Background. Chemistry 431. Pressure-Volume Work. The Conservation of Energy. Basic Definitions: System and Surroundings. Work and Heat 9/1/2008

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1 Chemistry 431 Lecture 5 The First Law of Thermodynamics NC State University Background Thermodynamics is a macroscopic science that could be used with reference to atoms or molecules. In the nineteenth century thermodynamics was developed to understand engines (e.g. steam engines) and other transformations of energy. However, our approach is to make constant reference to the microscopic world that we know exists based on many experiments. Thus, we will use an approach related to statistical thermodynamics. We are concerned with the average properties of a collection of a large number of atoms or molecules. The Conservation of Energy Energy in thermodynamics is the capacity to do work. The work occurs against an opposing force and can be defined as a force operating through a distance. When chemical change used to generate work there is often heat generated as well. We will show in this lecture that the sum of the heat generated and the work done is a constant amount of energy. Thus, the first law of thermodynamics is an expression of the conservation of energy. Internal Energy = Heat Exchanged + Work Done Pressure-Volume Work Students often wonder why we begin with pressure-volume work. After all, there are lots of kinds of work starting with the basic definition that work is a force acting through a distance. However, the work performed by an engine involves the expansion of a gas to drive a piston and it is this work that is of practical interest in the design of steam and internal combustion engines. It is also pressure-volume work that leads most directly to an understanding of the relationship between heat and work. In order to study the transfer of energy in the form of heat and work we define the system (the engine or mechanical device we are interested in) and the surroundings. Basic Definitions: System and Surroundings An open system is a system that can exchange both energy and matter with its surroundings. A closed system is a system that can exchange energy but not matter with its surroundings. An isolated system is a system that exchange neither matter nor energy with its surroundings. Work and Heat We consider a chemical reaction that generates both work and heat. For example, Zn(s) + 2 HCl(aq) ZnCl 2 (aq) + H 2 (g) Note that one mole of H 2 gas is produced for each equivalent of Zn consumed. If the reaction takes place in a container that is fitted with a piston the production of gas will cause the piston to rise. work If a system is isolated we say that it is adiabatic (Greek for not going through). Heat does not go through the walls of container of an isolated system. The opposite is diathermic. Reaction taking place heat 1

2 Work and Heat If we constrain the piston so that it cannot move, then the pressure in the container will rise. This will result in an increase in the temperature (PV = nrt). Thus, more heat will be released into the surroundings in this case. Although we have not accounted for the difference quantitatively, we know that energy must be conserved and so the amount of heat released must equal the work that would have been done. work Reaction taking place heat Exothermic and Endothermic Reactions A reaction that releases heat is exothermic. A reaction that absorbs heat is endothermic. The reaction we considered on the preceeding slide is exothermic. Endothermic reactions are much less frequent. We will see later that entropy must account for the fact that such reactions can be spontaneous. An example of an endothermic reaction is the dissolution of sodium nitrate in water. This reaction is the basis of instant cold packs that are included in first aid kits. The measurement of work Work is defined as the result of a force acting through a distance. This easily seen for gravitation. If an object of mass m is lifted to a higher elevation in the gravitational field of the earth the work done is w = mgh. dw = h 2 h 2 Fdh = mg dh = mg(h 2 h 1 ) h 1 h 1 The units of work are kg m 2 /s 2, which are also called Joules. Equivalent sets of units are listed below: 1 J =1Nm =1Pa m 3 =1kg m 2 s 2 Just as there is a sign associated with work of lifting (or letting something fall) in a gravitational field, there is a sign associated with work done and heat exchanged by the system. The Sign Convention We can summarize the sign convention in the following table. Energy + (increased) - (decreased) Work done (w) on system by system Heat flows (q) into system from system Alternatively, we can depict the convention as follows. w < 0 w > 0 q < 0 q > 0 The Sign Convention We can summarize the sign convention as follows: 1. Choose the correct answer A. Work done by the system is positive B. Work done by the system is negative 2. Choose the correct answer A. Heat absorbed by the surroundings is positive B. Heat absorbed by the surroundings is negative Internal Energy Now that we have given a symbolic form to both heat (q) and work (w) we can state the first law in terms of symbols. ΔU = w + q The symbol U represents the internal energy of the system. This is the energy that t is conserved. The first law states t thatt any change in the internal energy, ΔU, results either from work done or heat exchanged with the surroundings. We can also state the first law in differential form. du = δw + δq In this case there is an important difference between the differential written with d and that written with δ. 2

