Adiabatic Expansion/Compression Calculate the cooling in a the reversible adiabatic expansion of an ideal gas. P P 1, 1, T 1 A du q w First Law: Since the process is adiabatic, q = 0. Also w = -p ex d (definition) Since the process is reversible, p ex = p int and nrt du dwrev pd d P 2, 2, T 2 Recall from earlier that du = C dt by definition, and so we have We cannot integrate directly because T is changing with. So we simplify by dividing by T to obtain We can now integrate each side C T The last term arises because the gas is ideal. Note how we have used the given parameters. dt nr d CdT nrt d
Adiabatic Expansion/Compression P 1, 1, T 1 C T dt nr d P A a Recall that aln b ln b, so T f T i C T dt f i nr d and T ln ln ln f P C f i 2, 2, T 2 nr T i i f nr C T T f f i ln ln ln nr Ti Ti f C Exponentiating both sides, C T nr f i T i f
Adiabatic Expansion/Compression C T f f i Repeating, ln ln ln nr T i i f We could equally well have included the C/nR term with the volumes to obtain nr C T f nr f nr i i ln ln ln ln T C C i i f f nr C i and T T for a reversible adiabatic expansion f i f nr C f f nr C 3 Now T T i i i For an ideal gas, C nr, and so T T f i 2 f We knew for our expansion that i < f, so we have proven that T f < T i in a reversible adiabatic expansion. 2 3
Adiabatic and Isothermal Processes We again make use of the ideal gas behavior to obtain another relationship. If we use the ideal gas equation of state to replace the temperatures in 2 3 i T T f i f, then Adiabatic: 2 3 P P 2 2 1 1 1 nr nr 2 P P 2 2 1 1 or P P 2 1 2 2 1 1 nr nr 2 2 5 5 3 3 3 3 Now consider an isothermal expansion. In an isothermal expansion of an ideal gas, T 1 = T 2, and Boyle s Law gives P P P P 1 1 or This gives a way to unify things: 1 1 2 2 1 1 2 2 P P 1 1 2 2 where 1 for an isothermal expansion and 5 and for an adiabatic expansion. 3
Adiabatic Expansion/Compression We now can look back at the early picture with deeper understanding Ideal Gas Eq. of State Surface
Adiabatic Compression - Example Adiabatic: 2 3 i f i f T T If ideal, 5 5 3 3 P f f i i P Boulder can have windstorms, in which the temperature can rise substantially ( 50 F) in 30 minutes on a strong west wind. It is a bit like the jet stream blowing near treetop level. Can we estimate t the temperature t increase? Consider an adiabatic compression from the continental divide (P=0.5 Bar) to Boulder (P=0.8 Bar). What do we estimate for ΔT if T divide = 280 K? 5 3 5 5 3 5 f P i f P i P f f i i P P i f i f 3 3 P gives and 0.625 0.754 0.6 T f T i 2 3 i 1 0.754 f 0666 0.666 So if T i = 280K K, T f =307K K, 1.207 an increase of 49F
Exam 1 Review Ideal Gas Law Deduced from Combination of Gas Relationships: 1/P, Boyle's Law, Charles's Law n, Avogadro's Law Therefore, nt/p or P nt P = nrt =Nk B T where R = gas constant (per mole) or k B = gas constant (per molecule) The empirical Equation of State for an Ideal Gas
Ideal Gas Equation of State
Ideal Gas Law P = nrt where R = universal gas constant R = P/nT R = 0.08210821 atm L mol 1 K 1 R = 0.0821 atm dm 3 mol 1 K 1 R = 8.314 J mol 1 K 1 (SI unit) Standard molar volume = 22.4 L mol 1 at 0 C and 1 atm Real gases approach ideal gas behavior at low P & high T
Real Gases Compressibility P Pm Z nrt RT Z = 1 at all P, T Ideal Gas Behavior Ideal gas Now look at real gases at some temperature T Look at a broader 0 800 atm region
van der Waals equation of state Physically-motivated y y corrections to Ideal Gas EoS. For a real gas, both attractive and repulsive intermolecular forces are present. Empirical terms were developed to help account for both. 2 nrt an RT a 2 2 P nb b
General Principle!! Energy is distributed ib t d among accessible configurations in a random process. The ergodic hypothesis Consider fixed total energy with multiple particles and various possible energies for the particles. Determine the distribution that occupies the largest portion of the available Phase Space. That is the observed distribution.
