Stuff. ---Tonight: Lecture 3 July Assignment 1 has been posted. ---Presentation Assignment on Friday.

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Stuff ---Tonight: Lecture 3 July 0 ---Assignment 1 has been posted. Work from gravitational forces: h F gravity dx = h 0 0 mgh mg dx Where m (kg) and g is gravitational constant 9.

1 Stuff ---Tonight: Lecture 3 July 0 ---Assignment 1 has been posted. Work from gravitational forces: h F gravity dx = h 0 0 mgh mg dx Where m (kg) and g is gravitational constant 9.8 m/s ---Presentation Assignment on Friday. --Some more thermodynamics and then problem solving in class for Assignment #. Example: Let s consider an example where the mass may not be constant or where it varies with height (weight of the atmosphere). m = ρd F = ma = ρg d P = F ρ(h)g d A = A ρ(h)g A = dh = A h 0 ρ(h)g dh Hookeʼs Law: Work can come from springs either compressing or extending the spring (the side that yields determines the sign of + or -). xf x i f spring dx = xf x i k spring (x x 0 )dx x0 is the equilibrium starting position of the spring and is a constant. When the spring is compressed or extended the spring resists with a force in order Text to return to x0. Work from spring forces: compressing or extending a spring (the side that yields determines + or -). xf x i k spring (x x 0 )dx xf x i = k spring k spring (x x 0 )dx (xf x 0 ) (x i x 0 ) Work in An Electric Field: z 0 Q F electric dx F electric = QE(d) Q = charge (Coulombs) E = electric field The electric field is related to voltage: = Ed. Kinetic molecular theory relates the macroscopic bulk properties (P,,T) to the molecular properties of matter. KE = 1 mv = 3 RT (one mole of gas) R = J/K mol The average KE of a gas depends only on the absolute T of the gas. T = Kelvin It I t = R Charge is related to current: I = Q/time so I t = Q Ohm s Law = = IR 3RT v = v rms = M = 3kB T m Lighter gases have faster speeds for a given temperature! The internal energy of a gas only depends on T (not on P, )

du = du = q P d dt +,n i d T,n i Calculus tells us we can also write du like this: This is the total differential f i f i w = w = w f w i q = q = q f q i heat and work are path functions so we have

2 We can write the the first law in terms of differentials. Let s Play with the first law closed system and calculus U = q + w = q P du = q w du = U = U f U i for large changes for itty bitty changes To get the big change integrate the itty bitty. du = du = q P d dt +,n i d T,n i Calculus tells us we can also write du like this: This is the total differential f i f i w = w = w f w i q = q = q f q i heat and work are path functions so we have to integrate over the specific path. These paths must be known or given!! This term is important! It s is the heat capacity at constant volume! Let s review heat capcity. Observation: The amount of heat, q, transferred from an object at higher temperature to an object at lower temperature is proportional to the difference in temperature of the two objects. In math terms we write: q α T q = C T C = q T Th Thermom T Thermometer Th T Object 1 Object Heat q The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a quantity of the substance by one degree Celsius or Kelvin (units of J/ C or J/ K or cal/ C). ITS A MEASURE OF ENERGY STORAGE! Kinetic molecular theory relates the macroscopic bulk properties to molecular properties. KE = 1 mv = 3 RT (one mole of gas) v = v rms = R = J/K mol for 1 mole 3RT M = for 1 mole for 1 molecule 3kB T T = Kelvin Lighter gases have faster speeds for a given temperature! m The average KE of a gas depends only on the absolute T of the gas. The internal energy of a gas only depends on T (not on P, ) l A Derivation of how KE and T are linked. OA = v x + v y l v = OA + v z =(v x + v y)+v z Suppose a square box of length = l and Area = A = l and a single gas molecule moving at velocity = v as shown to the left. Gas pressure arises from change in momentum resulting from elastic collisions against the container. We want to express P in terms of molecular properties like velocity, KE. z mv x = mv x m( v x )=mv x roundtrip time = t = l v x F molecule = (mv) t x vz O vx vy v A = mv x = mv x l/v x l y x z vz v O vx vy A F molecule = (mv) t y P mole = F A = Nmv x (l )l P mole = Nmv 3 v = v 1 + v + v v N v x = v y = v z = v 3 = mv x = mv x l/v x l = Nmv x = Nmv = N 3 3 (1 mv )= N 3 KE P = nrt = N RT N A = N 3 KE KE = 1 mv = 3 RT = 3 N A k BT P mole = F A = Nmv x (l )l = Nmv x KE = 3 RT (one mole of gas)

