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Reversible Process

Grains of sand. In limit of infinitesmal changes, system moves through equilibrium states during transition-->reversible process. Reversible Process. Remove one grain of sand at a time. Allow system to equilibrate. Pressure decreases by very small amount.

candace-roberson
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Reversible Process

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  1. Grains of sand In limit of infinitesmal changes, system moves through equilibrium states during transition-->reversible process Reversible Process Remove one grain of sand at a time. Allow system to equilibrate Pressure decreases by very small amount Volume increases by very small amount. A reversible process is one where every step of the path is at equilibrium with its surroundings TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AA

  2. Examples of Reversible and Irreversible • Jumping into a pool? (from the perspective of the water) • Getting into a pool slowly? • Free expansion of a gas • Isothermal expansion (piston must move very slowly for bath) • Moving a glass slowly

  3. Isothermal Reversible Process: T constant: Ideal gas For an ideal gas U=q+w=0 w=-q In rev process Pext=Pgas=P=nRT/V w=-P1V -P2V- …

  4. Isothermal Reversible Process: T constant: Ideal gas For an ideal gas U= q + w =0 -w = q

  5. 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? TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAA

  6. 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) Thermal reservoir

  7. Change in State at Constant Volume 1) 2) Substitute in 1) into 2) 3) What is this at constant volume? How do we measure heat flow? (calorimeter) What is for an Ideal Gas?

  8. Free expansion Joule originally did a free expansion experiment and found that virtually 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 measured in surroundings so q=0 • Therefore DU =0, the internal energy is independent of pressure and volume. This is a calorimeter

  9. Joule originally did a free expansion experiment and concluded that virtually no heat was transferred. However, for non-ideal gas some temperature change occur. 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?

  10. 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?

  11. 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?

  12. heat qI heat qII C) qI > qII A) qI < qII B) qI = qII Consider Two Systems • Consider the two systems shown to the right. In Case I, the gas is heated at constantvolume; 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

  13. heat qI heat qII C) qI > qII A) qI < qII B) qI = qII U = q + w 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, WORK IS DONE by the system, since the volume is expanding, therefore w < 0 • In each case, temperature rise is the same, and so is DU. Therefore, more heat is required in Case II!

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