Energy Randomness is the basis of an exponential distribution of occupied energy levels n(e) A exp[-e/<e>] Average Energy <E> ~ k BT n(e) A exp[-e/k B T] This energy distribution is known as the Boltzmann Distribution.
Maxwell Speed Distribution Law 1dN Ndu 32 m 2 2 mu B B 4 ue 2 kt 2k T 1dN is the fraction of molecules per unit speed interval Ndu
Maxwell Speed Distribution Law N u du 1 u 2 u 1 N u du fraction of particles with u between u and u 0 1 2 Most probable speed, u dn 2 mp du 0 for u = u mp u mp kt m Average speed, <u> or ū u un u du 8kT m 0 m u 3kT 2 u 2 N u du m Mean squared speed, <u 2 > 0 Root mean square speed u rms u 2 3 3kT m
Distinguish between System & Surroundings
Internal Energy Internal Energy (U) is the sum of all potential and kinetic energy for all particles in a system U is a state function Depends only on current state, not on path U = U final -U initial
Internal Energy, Heat, and Work If heat (q) is absorbed by the system, and work (w) is done on the system, the increase in internal energy (U) is given by: U = q (heat absorbed by the system) + w (work done on the system)
Reversible and Irreversible Work Work done on the gas = also the min work required to compress the gas P 1 ( 2-1 )
ConcepTest #3 Two 10 kg weights sit on a piston, compressing the air underneath. One of the weights is removed, and the air underneath expands from 18.3 to 20.0L. Then the second weight is removed and the air expands from 20.0 to 22.0L. How does the amount of work done compare if instead both weights were removed at once? Assume the same total change in volume. a. More work is done removing one weight at a time than removing both weights at once. b. Less work is done removing one weight at a time than in removing both weights at once. c. The same amount of work is done removing one weight at a time, or if both are removed at once.
Internal Energy (2) U is a state function It depends only on state, not on path to get there U = U final - U initialiti This means mathematically that du is an f exact differential: U du For now, consider a system of constant composition. U can then be regarded as a function of, T and P. Because there is an equation of state relating them, any two are sufficient to characterize U. So we could have U(P,), U(P,T) or U(,T). i
Enthalpy Defined Enthalpy, H U + P At Constant P, H = U + P U = q + w q= q P = U - w, w = -P q P = U + P= H At constant, q = U = H
Comparing H and U at constant P H = U + P 1. Reactions that do not involve gases 0 and H U 2. Reactions in which n gas = 0 0 and H H UU 3 Reactions in which n 0 3. Reactions in which n gas 0 0 and H U
Heat Capacity at Constant olume or Pressure C = dq /dt = (U/T) Partial derivative of internal energy with respect to T at constant C P = dq P /dt = (H/T) P Partial derivative of enthalpy with respect to T at constant P Ideal Gas: C P = C + nr
Thermochemical h Equations CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) (a combustion reaction) c H = 890 kj Reaction must be balanced Phases must be specified H is an extensive property Sign of H changes when reaction is reversed
Standard State The Standard State of an element is defined to be the form in which it is most stable at 25 C and 1 bar pressure Some Standard States of elements: Hg (l) O 2 (g) Cl 2 (g) Ag (s) C (graphite) The standard enthalpy of formation ( f H ) of an element in its standard state is defined to be zero.
Enthalpies of Formation The standard enthalpy of formation ( f H ) of a compound is the enthalpy change for the formation of one mole of compound from the elements in their standard states. Designated by superscript o: H For example, CO 2 : C (graphite) + O 2 (g) CO 2 (g) rxn H = -393.5 kj/mol Appendix D f H CO 2 (g) = -393.5 kj/mol
Enthalpies of Reaction The enthalpy of reaction can be calculated from the enthalpies of formation of the reactants and products rxnh = f H (Products) - f H (Reactants)
Example: Find rxn H (using Standard Enthalpies of Formation) CH 4 (g) + 2 O 2 (g) CO 2 (g) +2HO 2 (l) f H (from Appendix D, text): CH 4 (g) -74.6 kj/mol O 2 (g) 0 CO 2 (g) -393.5 H 2 O (l) -285.88 rxn H = -393.5-2 (285.8) 0 (-74.6) kj/mol Therefore, rxn H = -890.5 kj/mol
Exam 1 ~ 5/6 problems (weights given) budget your time Closed book Don t memorize formulas/constants You will be given things you need Exam will not be heavily numeric, but will emphasize concepts If a question seems lengthy, do another problem & come back later Understanding homework will be useful