General Deductions

Excerpts

• General Deductions

  Our first application of the principle of the entropy which was expressed in its most general form in the preceding chapter, will be to Carnot’s cycle, described in detail for perfect gases in 90.

• General Deductions

  Observe, however, that the expressions % [eqn:(64)](64)% for the change of the entropy of the reservoirs are still correct, provided we assume that any changes of volume of the substances used as reservoirs are reversible.

• General Deductions

  In this case the cyclic process results in the transference of heat ($Q_{2}$) from the reservoir of temperature $\theta_{2}$ to that of temperature $\theta_{1}$, and the inequality means that this flow of heat is always directed from the hotter to the colder reservoir.

• General Deductions

  This means that the amount of work, $W'$, to be gained by means of a cyclic process from the transference of the heat, $Q_{1}'$, from a hotter to a colder reservoir, is always smaller for an irreversible process than for a reversible one. Consequently the equation % [eqn:(66)](66)% represents the maximum amount of work to be gained from any cyclic process between heat-reservoirs at the temperatures $\theta_{2}$ and $\theta_{1}$.

• General Deductions

  This time, the system operated upon may be of any character whatsoever, and chemical reactions, too, may take place, provided they are reversible.

• General Deductions

  This leads to the proposition that chemical reactions, in which there is no external work, take place in such a manner as to give the greatest heat effects (Berthelot’s principle).

• General Deductions

  Among all the states of the system which can proceed from one another by adiabatic processes, the state of equilibrium is distinguished by a maximum of the entropy.

• General Deductions

  It cannot in general be integrated, since the left-hand side is not, in general, a perfect differential.

• General Deductions

  has been called by H. v. Helmholtz the *free energy* (freie Energie) of the system.

• General Deductions

  The second law, then, does not lead to a general statement with regard to finite changes of a system taken by itself unless something be known of the external conditions to which it is subject.

• General Deductions

  For finite reversible isothermal changes the total work done on the system is equal to the increase of $F$; or, the entire work performed by the system is equal to the decrease of $F$, and, therefore, depends only on the initial and final states of the system. Where $F_{1} = F_{2}$, as in cyclic processes, the external work is zero.

• General Deductions

  Hence, any reversible transformation of the system from one state to another yields the maximum amount of work that can be gained by any isothermal process between those two states.

• General Deductions

  Such a process, as here described, is composed only of states of equilibrium. Hence it is reversible, and the external work thereby gained represents at the same time the decrease of the free energy, $F_{2} - F_{1}$, which takes place on directly mixing the solution and the water.

• General Deductions

  Dividing this quantity by the number of oxidized molecules of hydrogen, we obtain a measure of the force with which a molecule of hydrogen tends to become oxidized. This definition of chemical force, however, has only a meaning in so far as it is connected with that work.

• General Deductions

  States of equilibrium of this description are always unstable. Often a very small disturbance, not comparable in size with the quantities within the system, suffices to produce the change, which under these conditions often occurs with great violence. We have examples of this in overcooled liquids, supersaturated vapour, supersaturated solutions, explosive substances, etc.

• General Deductions

  *i.e.* among the states which can proceed from one another by isothermal processes, without the performance of external work, the state of most stable equilibrium is distinguished by an absolute minimum of the free energy.

Equations

• General Deductions

  \frac{Q_{1}}{\theta_{1}} + \frac{Q_{2}}{\theta_{2}} < 0

  For a cycle with reservoir entropy changes allowed to be reversible in volume, the sum of Q/theta over the reservoirs is negative for an irreversible cycle.

• General Deductions

  Q_{2} = W' + Q_{1}'

  In a cyclic process the heat given out by the hotter reservoir equals the work done by the system plus the heat received by the colder reservoir.

• General Deductions

  Q_{1} + Q_{2} + W = 0

  Over a complete cycle the heat exchanged with the two reservoirs and the work done on the system sum to zero (energy equation).

• General Deductions

  \frac{Q_{1}}{\theta_{1}} + \frac{Q_{2}}{\theta_{2}} = 0

  For a reversible cycle the total entropy of the two heat reservoirs is unchanged.

• General Deductions

  Q_{1} : Q_{2} : W = (-\theta_{1}) : \theta_{2} : (\theta_{1} - \theta_{2})

  For a reversible Carnot cycle the heats and work are in the ratio of the temperatures given by the second law, for any working substance.

• General Deductions

  Q_{1}' = \frac{\theta_{1}}{\theta_{2} - \theta_{1}} W'

  The heat that must pass from the hotter to the colder reservoir in a reversible Carnot cycle to gain a given work W' is fixed by the two temperatures.

