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Work, Energy & Heat

Work, Energy & Heat. The First Law: Some Terminology. System: Well defined part of the universe Surrounding: Universe outside the boundary of the system Heat (q) may flow between system and surroundings. Closed system: No exchange of matter with surroundings

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Work, Energy & Heat

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  1. Work, Energy & Heat The First Law: Some Terminology System: Well defined part of the universe Surrounding: Universe outside the boundary of the system Heat (q) may flow between system and surroundings Closed system: No exchange of matter with surroundings Isolated System: No exchange of q, or matter with surroundings Isothermal process: Temperature of the system stays the same Adiabatic: No heat (q) exchanged between system and surroundings

  2. THE CONCEPT OF REVERSABILTY Irreversible processes: Warm Hot Temperature equilibration Cold Warm Mixing of two gases P = 0 Expansion into a vacuum P = 0 Evaporation into a vacuum

  3. THE CONCEPT OF REVERSABILTY Reversible processes: Tiny weight Po Po Po + P Condensation (pressure minimally increases by adding tiny weight) Evaporation (pressure minimally decreases by removing tiny weight)

  4. Pext Pext V2 V1 P P, V1 P, V2 Pext w = -Pext(V2 – V1) V1 V2 V IRREVERSIBLE EXPANSION

  5. Pext = 1 atm Pext = 1 atm REMOVE PINS (1) pins P = 1 atm Irreversible Expansion P = 2 atm Pext = 1 atm Pext = 1.998 atm Pext = 1.999 atm Pext = 2 atm Step 2 Step 1 Infinite number of steps (2) Reversible Expansion P = 1 atm P = 1.999 atm P = 1.998 atm P = 2 atm THE CONCEPT OF REVERSABILTY

  6. REVERSIBLE EXPANSION Pext = Pressure of gas. If the gas is ideal, then Pext = nRT/V How does the pressure of an ideal gas vary with volume? This is the reversible path. The pressure at each point along curve is equal to the external pressure. P V

  7. REVERSIBLE EXPANSION IRREVERSIBLE EXPANSION A A The reversible path The shaded area is Pi Pi Pext = Pf P P B B Pf Pf Vi Vf V Vi Vf V The shaded area is Reversible expansion gives the maximum work

  8. REVERSIBLE COMPRESSION IRREVERSIBLE COMPRESSION Pext = Pf B B The reversible path The shaded area is Pf Pf P P A A Pi Pi Vf Vi V Vf Vi V The shaded area is Reversible compression gives the minimum work

  9. A system from a state 1 (or 2) to a new state 2 (or 1). Regions A, B, C, D, and E correspond to the areas of the 5 segments in the diagram. 1. If the process is isothermal reversible expansion from state 1 to state 2, the total work done by the system is equal to A. Area C + Area E B. Area C only C. Area E only D. Area A + Area C + Area E E. Area A only 2. If state 2 undergoes irreversible compression to state 1 against an external pressure of 5 atm, the work done by the surroundings on the system is equal to A. Area A + Area C + Area E B. Area C only C. Area A + Area C E. Area E only F. Area A only 3. If the process in question 1 was carried out irreversibly against a constant pressure of 2 atm, the total work done by the system is equal to A. Area C + Area E B. Area C only C. Area A + Area C + Area E D. Area E + Area D E. Area E only 4. For a reversible adiabatic expansion of a gas, which one of the following is correct? A. Heat flows to maintain constant temperature B. The gas suffers a maximum drop in temperature C. The gas suffers a minimum drop in temperature D. The work done is a positive quantity E. There is zero change in internal energy 5. The heat capacity (Cp) for a solid at low temperatures is approximately represented by Cp = AT3, where A is a constant. Using the equation for Cp, the change in entropy (S) for heating a solid from 0K to 1K is A. A/4 C. A/2 B. A/3 D. A 6. Which one of the following is not true? A. There is no heat flow between system and surrounding for a reversible adiabatic process B. Work done by the gas in a reversible expansion is a maximum C. Work done by the gas in a reversible expansion is a minimum D. Work done by the gas in a reversible compression is a minimum E. Work done by the gas in a reversible expansion is not the same as the work done against a constant external pressure

  10. Total number of microstates = 16 1 way 4 ways 6 ways 4 ways 1 way The most probable (the “even split”) Statistical Entropy Consider four molecules in two compartments: If N   the “even split” becomes overwhelmingly probable

  11. Boltzmann S = kB lnW Consider spin (or dipole restricted to two orientations) 1 particle or W = 2, and S = kB ln2 2 particles W = 4, and S = kB ln4 , , , 3 particles W = 8, and S = kB ln8 1 mole W =2NA, and S = kB ln(2NA) = NA kB ln2 = Rln

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