1.

An ideal gas expands against a constant pressure of 2.026 × 105 Pa from 5 dm3 to 15 dm3. If the change in the internal energy is 418 J, calculate the change in enthalpy.

Answer»

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

= 2.444 kJ

∴ Change in enthalpy = 2.444 × 103 J

= 2.444 kJ



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