0,245 kg of gas with a density of 1,2 kg/m is compressed adiabatically from a pressure of 150 kPa to a pressure of 800 kPa. R = 0,293 kJ/kg. K and Cp = 1,005 kJ/kg. K. Calculate the following: 2.1 The adiabatic index 2.2 The original volume and temperature 2.3 The final absolute temperature 2.4 The work done in kJ/kg 2.5 The change in internal energy

a) ∝ = 2.43

b) V = 0.2042 m³; T = 426.69 K

c) T final = 695 K

d) W = 340.88 KJ/Kg

e) ΔU = 191.04 KJ/Kg

Explanation:

b) PV = nRT

Volume of gas = 0.245 g * (m³ / 1,2 Kg) = 0.2042 m³ (204.16 L)

Temperature:

∴ T = PV / nR = ((150 KJ / m³) * (0.2042 m³)) / ((0.245 Kg) * (0.293 KJ/Kg. K))

⇒ T = 426.69 K

a) adiabatic index (∝):

∝ = Cv / Cp

∴ Cv = Cp - nR ... ideal gas ... n = 1

⇒ Cv = 1.005 KJ/Kg. K - 0.293 KJ/Kg. K

⇒ Cv = 0.712 KJ/Kg. K

⇒ ∝ = 0.712 / 0.293 = 2.43

c) final temperature (T2):

∴ (T2/T1) ∧ ((R+Cv) / R) = (P2/P1) ... ideal gas compressed

⇒ (R + Cv) / R = (0.293 + 0.712) / (0.293) = 3.43

⇒ (T2 / T1) ∧ (3.43) = (P2/P1) = 800 / 150 = 5.33

⇒ (T2 / T1) = (5.33) ∧ (1 / 3.43) =

⇒ T2 = 1.628 * 426.69 K

⇒ T2 = 695 K

d) W = - nRTLn (P1/P2) ... n=1

⇒ W = - (0.293 KJ/Kg. K) * (695 K) * Ln (150/800)

⇒ W = 340.88 KJ/Kg

e) ΔU = Cv ΔT ... constant volume

⇒ ΔU = (0.712 KJ/Kg. K) * (695 - 426.69) K

⇒ ΔU = 0.712 * 268.31 = 191.04 KJ/Kg