InterviewSolution
Saved Bookmarks
| 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 J = 2.444 kJ ∴ Change in enthalpy = 2.444 × 103 J = 2.444 kJ |
|