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Company LOGO THERMODYNAMICS 1 The First Law and Other Basic Concepts (part 1) Department of Chemical Engineering, Semarang State University Dhoni Hartanto S.T., M.T., M.Sc. Joule’s Experiments Understanding of heat and its relation to work developed during the last half of the nineteenth century James P. Joule carried out the experiment in Manchester, England. His experiment is simple enough with good accuracy. He placed known amounts of water, oil, and mercury in an insulated container and agitated the fluid with a rotating stirrer. The work done on the fluid by stirrer were accurately measured and the temperature changes of fluid were carefully noted. Joule found for each fluid fixed amount of work was required per unit mass for every degree of temperature rise caused by the stirring, and that the original temperature of the fluid could be restored by the transfer of heat through simple contact with a cooler object Joule concluded that heat is a form of energy Internal Energy In Joule’s experiment, we can see that energy is added into a fluid, and it is transferred from the fluid as heat. Why ?? Another form called internal energy is contained in the fluid. The internal energy of a substance does not include energy that it may possess as a result of its macroscopic position or movement. Internal energy refers to energy of the molecules internal to the substances All molecules possess kinetic energy of translation, except monoatomic molecules which also have kinetic energy of rotation and internal vibration. The addition of heat or work to a substance increases molecular activity and then increases internal energy Internal Energy (cont.) The internal energy of a substance also includes the potential energy resulting from intermolecular forces. On a submolecular scale energy is associated with the electrons and nuclei of atoms, and with bond energy resulting from the forces holding atoms together as molecules. This form of energy is named internal to distinguish it from the kinetic and potential energy associated with a substance because of its macroscopic position or motion, which can be thought of as external forms of energy. Internal energy, has no concise thermodynamic definition. It is a thermodynamic primitive. It cannot be directly measured; there are no internal-energy meters. As a result, absolute values are unknown. However, this is not a disadvantage in thermodynamic analysis, because only changes in internal energy are required The First Law of Thermodynamics “Although energy assumes many forms, the total quantity of energy is constant, and when energy disappears in one form it appears simultaneously in other forms” In application of the first law of the given process, it divided into two parts : 1. System 2. Surrounding In its most basic form, the first law requires : ∆(Energy of the systems) + ∆(Energy of the surroundings) = 0 The system may change in its internal energy, in its potential or kinetic energy, and in the potential or kinetic energy of its finite parts. The First Law of Thermodynamics (cont.) In the thermodynamic, heat and work refer to energy in transit across the boundary which divides the system from its surroundings These forms of energy are not stored, and are never contained in a body or system Energy is stored in its potential, kinetic, and internal forms; these reside with material objects and exist because of the position, configuration, and motion of matter Energy Balance for Closed Systems Closed systems : no transfer of matter between systems in its surroundings no streams enter or leave in systems no internal energy is transported accros the boundary of the systems surroundings systems Closed Systems surroundings energy systems energy mass Open Systems All energy exchange between a closed system and its surroundings then appears as heat and work Total energy change of equals the net energy transferred to or from it as heat and work Energy Balance for Closed Systems (cont.) ∆(Energy of the surroundings) = ± Q ± W Sign positive means transfer into systems from surroundings The equation become ∆(Energy of the surroundings) = Q + W The equation means that the total energy change in closed system equals to the net energy transferred into it as heat and work. Energy Balance for Closed Systems (cont.) Closed systems often happened in the systems which no change occurred other than its internal energy. ∆Ut = Q + W Where Ut total internal energy of the system in finite changes. For differential changes, the equations yield dUt = dQ + dW Energy Balance for Closed Systems (cont.) Properties such as Vt and Ut depend on the quantitiy of material in a systems, it is called extensive properties. Properties in homogeneous fluid such as temperature and pressure are not depend on the quantitiy of the materal in a systems, it is called intensive properties. Although Vt and Ut for a homogeneous systems of arbitrary size are extensive properties, specific and molar volume V (or density) and specific and molar internal energy U are intensive properties. For closed system of n moles, ∆(nU) = n ∆U = Q + W d(nU) = n dU = dQ + dW When n = 1 mole, thus ∆U = Q + W dU = dQ + dW Energy Balance for Closed Systems (cont.) The equation connect the internal energy to measurable quantities. It does not represent a definition of internal energy nor lead to absolute values of internal energy, it is just provide calculation of internal energy changes Example Water flows over a waterfall 100 m in height. Take 1 kg of the water as system and assume that it does not exchange energy with its surroundings a) What is the potential energy of the water at the top of the falls with respect to base of the falls? b) What is the kinetic energy of the water just before it strikes