Question

The following problems summarize this section: Problem 3.4 - A frictionless piston-cylinder device contains 0.2 kg of air at 100 kPa and 27°C. The air is now compressed slowly according to the relation P VK = constant, where k = 1.4. until it reaches a final temperature of 77°C. . a) Sketch the P-V diagram of the process with respect to the relevant constant temperature lines, and indicate the work done on this diagram. • b) Using the basic definition of boundary work done determine the boundary work done during the process [-7.18 kJ]. . c) Using the energy equation determine the heat transferred during the process [0 kJ], and verify that the process is in fact adiabatic. Derive all equations used starting with the basic energy equation for a non-flow system, the equation for internal energy change or an ideal gas (Au), the basic equation for boundary work done, and the ideal gas equation of state [P.V = m. R.T]. Use values of specific heat capacity defined at 300K for the entire process.

Answer 1

solution air in Initial state of piston cylinder m= 0.2 kg P = 100 kPa + T = 27°C = 273 15+ 272 = 300: 15k. how by ideal gas equation PV - MRT R = 0.289 kJ/kak V . MRT 0.2 x 287 x 18 x 300.15 100X103 -0.17228 m3 nt lotostrada = Initial volume = 0.17 228 m². slowly since air is compressed to temperature 77° c otori 273:15+ 22 - 350.15 K. . B » 1 2 = MRT₂ P = 2X. 2.87 Xlo?x350.18 = * Pz V2

> ļ = 20098.6) also it process is given that follows law prk = constant A + K = 1. 4 tre 1.4 = 100 710x10 17228) 2009 8.61 XV 100 x 10+ x( 18220)"9. Long a 20098.61 ✓ = (0.42415) 0-1179 m2. 20098.61 0.11719 = 171504.4799. ~ 171. SO4Krau

171.504 100 -. work donu. "11719 1998 v m? (b) Now we know that boundary work abo puka c m P = ccrrrroos » Wir ser du V du can sf cs , * = 100X10% (19228)4 - 880410 = 8525.675

W-2 = 8525.695,- 1._ , '-luJ 1-1.4 ay 18525.675 2 520s content of the app . 228) .4 == 8525.675 = -8525.675 -112,599.19928""] 2 · 11719 - 19228 4 = -7177. 283 J RIMO RDINADOR = -7-177 KJ ^ -7.18 kJ (c) from first law of thermodynamics we have so = swt Du = 718 kJ / kgk at 300k. sa=-7.18+ mo DT E -7.18 + :2 X.728 [ 350:15- 3000157 = -7.18 + 7.18 = 0 » Adiabatic procen :