3.1 State functions 1). Exact and inexact differential Internal In path 2. the initial and final states are energy, U Path 2 the same but in which the expansion W≠0,q≠0 Path 1 W≠0,q=0 not adiabatic. In this path an energy q enters the system as heat and the work w'is not the same as w. The work and Temperature.T the heat are path functions. q i, path Volume, V 版权所有:华东理工大学物理化学教研
版权所有：华东理工大学物理化学教研室 6 In Path 2, the initial and final states are the same but in which the expansion is not adiabatic. In this path an energy q' enters the system as heat and the work w' is not the same as w. The work and the heat are path functions. 1). Exact and inexact differential = f i q q ,path d 3.1 State functions
3.1 State functions 1). Exact and inexact differential AU, independent of the path, an exact differential. An exact differential is an infinitesimal quantity which, when integra ted, gives a result that is independent of the path between the initial and final states g or w, dependent on the path, an inexact differential. When a system is heated, the total energy transferred as heat is the sum of all individual contributions at each point of the path q dq 版权所有:华东理工大学物理化学教研
版权所有：华东理工大学物理化学教研室 7 △U, independent of the path, an exact differential.An exact differential is an infinitesimal quantity which, when integrated, gives a result that is independent of the path between the initial and final states. q or w, dependent on the path,an inexact differential.When a system is heated, the total energy transferred as heat is the sum of all individual contributions at each point of the path. 1). Exact and inexact differential △q qf - qi q dq dq 3.1 State functions
D) Example -Calculating work, heat, and internal energy Consider a perfect gas inside a cylinder fitted with a piston. Let the initial state be t, vi and the final state be t, vf. The change of state can be brought about in many ways, of which the two simplest are following: Path 1, in which there is free expansion against zero external pressure; Path 2, in which there is reversible isothermal expansion. Calculate w, q, and Au for each process. 版权所有:华东理工大学物理化学教研
版权所有：华东理工大学物理化学教研室 8 Consider a perfect gas inside a cylinder fitted with a piston. Let the initial state be T, Vi and the final state be T, Vf . The change of state can be brought about in many ways, of which the two simplest are following: Path 1, in which there is free expansion against zero external pressure; Path 2, in which there is reversible, isothermal expansion. Calculate w, q, and U for each process. Example -Calculating work, heat, and internal energy
D) Example -Calculating work, heat, and internal energy Method: To find a starting point for a calculation in thermodynamics, it is often a good idea to go back to first principles, and to look for a way of expressing the quantity we are asked to calculate in terms of other quantities that are easier to calculate. because the internal energy of a perfect gas arises only from the kinetic energy of its molecules, it is independent of volume; therefore, for any isothermal change, in general△U=q+w 版权所有:华东理工大学物理化学教研
版权所有：华东理工大学物理化学教研室 9 Method: To find a starting point for a calculation in thermodynamics, it is often a good idea to go back to first principles, and to look for a way of expressing the quantity we are asked to calculate in terms of other quantities that are easier to calculate. Because the internal energy of a perfect gas arises only from the kinetic energy of its molecules, it is independent of volume; therefore, for any isothermal change, in general U = q+w. Example -Calculating work, heat, and internal energy
D) Example -Calculating work, heat, and internal energy Answer: Since△U=0 for both paths and△U=q+w. The work of free expansion is zero in Path 1, w=0 and g=0; for Path 2, the work is given by w=-nRr/ dv Jy v=-nRTInkr SO w=- nRT In Vr/vi, and g=nRTIn (r/vi. 版权所有:华东理工大学物理化学教研
版权所有：华东理工大学物理化学教研室 10 Answer: Since U = 0 for both paths and U = q+w. The work of free expansion is zero, in Path 1, w = 0 and q = 0; for Path 2, the work is given by so w = - nRT ln (Vf /Vi ), and q = nRT ln (Vf /Vi ). i f ln f d i V V nRT V V w nRT V V = − = − Example -Calculating work, heat, and internal energy