3 State Functions and Path Functions We shall show that the internal energy is state function. A state function is not dependent on the path used to get to a certain set of parameters. For a given value of the variables T and V the value of internal energy, ΔU, is fixed and ddoes not tdepend don the path. Pressure-Volume Work The work of expansion when a piston moves inside a cylinder is an example of pressure-volume work. We can show that pressure-volume work is related to the standard definition of a force operating through a distance. δw = Fdx = F = A Adx PdV On the other hand the work and the heat are not state functions. The magnitude of each of them depends on the particular path in a given process. We express this as follows: U 2 du = U 2 U 1 = ΔU, δw = w, δq = q U 1 A dx It is important that we define the work in this form since it depends on the path. Let s consider two paths: 1. Reversible (very slow, equilibrium) 2. Irreversible (fast, sudden jump) Reversible For the reversible path we will consider a very slow process where the forces are balanced on either side of the piston as it expands. Since the pressure on the inside and outside are evenly balanced, the system in equilibrium and we can apply the ideal gas law at all times, P = nrt/v. A dx δw = w = = nrt V 2 PdV = V 2 nrt V dv V 2 dv = nrt ln V 2 V Note that temperature is held constant for this path. Therefore, a reversible path is also called an isothermal path. Irreversible For the irreversible path the expansion is very fast and the external pressure is a constant. In this case we no longer treat P as a variable (as we did for the reversible) case and we call it P ext. V 2 V 2 δw = w = P ext dv = P ext dv A dx = P ext V 2 = P ext ΔV P ext must be a smaller pressure than the starting pressure. Since the forces are not balanced, less work is obtained for the irreversible expansion than for the reversible expansion. 3

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5 Since the work is the integral in P-V space it can represented graphically as the area under the curves. For the reversible work this is the area under the blue curve. For the irreversible work this is the rectangle under the violet line. Since the work is the integral in P-V space it can represented graphically as the area under the curves. For the reversible work this is the area under the blue curve. For the irreversible work this is the rectangle under the violet line. Since the work is the integral in P-V space it can represented graphically as the area under the curves. For the reversible work this is the area under the blue curve. For the irreversible work this is the rectangle under the violet line. Clearly the work of expansion is greatest for the reversible path. This is always true. There are many possible nonreversible paths and the system performs less work along any of those paths than it does along the reversible path. 5

6 Work of Compression The two paths shown below represent the reversible (blue) and single-step irreversible (red and violet) paths for a compression. For the single-step path the external pressure must equal the final internal pressure and it is still a constant. Work of Compression Thus, the irreversible path requires the surroundings to do more work on the system (w > 0) than the reversible path. Again, we can see the work graphically as the area under the curve. Work of Compression Thus, the irreversible path requires the surroundings to do more work on the system (w > 0) than the reversible path. Isothermal Processes For an ideal gas the internal energy is the kinetic energy. There is no potential energy for an ideal gas so this must be true. Therefore, U = 3 2 nrt, U m = 3 2 RT For an isothermal process T 2 = T 1 = constant so ΔU = U 2 U 1 = 3 2 nrt nrt 1 =0 The internal energy change is always zero for an isothermal process. If we return to the first law we see that ΔU = w + q =0 q = w The system absorbs heat from the surroundings with an energy equal to the work done. Isochoric and Isobaric Processes Isochoric means constant volume. The red line in the preceeding figures is a constant volume process. In such a process w = 0 since dv = 0 (Recall that δw = -PdV). Therefore, ΔU = q V for an isochoric process. The V subscript means that V is held constant. Isobaric means constant pressure. In a constant pressure expansion such as we saw in the preceeding frames we know how to calculate the work, w = -P ext ΔV. Moreover, we know that ΔU for the isochoric process is -q V. Since the end point of the isochor and isobar is the same as that for the isotherm, we know that overall ΔU = 0. Therefore, q P = q V + P ext ΔV. The system absorbs heat from the surroundings with an energy equal to the work done. The relationship of q V and q P We can decompose the internal energy for the constant V and constant P steps into two parts (ΔU 1 and ΔU 2 ). Since the overall ΔU = 0 = ΔU 1 + ΔU 2 = -q V + q P -PΔV we can solve for q P to obtain: q P = q V + PΔV. ΔU 1 ΔU 2 6