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State Variables

State Variables . The Physical State of a system is completely described by a small number of Macroscopic Variables (N,V,T,U, concentrations… we will add more later on) These variables are called state variables and include pressure, volume, temperature, amount of substance, Internal Energy

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State Variables

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  1. State Variables • The Physical State of a system is completely described by a small number of Macroscopic Variables (N,V,T,U, concentrations… we will add more later on) • These variables are called state variables and include pressure, volume, temperature, amount of substance, Internal Energy • In a chemical reaction it is the amounts (concentrations) of the products and reactants, their phases (gas, liquid, solid), Temperature and Pressure.

  2. Free Expansion the same final state the same initial state Internal energy of an ideal gas Partition is broken and gas is Allowed to expand in an adiabatic (no heat transfer container) Controlled expansion (Isothermal Reversible) Work done No work done Different paths Heat exchanged No heat exchanged Is DU is the same for both processes?

  3. c b f a i State Functions for a Mountain Trekker

  4. P2 P1 V1 V2 Is work a state function? i Ideal Gas f Imagine two different transformations from the same initial state (i) to the same final state f. One tr Is the amount of work done on each path the same?

  5. P2 P1 V1 V2 • Work on red path is wred = -p2(V2-V1) . • Work on blue path is wblue = -p1(V2-V1) Is work a state function? i Ideal Gas f • Thus wred < wblue . • Work done on system differs for different paths from i to f. Work is not a state function

  6. Equilibrium States State space f i • Thermodynamics is concerned with equilbrium states. • An equilibrium state is one in which macroscopic properties • (E,V,P,T,n1,n2 ,...) do not change with time. Properties are • Either extensive (E,V,n) or intensive (T,P)

  7. f a i b Energy is a State Function if and only if ¢ Ua = ¢ Ub • Assume the contrary and take ¢Ua > ¢Ub. • Go forward on a (gaining ¢ Ua) and backward along b (losing -¢Ub) • During each cycle there is a net gain in energy (¢Ea - ¢Eb ) > 0, available to do work. • How nice! This would solve energy crisis! Perpetual motion machine! • Not possible, therefore, ¢ Ua = ¢ Ub. Energy is conserved. • Energy is a state function

  8. f a i b Is Heat a state function? ¢ U is a state function but w is not, thus q is not a state function.

  9. Different paths in the Expansion of a gas Consider the expansion of an ideal gas with various coupling between it (the system) and the surroundings • Free expansion (No external Pressure) • Isothermal (Constant Temperature) • Adiabatic (No heat transfer) • Isobaric (Constant pressure) • Isochoric (No work) Each of these represents different coupling Between the system and the surroundings. Thermal reservoir

  10. Free expansion Joule originally did a free expansion experiment and concluded that no heat was transferred between the system and its surroundings. • U = KE + PE • U = q + w • w= pextdV , pext=0, w=0 • No temperature change in surroundings so Joule measured that q=0 • Therefore DU =0, the internal energy is independent of pressure and volume and is only dependent on Temperature

  11. Free expansion Joule originally did a free expansion experiment and found that no heat was transferred between the system and its surroundings. • Joule concluded that there was no change in energy. Is this true for an ideal gas? • He was clearly working on a real gas. • What could have gone wrong? U(r) U(r) U(r) r van der Waals attraction

  12. Joule did his experiments with relatively dilute gases so that their behavior was similar to an ideal gas. How would you expect a denser gas to behave? When thinking about this, think of the Vanderwaals attractions between the particles. Would they become more or less favorable? If the expansion was done in an adiabatic container what would happen? Would the final temperature be higher, lower or remain the same? If the Temperature was controlled with a bath would heat be transferred into or out of the system?

  13. Joule originally did a free expansion experiment and concluded that virtually no heat was transferred. However, for non-ideal gas some temperature change occurs. The internal energy U is the sum of the kinetic and potential energies for all the particles that make up the system. Non-ideal gas have attractive intermolecular forces. So if the internal energy is constant, the kinetic energies generally decrease (think of the vanderwaals potential). Therefore as the temperature is directly related to molecular kinetic energy, for a non-ideal dilute gas a free expansion results in a drop in temperature. U(r) Internal energy of Real Gas r van der Waals interaction What about for a dense gas?

  14. heat QI heat QII C) QI > QII A) QI < QII B) QI = QII Two Systems • Consider the two systems shown to the right. In Case I, the gas is heated at constant volume; in Case II, the gas is heated at constant pressure. Compare QI , the amount of heat needed to raise the temperature 1K (= 1ºC) insystem I to QII, the amount of heat needed to raise the temperature 1ºC insystem II. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAA

  15. heat QI heat QII C) QI > QII A) QI < QII B) QI = QII U = Q - Wby ACT 3 Solution • Consider the two systems shown to the right. In Case I, the gas is heated at constant volume; in Case II, the gas is heated atconstant pressure. Compare QI , the amount of heat needed to raise the temperature 1ºC insystem I to QII, the amount of heat needed to raise the temperature 1ºC insystem II. • Apply the First Law of Thermodynamics to these situations! • In Case I, NO WORK IS DONE, since the volume does not change. • In Case II, POSITIVE WORK IS DONE by the system, since the volume is expanding, therefore Wby > 0 • In each case, temperature rise is the same, and so is DU. Therefore, more heat is required in Case II!

  16. = U 3 Nk T Internal Energy of a Classical ideal gas • “Classical” means Equipartition Principle applies: each molecule has average energy ½ kT per quadratic modein thermal equilibrium. At room temperature, for most gases: • monatomic gas (He, Ne, Ar, …) • 3 translational modes (x, y, z) U= 3/2 NkT diatomic molecules (N2, O2, CO, …) 3 translational modes (x, y, z) + 2 rotational modes (wx, wy) U= 5/2 NkT • non-linear molecules (H2O, NH3, …) • 3 translational modes (x, y, z) • + 3 rotational modes (wx, wy, wz) • For an ideal gas, the internal energy only depends on the Temperature: U= ® NkT = ® nRT • (® (often written f) depends on the type of molecule) TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAA

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