First take the case where the heat is lost from a constant volume system Then dQv can be identified with dU if there is no non-pV work involved. Putting this into eq gives dS-dUT全0.0r0全dU-TdS du<TdS (const. V, no non-pV work) When the heat is lost from the system under conditions of constant pressure and when no non )V ork is involved 4上一内容下一内容◇回主目录 返回 2021/2/21上一内容 下一内容 回主目录 返回 2021/2/21 First take the case where the heat is lost from a constant volume system. Then dQv can be identified with dU if there is no non-pV work involved. Putting this into eq.gives dS - dU/T ≧0, or 0 ≧ dU- T dS dU≤TdS (const. V, no non-pV work) When the heat is lost from the system under conditions of constant pressure and when no non￾pV work is involved