Oxygen at 100kPa and 250 degrees C is compressed to half its

Oxygen at 100kPa and 250 degrees C is compressed to half its initial volume. Determine the final state in terms of pressure (p2) and temperature (T2). Use the PG model for oxygen.

a) the compression is carried out in an isobaric manner

b) the compression is carried out in an isothermal manner

c) the compression is carried out in an isentropic manner

Solution

Initial pressure P = 100 x10 3 Pa

Initial temprature T = 250 o C

                           = 250 + 273

                          = 523 K

Initial volume V = V

Final volume V \' = V/2

(a).In isobaric process ,

final pressure P \' = P       Since in isobaric process ,pressure remains constant.

Final temprature T \' = ?

At constant pressure V \' / V = T \' / T

From this T \' = ( V \' / V ) T

                    = (1/2) T

                    = 261.5 K

(b). Isothermal process:

In isothermal process , temprature remains constant .

So, final temprature T \' = T = 523 K

At constant temprature , PV = P \' V \'

From this final pressure P \' = PV / V \'

                                       = P(V/(V/2))

                                       = 2P

                                       = 200 kPa

(c).Isentropic process:

In this process PV ^r = P \'V \' ^r

Where r = ratio of specific heats = 1.4    Since oxygen is diatomic

             P \' = P (V/V \') r

                 = 100 kPa (2) ^1.4

                 = 100kPa (2.639)

                 = 263.9 kPa

Aslo ,TV ^r-1 = T \' V \' ^r-1

     T \' = T(V/V \') ^r-1

         = 523 (2)^0.4

         = 523(1.319)

         = 690.1 K