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Work of Expansion System not in mechanical equilibrium w/environment will expand or contract

Ch 4: 1 st Law of Thermodynamics. Work of Expansion System not in mechanical equilibrium w/environment will expand or contract Assume S is surface of our system that expands infinitesimally to S ′ in direction ds

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Work of Expansion System not in mechanical equilibrium w/environment will expand or contract

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  1. Ch 4: 1st Law of Thermodynamics • Work of Expansion • System not in mechanical equilibrium w/environment will expand or contract • Assume S is surface of our system that expands infinitesimally to S′ in direction ds • The surface element d has performed work against external pressure as follows: where (dV)d is the cylindrical volume element swept by d moving from S to S′ • If pressure p exerted by environment is constant over all the surface S, we can integrate over d where dV is the whole incremental volume change of the system

  2. Ch. 4 – 1st Law of Thermodynamics • Work of Expansion • For a finite expansion from i  f the integral is • For a cycle (at right) we have • This is = to the enclosed area and it is not =0  dW is not an exact differential. • Work is +ve if done by the system, and –ve if done on the system by the environment Work of expansion is the only kind of work that we shall consider in our atmospheric systems.

  3. Example 4.1:One mole of a gas expands from a volume of 10 liters and a • temperature of 300 K to: • A volume of 14 liters and a temperature of 300 K, and • A volume of 14 liters and a temperature of 290 K. • What is the work done by the gas on the environment in each case? • From the definition of work • (b) In this case:

  4. Example 4.1 – continue Consider state (a); the pressure in this state is given by the equation Of state:

  5. Example 4.1 – continue Consider state (b); the pressure in this state is given by the equation Of state:

  6. Example 4.1 – continue The work done can be approximated as:

  7. Example 4.2:Consider an initial, i, and a final, f, state of a gas on an Isotherm with Vf > Vi . The gas goes from i to f via an isobaric expansion and subsequent isochoric cooling; sketch on a (p,V) diagram the complete transformation and find the work done. If instead the gas goes from i to f via an isochoric cooling followed by an isobaric expansion, will the work done be different from before?

  8. Example 4.2 – continue

  9. Ch. 4 – 1st Law of Thermodynamics: conservation of energy for the thermodynamic system; energy cannot be created or destroyed. It can only change from one form to another. • Internal Energy  Assume an adiabatic container (closed system) – allows no heat transfer in/out;  A system undergoes a change while in this container  Change may be affected by different processes, e.g., • Increase T of mass of water by stirring with paddles • Increase T by passing current through wire in the water • Both of these involve external forces performing work upon the system

  10. Internal Energy Work done by an external force adiabatically on system, -Wad, in order to bring on a change in state, is path independent, i.e., Wad has the same value for every adiabatic process causing the same change and depends only on i and f states

  11. Ch. 4 – 1st Law of Thermodynamics • Internal Energy Then Wad can be expressed by the difference in a state function The state function, U, is the internal energy of the system (and it is an exact differential) Internal energy is defined by the above equation except for an arbitrary constant, which is set by choosing a reference state Thermodynamics only considers the variations of U, ΔU, rather than its absolute value, so the indetermination unimportant For constant to be unique, it is necessary that any state may, in principle, be related to the same reference state through an adiabatic process. This can always be done.

  12. Ch. 4 – 1st Law of Thermodynamics • Heat • If we consider a non-adiabatic process causing the same change, we will find that the work performed on the system will not be the same, i.e., W  Wad • To account for the difference, define the heat, Q, absorbed by the system from the environment as the difference • For an infinitesimal process • With Q = 0 for the adiabatic case, by definition, and for a cyclic process, W = Q. The above is the mathematical expression for the 1st Law • Since W = pdV, we can write the 1st Law as • For unit mass the 1st Law becomes

  13. Ch. 4 – 1st Law of Thermodynamics • Internal Energy • Exchanging heat is, like performing work, a way of exchanging energy • Joule’s experiments determined the equivalence of mechanical work and heat • Experiment showed that 1 cal (15°C) = 4.1855 J, known as the mechanical equivalent of heat

  14. Ch. 4 – 1st Law of Thermodynamics • Heat • Determination of Q for processes with V = const, or p = const, in which no work is performed on or by the system (except expansion term with p = const), subject of Calorimetry. Results summarized: • If a process with p = const has no change in physical state or chemical reactions, heat absorbed  mass and change in T, i.e., where sub-index, p, indicates constant pressure p, and cp is proportionality factor called specific heat capacity at constant pressure

  15. Ch. 4 – 1st Law of Thermodynamics • Heat • Calorimetry results (continued) • If a process with V = const has no change in physical state or chemical reactions, heat absorbed also  mass and ΔT: where sub-index, V, indicates constant V, and cV is proportionality factor called specific heat at constant volume • If effect of absorption of heat at p = const is a change of physical state (at T = const, too), Q  to the mass that undergoes change

  16. Ch. 4 – 1st Law of Thermodynamics • Heat • Calorimetry results (continued) • The proportionality factor, l, is called the latent heat of the change of state (whatever that change might be) • If effect of heat absorption at either p or V = const is a chemical reaction, Q  mass of reactant that reacts and proportionality factor is called heat of reaction (at const. p or const. V) • Heat capacities and latent heat can be referred to the mole rather than kg (or gram). Then are called molar heat capacities.

  17. Example 4.3:Calculate for the following processes: • Isothermal reversible compression from state i=(pi,Vi) to state f=(pf,Vf); • Adiabatic reversible compression from state i=(pi,Vi) to state m=(pf,Vm) • And a subsequent isobaric reversible compression to state f=(pf,Vf); • (c) Reversible isochoric increase of the temperature from state i=(pi,Vi)to state • m’=(pf,Vi) and a subsequent reversible isobaric decrease in temperature to • State f=(pf,Vf) • (a)

  18. Example 4.3: continue (b)

  19. Example 4.3: (b) continue Tm and Ti are related (adiabatic transformation)

  20. Example 4.3: (b) continue m  f isobaric transformation Vm and Viare related via Poisson’s Equation

  21. Example 4.3: (b) continue m  f isobaric transformation (c) i  m’ isochoric  m’  f isobaric  Again, Ti=Tf=T, 

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