As the expansion takes place at a constant pressure, the change in enthalpy is given by
ΔH = ΔU + P(V2 – V1)
ΔH = Change in enthalpy = ?
ΔU = Change in internal energy = 418 J
P = Constant pressure = 2.026 × 105 Pa
V2 = 15 dm3 = 15 × 10-3 m3
V1 = 5 dm3 = 5 × 10-3 m3
∴ ΔH = 418 + 2.026 × 105 × (15 × 10-3 – 5 × 10-3)
= 418 + 2026
∴ ΔH = 2.444 × 103 J
= 2.444 kJ
∴ Change in enthalpy = 2.444 × 103 J
= 2.444 kJ
