CH /2015. Tutorial: Thermodynamics

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1 CH /2015 Tutorial: Thermodynamics

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3 Energetic transformations take place within regions of space. A specified collection of material particles enclosed in a specified real or hypothetical boundary is said to constitute a thermodynamic system. The surroundings or environment is labelled as any region outside the system. We do our measuring in the surroundings. The system and surroundings are separated by a boundary. In an isolated system there is no transfer of either energy or matter between the system and the surroundings. In a closed system there is transfer of energy but not of matter across the system/environment boundary. In an open system both energy and matter can be transferred between the system and the surroundings.

4 First Law of Thermodynamics: Internal energy U or E. Internal energy (U or E) : is the total energy of the system at any given time. Comes from the total kinetic and potential energy of molecules which compose the system. Change in internal energy (U or E) : energy change as system goes from an initial state with energy U i to a final state with energy U f. Hence U = U f U i. We suppose that a closed system undergoes a process by which it passes from a state A to a state B. Then the change in internal energy U = U B U A is given by DU = q +W. The first law may be cast in differential form corresponding to the situation here there is an infinitesimal change in internal energy du caused by the addition of an infinitesimal quantity of heat d q and the performance of an infinitesimal amount of work d w on the system. Hence We can write: du = d q + d w. We note that d q and d w are inexact differentials they cannot be evaluated from a knowledge of the initial and final states alone. q U System W Surroundings We area led therefore to the important conclusion that the energy of an isolated system is constant. In an isolated system there is no transfer of heat or work and so q = 0 and W = 0. Hence U = 0 hence U = constant. U qw Increase in internal energy Heat transfer to system Work done on system This is a mathematical statement of the First Law of Thermodynamics.

5 Heat Capacity Increasing temperature increases the internal energy of a system. The exact increase depends upon the heating conditions. Heat cannot be detected or measured directly. There is no heat meter. One way to determine the magnitude of a heat transfer is to measure the work needed to bring about the same change in the thermodynamic state of a system as was produced by heat transfer. Another approach is to deduce the magnitude of a heat transfer from its effects: namely, a temperature change. When a substance is heated the temperature typically rises. For a specified energy q transferred by heating, the magnitude of the resulting temperature change T depends on the heat capacity C of the substance. We can therefore simply measure the heat absorbed or released by a system : We determine the temperature change and use the appropriate value of the heat capacity of the system. Units of C are J K -1. Heat capacity is an extensive property. C q T q CT Can also define specific heat capacity c according to c = C/m where m = mass of substance with units J K -1 g -1 or the molar heat capacity C m as C m = C/n with units: J K -1 mol -1. The heat capacity depends on whether a sample is maintained at constant volume (C = C V ) or constant pressure (C = C P ). The respective molar quantities are C V,m and C P,m. Loss of energy into the surroundings can be detected by noting whether the temperature changes as the process proceeds. This is the principle of a calorimeter. Specific heat capacity c = quantity of heat Required to change the temp of 1g of a substance By 1 K. c q m T Unit : Jg 1 K 1 q cm T Molar heat capacity C m =quantity of heat required to change the temp of 1 mol of substance by 1 K. C m q Unit : Jmol 1 K 1 n T c(h 2 O(l)) = J g 1 K 1 C m (H 2 O(l)) = 75.4 J mol 1 K 1

6 Enthalpy H The change in internal energy is not equal to the energy transferred as heat when the system is free to change its volume. Under such circumstances some of the energy supplied as heat to the system is returned to the surroundings as expansion work. We shall show that under constant pressure conditions, the energy supplied as heat is equal to the change in another thermodynamic property of the system, called the enthalpy. From the first law of thermodynamics for a finite change at constant pressure U qwqp PV Now Hence U U F U I V V F V I U q P F U I qp PV F VI qp PVF PVI U PV U PV H H H F F Here we have introduced the enthalpy function H as H I U PV Hence the heat absorbed at constant pressure equals the increase in enthalpy of the system. I F I q P H For an exothermic reaction U < 0, H < 0, whereas for an endothermic reaction, U > 0 and H > 0.

7 Relationship between U and H. The term V is significant for gases. The enthalpy is given by H U PV For a finite change at constant pressure H Since U PV U PV U PV VP U PV H q q P P q V U q PV V we note that Hence for gas phase species Hence the heat absorbed at constant pressure Exceeds that absorbed at constant volume by the H U Amount PV. For condensed phases (liquids, solids) V= V products -V reactants Will be very small and so V ~0. Hence Condensed phases H U P=0 Constant P Temperature variation of enthalpy. Assume ideal gas behaviour. At constant T and P we have PV nrt For reactants and products Hence P PV PV Rx Pr n n Rx Pr RT RT V V n n RT nrt Pr where C P Rx Pr n n Pr Rx n Rx nrt qp H T T H dh H CV Lim T 0 T dt T P

8 The specific heat capacity of liquid water is 4.18 JK -1 g -1. Calculate the energy required to heat 1.0 mol of water from 298 K to 363 K. c q m T q cmt q 1.0 mol 18g mol 4.18 JK g ( ) K 4891J 4.9kJ The standard enthalpy change of vaporization of ethanol C 2 H 5 OH is 43.5 kjmol -1. Calculate the enthalpy change when 0.5 g of ethanol vaporizes at its boiling point at 1 bar pressure. Enthalpy change when 1 mol of ethanol C 2 H 5 OH vaporizes = 43.5 kjmol -1. We calculate amount in mole corresponding to 0.5 g EtOH. 1 M gmol CHOH 2 5 1mol C2H5OH 46.07g 1 2 1gC2H5OH mol gCHOH mol vap ( 2 5 ) (43.5 ) ( ) H C H OH kjmol mol kj

9 The change in internal energy for the combustion of naphthalene at 298 K which proceeds according to C 10 H 8 (s) + 12 O 2 (g) 10 CO 2 (g) + 4 H 2 O (l) was measured in a bomb calorimeter as U 0 = kj mol -1. Calculate the enthalpy of combustion of naphthalene H 0 at 298 K. 0 0 H U nrt C 10 H 8 (s) + 12 O 2 (g) 10 CO 2 (g) + 4 H 2 O (l) Products Reactants We only take gas phase reactants n n n U 5142 kj mol 0 1 R Jmol K T 298K 1 1 & products into account. H J mol (2 mol) 8.314Jmol K 298K J mol J mol Jmol 5147 kjmol 3 1 1

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