3 The internal energy of an ideal monoatomic gas has only depends on temperature and nothing else! U = KE = 1 mv = 3 The internal energy RT (one mole of gas) depends only on the absolute T of the gas. Isothermal processes (dt = 0) for ideal gases U = 0 R = J/K mol T = Kelvin Atoms have different ways to store energy. They can translate, vibrate, rotate. Many properties depend on these degrees of freedom Equipartition Theorem: All degrees of freedom have the same available energy 1/kT per molecule. axes of rotation linear v = v rms = for 1 mole 3RT M = for 1 molecule 3kB T m Lighter gases have faster speeds for a given temperature! The internal energy of a gas only depends on T (not on P, ) 3 axes of rotation non-linear If we look at the results of the equipartion theorem we find the internal energies for linear and non-linear molecules. U = 3 RT U = 3 RT + RT +(3N 5)RT U = 3 RT + 3 RT +(3N 6)RT C v = 3 R C v = 3 R + R +(3N 5)R C v = 3 R + 3 R +(3N 6)R (monoatomic) (linear molecule) (non-linear molecule) (monoatomic) (linear molecule) U C v (non-linear molecule) KMT and the equipartition of energy links internal energy U and heat capacity, C of ideal gases. U = U = 3 nrt C v = qv qv n moles of monoatomic gas ( 3 = nrt ) IMPORTANT: This equation says that the internal energy of a gas only depends on T and nothing else! We can take the derivative and link the heat capacity to bulk properties = 3 nr Using the energies arising from equipartition of energy, show that the heat capacity at constant volume, Cv are the given values on the previous slides. qv U = U = 3 RT U = 3 RT + RT +(3N 5)RT (monoatomic) (linear molecule) Heat Capacity C = q dt C = q T Units: J/K q = T Don t forget our notation: big changes are represented by q and small changes as!q. Also, the partial derivative means that heat is path-dependent--we need to have knowledge of the process variables or a function describing q. We write C(T) just in case C is not a constant. It might be and if it is between T1 and T we can pull it outside the integral sign. C(T )dt U = 3 RT + 3 RT +(3N 6)RT (non-linear molecule) Molar Heat Capacity C = C n = C n where n is the number of moles of substance. We can normally look these up in a table.

4 Heat capacity of gases depends on P and we define: Key Linkage to the First Law of Thermodynamics from large change ===> small change C v = q v T = U T C p = q p T = H T q v = T C v dt C v = C p = q p = qv qp T,n P,n C p dt Heat Capacity Constant olume Heat Capacity Constant Pressure du = du = q P d dt +,n i First Law Differential Form d T,n i du = C v dt + d T,n i q v = C v dt = Total Differential,n Total Differential Common Restricted Paths In Thermodynamics (Plays on words to solve problems) 1. Free Expansion ==> P = Pext = 0 pressure. Constant olume (Isochoric) ==> f - i = 0 = d = 0 3. Expansion Against Constant P (Isobaric) => P = Pext = Constant 4. Reversible Expansion/Compression => Pext = Pin (ideal gas) 5. Ideal Gas use P = nrt 6. Non-Ideal Gas use van der waals equation 7. Irreversible Expansion P = Pfinal = Constant 1. Free Expansion: P = Pext = 0 pressure W = P ( )d = (0)d =0. Constant olume (Isochoric) ==> f - i = 0 = d = 0 W = P ( )d = P ( )(0) = 0 3. Expansion Constant P (Isobaric) => P = Pext = Constant W = P ( )d = P ext d f = P ext d = P ext ( f i ) i 5. Isothermal reversible Expansion/Compression => Pext = Pin with dn and dt = constant This case represents an infinitesimal change that is conceptual only. Pext = Pint. It does not mean Pext = constant--instead it will follow an equation of state like the idea gas Same mechanics as before: W = P ( )d = f P ext d = i Start with: P = nrt use ideal gas, T constant nrt f 1 d = nrt i d = nrt ln f i 6. Non-Ideal Gas use van der waals equation isothermal P + an ( bn) =nrt P = nrt ( bn) an W = P ( )d = P int d = This is a homework problem rearranging same basic set-up

