Question

Two moles of an ideal gas undergo an isothermal expansion at 565 K from a pressure of 12.5 Bar to a final pressure of 1.50 Bar. Calculate AU, AH, and AS for the process if Cy = R. The same ideal gas undergoes an adiabatic expansion from the same initial pressure to the same final pressure (and the same initial temperature). Calculate the final temperature, AU, AH, and AS for the process.

Answer 1

P2 Answer to the following question or n = 2 moles , c = ER. Initial pressure (4) = 12.5 Bar (Iso thermal expansion) Final Pressure (P2) Temp constant s = 1:50 Bar change of internal energy (au) = 0 au - Cviar Gr. 0 = 0 lot as= nepen I2 + MR en b Pe ( As temp constant, so, Ti=tz e as = nhen & (R=0.083 " L.Bar.moilk) - 2x 0.083 en 12.5 45 = 0.351 L.Bar. K'. Now we know I atm = 1.013 bar So, as = 0.352 (l.atm) KI - 0.352 x 101.325 Š /K. s 35.208 J/K (I hatm = 101 3255) As the process undergoes via iso thermal expansion, change of temperature (87) =0 AH = (p AT = (pio =0. so, AH = 40 =0, AS = 350208 g/k MOTO) HOMI L 1-013 12013

2) Ideal gas undergoes an adiabatic expansion. So, dq = 0, so as = ds = (dal = 0. Initial Temperature (t) = 565k. Now Cv = 58 cp = Gt R = R + R = 1/R ar cel = 4/5 Now for Adiabatic process to p. vr = constal = 12.5 bar =) P. (RT) = const IT, = 565k P2 = 1.5 bar =) pl-. t = const. The final Temp = ? => 0 ) - -0.) = +2 = 565x (41 ) 308-28k & Final Temperative – 308-28 K. " Au znow. AT = 2 x R. (308:28-565) - 5*8.314 x ( 308.28 - 565) = - 10671.85 j / moi = – 10. 67185 kj/mol 4H ZnCp. 4T = xx 7 x 8 +3 14 x (308-28-565) 7 748-314x (308-28 – 565) = - 14940-59 5 (mos - - 14,94059 Kj/mol