Question

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

Answer 1

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

Step-by-step 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