7. Ideal Isothermal Irreversible Expansion P = Pfinal = Constant This case represents Pext = constant Same mechanics as before: Start with: W = P ( )d = P ext d = P ext f = nrt f ( f i ) i d = P ext

5 7. Ideal Isothermal Irreversible Expansion P = Pfinal = Constant This case represents Pext = constant Same mechanics as before: Start with: W = P ( )d = P ext d = P ext f = nrt f ( f i ) i d = P ext ( f i ) same basic set-up using ideal gas law constant n and T Solving Thermodynamic Problems 1. Realize:!U = U - U1 and!h = H - H1 depends only the state variables P,, T, n. We connect PT with U and H and P1T11 with U1 and H1. We look for simplifications like dt, dp, d = 0 (for solids and liquids d = 0). Relationship between Cp and Cv. We can solve for a irreversible process that is impossible to solve by finding a reversible path that leads to the same starting and end points of the variables of state. Write the total differential of for the equation of state of an ideal gas P(n,,T)? Write the total differential of for the equation of state of an ideal gas P(n,,T)? Write the total differential of for the function: z = x - 3xy. Write the total differential of for the function: z = x - 3xy. A) Calculate the work done on a closed isothermal system consisting of g of Argon when it expands reversibly from 5.000L to 10.00L at 88.15K B) Write the total differential for P(,T) and see if you can set up the integral for Calculate the work done on a closed isothermal system consisting of g of Argon when it expands reversibly from 5.000L to 10.00L at 88.15K Write the total differential for P(,T) and see if you can set up the integral for

Write the total differential for P(,T) and see if you can set up the integral for Consider the following heating or cooling of water at constant atmospheric pressure.

06 atm q p = T C p dt If Cp is constant over dt then: T q p = C p dt Integrating we get something familar. q p = C p T = C p (T ) If moles and molar Cp is used q p = C p T = nc p (T ) 1.

6 Write the total differential for P(,T) and see if you can set up the integral for Consider the following heating or cooling of water at constant atmospheric pressure. n mole HO(l) at P11T1 =>> n mole HO(l) at P1T 1. We might ask what is: qp = heat at constant pressure or enthalpy =? = atm q p = T C p dt If Cp is constant over dt then: T q p = C p dt Integrating we get something familar. q p = C p T = C p (T ) If moles and molar Cp is used q p = C p T = nc p (T ) 1. We might ask what is: qv = heat at constant volume =? n mole HO(l) at P11T1 =>> n mole HO(l) at P1T q v = C v T w p = P ext d = WE are playing with liquid water which for all intensive purposes can not change volume. d = 0 P ext (0) = 0 For liquids and solids in a rigid container, d = 0 Internal energy is hard to measure the lab, so chemists define a new lab-friendly function and give it the name: enthalpy, H. H = U + P!H =!U +!(P) =!U + (Pff - Pii)!H =!U + P! +!P +!P!!H = q - Pd + P! +!P!H = q +!P!H = q P =!U + P! Imposing constant pressure conditions: Enthalpy is the heat gained or lost by a system under conditions of constant pressure. Let s write the enthalpy in terms of its total differential using T and P as the natural variables.!h = q P =!U + P! dh = dh = q p = C p = dt + P qp = p dh = C p dt + dt P dp P T P dp P T,n i Large changes Total Differential At dp = 0 By definition Recasting Chemists measure the enthalpies many chemical reactions, give them names and tabulate these values in Handbooks. They are useful in the real-world.

7 Calculate!U and!h in joules for heating 1 mol of water from 1 atm and 0 C to 100 C and 10 atm. Consider the density of water at this 0.98 g/ml From 180K to 310K, the Cp in J/K mol of CS(l) at 100 kpa fits the empirical equation, Cp (J/mol) = X 10 - T x 10-5 T Calculate!H for 1 mole of CS when it is heated from 180K to 310K at 100kPa. H = q p = H = H = 310 T H = 9800J C p (T )dt [A BT + CT ] dt [A BT + CT ] dt = AT + 1 BT CT Problems From Tinoco Chapter Problems From Tinoco Chapter Problems From Tinoco Chapter Problems From Tinoco Chapter

8 Problems From Tinoco Chapter Problems From Tinoco Chapter

Rate of Heating and Cooling

Rate of Heating and Cooling 35 T [ o C] Example: Heating and cooling of Water E 30 Cooling S 25 Heating exponential decay 20 0 100 200 300 400 t [sec] Newton s Law of Cooling T S > T E : System S cools