Thermodynamics Question 14 Follow Link Below for question pl

Thermodynamics

Question 14: Follow Link Below for question please

http://imgur.com/rugJywB

A rigid tank contains 1.86 kg of R134a at -32degreeC. The volume of the tank is 0.22919 m The R134a is heated until its pressure reaches 200 kPa. Determine the initial pressure and final temperature.

Solution

Mass m = 1.86 kg

Molecular weight M = 102.03 g / mol

                             = 0.10203 kg / mol

Number of moles n = 18.229 mol

Initial temprature T = -32 ^o C

                            = -32 + 273

                            = 241 K

Initial volume V = 0.22919 m ³

Initial pressure P = nRT / V

where R = Gas constant = 8.314 / mol K                 

Substitute values you get P = (18.229)(8.314)(241) /(0.22919)

                                         = 159.36 x10 ³ Pa

Final pressure P \' = 200 kPa = 200 x10 ³ Pa

At constant volume , T \' / T = P \' / P

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

                    = 1.254 x 241

                   = 302.5 K

                   = 302.5 -273

                   = 29.5 ^oC