PR Om Aad

Satu kilogram udara dipanaskan secara reversibel pada tekanan konstan pada keadaan awal 300K dan 1 bar sampai volumenya menjadi tiga kali dari semula. Hitunglah W, Q, ∆U, dan ∆H untuk proses itu! Asumsikan untuk udara PV/T = 83,14 bar.cm3mol-1K-1 dan Cp = 29 mol-1K-1.
 
            Penyelesaian
*  Diketahui   :  m  = 1 kg  = 1000 g
                          P   =  konstant
                          T1  = 300 K
                          P1   = 1 bar
                           V2   = 3 V1  
                          PV/T = 83,14 bar.cm3mol-1K-1
                           Cp = 29 mol-1K-1
                          BM udara 28,97 g/gmol
*  Ditanya  :  W, Q, ∆U, dan ∆H
*  Jawab      : 
                        Pada keadaan 1
                                    PV/T = 83,14 bar.cm3mol-1K-1
                                    1(bar)* V1 / 300 (K)   = 83,14 bar.cm3mol-1K-1
                                                >> V1   = 24.942 cm3/mol
 
                                    Jadi   V2   = 3 V1  
                                                     =  3* 24.942
                                                     =  74.826 cm3mol-1
 
                        Pada keadaan 2
                                    PV/T = 83,14 bar.cm3mol-1K-1
                                    1(bar)* 74.826 (cm3mol-1) / (T2) (K)   = 83,14 (bar.cm3mol-1K-1)
                                                >> T2 =  900 (K)
 
                        Jumlah mol udara
                                    n  = m/BM
                                        = 1000/28,97
                                        =  34,52 mol
 
a.  W = – dV  = – dV =  – ( V2-V1)   =  – ( 74.826 – 24.942) =  49.884 barcm3mol-1
      maka untuk 1000 g udara >> Wt =  n . W
                                                           =  34,52 (mol) * 49.884 barcm3mol-1
                                                           =  1.721.995,68 barcm3
                                                =  1.721.995,68 (bar)* (105Nm-2)/ 1 bar* 1 cm3(10-6m3)
                                                = 17.219.956,8 Nm
                                                =  17.219.956,8 J 
 
b.  ∆H = dT  =  dT  = 29 ( T2 – T1) = 29 ( 900-300)  = 17.400 mol-1
     maka  untuk 1000 g udara ∆Ht = n. ∆H
                                                       = 34,52 * 17.400
                                                       = 600.648 J
 
c.   Q  = n. ∆Ht
            = 34,52 * 600.648
            =  20,73.106 J
 
d.  ∆U  = Q + W
            =  20,73.106 + 17.219.956,8 J 
            = 37,95.106J

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