT_{1}}{P_{1}}$            (as PV = nRT)

                 = $(\frac{2.5}{5})^{\frac{1}{1.33}} \times \frac{58.82 \times 8.314 J/mol K \times 303 K}{2.5 \times 10^{5}}$

                = 0.352 $m^{3}$

Also,       w = $\frac{P_{1}V_{1} - P_{2}V_{2}}{\gamma - 1}$

                  = $\frac{2.5 \times 10^{5} \times 0.5927 - 5 \times 10^{5} \times 0.352}{1.33 - 1}$

                  = -84318.2 J

As 1 kJ = 1000 J. So, -84318.2 J = -84.318 kJ

Hence, the work required in kJ is -84318.2 J.

(b)    It is known that for adiabatic system Q = 0,

                           $\Delta U = Q - w$

                          $nC_{v}dT$ = -w

                             dT = $\frac{-w}{nC_{v}}$

                                  = $\frac{84318.2}{58.82 \times 1.6}$

                                  = 895.93 K

We known that dT = $T_{1} - T_{2}$

so,                   895.93 = 303 K - $T_{2}$                            

                         $T_{2}$ = (895.93 - 303)K

                                     = 592.93 K

                                     = (592.93 - 273.15)^{o}C

                                     = $319.78^{o}C$

Hence, the final temperature is $319.78^{o}C$.