bottom? c) After 1 kg of water enters the river below the falls, what change has occured in its state? Energy Balance for Closed Systems (cont.) Solution : No exchange with surroundings ∆(Energy of the system) = 0 Energy : internal, kinetic, and potential ∆U + ∆Ek + ∆Ep = 0 a) Ep = m g z = 1 kg x 100 m x 9.8066 m s-2 = 980.66 kg m2/ s2 = 980.66 N m = 980.66 J b) No mechanism exist for conversion of potential or kinetic energy into internal energy during free fall (∆U must be zero) ∆Ek + ∆Ep = Ek2 – Ek1 + Ep2 – Ep1 = 0 Assumtion Ek1 = Ep2 = 0, then Ek2 = Ep1 = 980.66 J Energy Balance for Closed Systems (cont.) Solution (cont.) : c) When 1 kg of water strikes bottom and mixed with other falling water, kinetic energy is converted into internal energy. During this process, ∆Ep = 0. ∆U + ∆Ek = 0 or ∆U =Ek2 – Ek3 The river velocity is small, so that Ek3 is negligible. Thus ∆U =Ek2 = 980.66 J The temperature increase occured. (Energy required for a temperature rise of 1 oC is 4184 J kg-1). Thus the temperature increase is 980.66/4184 = 0.234 oC (assumption : no heat transfer occured with surrounding) Thermodynamic state and state functions The internal energy term on the left side of equation reflect changes in the internal state or the thermodynamic state of the system. This state is reflected by its thermodynamic properties, among which are temperature, pressure, and density. For a homogeneous pure substance, fixing two of these properties automatically fixes all the others i.e Nitrogen gas at temperature 300 K and a pressure 105 kPa (1 bar) has a fixed specific volume or density and fixed molar internal energy. If this gas is heated or cooled, compressed or expanded, and then returned to its initial temperature and pressure, its intensive properties are restored to their initial values. Thermodynamic state and state functions (cont.) Such properties do not depend on the past history of the substance nor on the means by which it reaches a given state They depend only on present conditions, and it is called state functions. State functions such as internal energy is a property that always has a value, it may be expressed mathematically as a function of other thermodynamics properties such as temperature and pressure, or temperature and density. The right side of equation represents heat and work which is not properties. They account for the energy changes that occur in the surroundings and appear only when changes occur in a system. Thermodynamic state and state functions (cont.) The differential of a state function represent as infinitesimal change in its value. In example : P2 P1 dP P2 P1 P V2 and V1 dV V2 V1 V The differential of heat and work are not change, but are infinitesimal amounts. When integrated, these differentials give not finite changes, but finite amounts. Thus, dQ Q and dW W For closed system, the amounts of heat and work required differ for different process, But that the sum of Q +W is the same for all processes. This is the basis for identification of internal energy as a state function. Thermodynamic state and state functions (cont.) Example : A gas is confined in a cylinder by a piston. The initial pressure of the gas is 7 bar, and the volume is 0.10 m3. The piston is held in place by latches in the cylinder wall. The whole apparatus is placed in a total vacuum. What is the energy change of the apparatus if the restraining latches are removed so that the gas suddenly expands to double its initial volume, the piston striking other latches at the end of the process? Solution : System : the gas, piston, and cylinder. No work done during the process (no force external moves to the system, no heat is transferred) Q and W are 0 Total energy of the system doesn’t change It can not explain more about the distribution of energy among the parts of the system when further information are not available Thermodynamic state and state functions (cont.) Example : If the process in previous example is repeated, not in vacuum but in air at atmospheric pressure of 101.3 kPa, what is the energy change of the appratus? Assume the rate of heat exchange between the apparatus and the surrounding air is slow compared with the rate at which the process occurs. Solution : System : the gas, piston, and cylinder. Work is done by the system in pushing back the atmosphere Work is given by the product of the force exerted by atmospheric pressure on the back side of the piston and the displacement of the piston. Thermodynamic state and state functions (cont.) Solution (cont.): Work done by system : F ∆l= Patm ∆Vt = (101.3 kPa) (0.2-0.1) kPa m3 = 10.13 kNm Since work done on the system, so that W = -10.13 kNm = -10.13 J Heat transfer is possible occur on the system, but the problem is worked for the instant before heat transfer take place. Thus Q is assumed to be 0. ∆(Energy of the system) = Q + W = 0 – 10.13 = -10.13 J Total energy of the system has decreased by amount equal to work done on the surrounding Thermodynamic state and state functions (cont.) Example : When a system is taken from state (a) to state (b) in Figure below along path (acb), 100 J of heat flows into the system and the system does 40 J of work. a) How much heat flows into the system along path (aeb) if the work done by the system is 20 J? b) The system returns from (b) to (a) along path (bda). If the work done on the system is 30 J, does the system absorb or liberate the heat? How much? Thermodynamic state and state functions (cont.) Solution: Assume the system change only in internal energy. t For path (acb) U ab Qacb Wacb 100 40 60 J The internal energy change from (a) to (b) by any path a) Path (aeb) t U ab 60 Qaeb Waeb Qaeb 20 Qaeb 80 J b) Path (bda) t t U ba U ab 60 Qbda Wbda Qbda 30 Qbda 60 30 90 J Heat is transferred from the system to the surroundings Thank you