Question

3.0 kg of nitrogen gas in a piston cylinder device initially at a temperature of 300 K and a pressure of 100 kPa is compressed to a volume equal to 90% of the initial volume and at a pressure of 140 kPa. Determine the change in specific internal energy of the nitrogen during the process, assuming that the nitrogen acts as an ideal gas with constant specific heats.

Answer 1

ΔU = 103.54 KJ

Step-by-step explanation:

∴ ΔU = Q + W

ideal gas:

∴ PV = nRT

∴ R = 8.314 L.KPa/K.mol

∴ n = 3000 g N2 * ( mol/28.0134g N2) = 107.143 mol N2

∴ m N2 = 3.0 Kg

∴ T1 = 300 K

∴ P1 = 100 KPa

∴ V1 = nRT1/P1 = 2672.36 L = 2.67 m³

⇒ V2 = 0.9*V1 = 2405.12 L = 2.41 m³

∴ P2 = 140 KPa

⇒ T2 = P2.V2/n.R = 377.99 ≅ 378 K

⇒ W = P1V1 - P2V2

⇒ W = ((100KPa)*(2.67m³)) - ((140KPa)*(2.41m³))

⇒ W = - 70.164 KJ

∴ Q = nCpΔT

∴ Cp = (5/2)*R = 20.785 J/mol.K ....ideal gas

⇒ Q = (107.143mol)*(20.785 J/mol.K)*(378 - 300)

⇒ Q = 173703.446 J = 173.703 KJ

⇒ ΔU = 173.703 KJ - 70.164 KJ

⇒ ΔU = 103.54 KJ