• General Deductions

  W' = \frac{\theta_{2} - \theta_{1}}{\theta_{1}} Q_{1}'

  The work obtainable from heat transferred between two reservoirs by a reversible cycle is the maximum possible for any cyclic process between them.

• General Deductions

  - \frac{Q_{1}}{\theta_{1}} - \frac{Q_{2}}{\theta_{2}} > 0

  For an irreversible cycle the sum of the entropy changes of the reservoirs is positive.

• General Deductions

  W' < \frac{\theta_{2} - \theta_{1}}{\theta_{1}} Q_{1}'

  The work gained by an irreversible cycle is smaller than the reversible maximum for the same heat transfer.

• General Deductions

  Q_{2} \left(\frac{1}{\theta_{2}} - \frac{1}{\theta_{1}}\right) < 0

  In a cycle with no work, heat can only flow from the hotter to the colder reservoir.

• General Deductions

  W + Q = 0

  For a process using one reservoir of constant temperature, the work done on the system and the heat absorbed by it cancel.

• General Deductions

  -\frac{Q}{\theta} \geq 0

  The entropy change of the constant-temperature reservoir is non-negative.

• General Deductions

  Q \leq 0

  Heat is added to the reservoir, not taken from it, in any such process.

• General Deductions

  W \geq 0

  Work must be expended on the system in any cycle using a single reservoir; in the reversible limit it vanishes.

• General Deductions

  dU = Q + W

  For any infinitesimal change the increase of internal energy equals the heat absorbed plus the work done on the system.

• General Deductions

  d\Phi + d\Phi_{0} \geq 0

  The total entropy change of the system and its surroundings is non-negative for any natural process.

• General Deductions

  d\Phi_{0} = -\frac{Q}{\theta}

  The entropy change of the surrounding medium equals minus the heat absorbed by the system divided by the temperature.

• General Deductions

  d\Phi_{0} = -\frac{dU - W}{\theta}

  Substituting the first law, the entropy change of the surroundings is minus the energy change less work, over the temperature.

• General Deductions

  d\Phi - \frac{dU - W}{\theta} \geq 0

  The criterion for any natural change at fixed temperature combining both laws.

• General Deductions

  dU - \theta\, d\Phi \leq W

  The fundamental relation (70) between energy, entropy, temperature and work for any change at a common temperature.

• General Deductions

  dU = W

  In an adiabatic process with no heat exchange the energy change equals the work done on the system.

• General Deductions

  d\Phi \geq 0

  In an adiabatic process the entropy of the system increases or remains constant.

• General Deductions

  d(U - \theta\Phi) \leq W

  In an isothermal process the increment of U minus theta times the entropy is at most the work done on the system.

• General Deductions

  U - \theta\Phi = F

  Defines the free energy F as internal energy minus temperature times entropy.

• General Deductions

  dF = W

  For a reversible isothermal change the increment of free energy equals the work done on the system.

• General Deductions

  F_{2} - F_{1} = \tsum W

  For finite reversible isothermal changes the free energy change equals the total work done on the system.

• General Deductions

  F_{2} - F_{1} < \tsum W

  For irreversible isothermal changes the free energy increases by less than the work done on the system.

• General Deductions

  dF < W

  In an irreversible isothermal process the free energy increment is less than the work done on the system.

• General Deductions

  U - F = \theta\Phi

  Defines the latent energy as the difference of total energy and free energy, equal to temperature times entropy.

• General Deductions

  \tsum W = 0

  When the work during an isothermal process vanishes, the sum of work is zero.

• General Deductions

  F_{2} - F_{1} < 0

  With no external work in an isothermal process, the free energy decreases.

• General Deductions

  dF = dU - \theta\, d\Phi - \Phi\, d\theta

  The differential of the free energy in terms of the differentials of energy, entropy and temperature.

• General Deductions

  dF \leq W - \Phi\, d\theta

  For any physical or chemical process the free energy change is bounded by the work less entropy times temperature change.

• General Deductions

  U = Mu = M(c_{v} \theta + \const)

  The internal energy of a perfect gas is linear in temperature, up to an arbitrary constant.

• General Deductions

  \Phi = M\phi = M(c_{v} \log \theta + \frac{R}{m} \log v + \const)

  The entropy of a perfect gas as a function of temperature and specific volume, up to an arbitrary constant.

• General Deductions

  F = M\{c_{v} \theta (\const - \log \theta) - \frac{R\theta}{m} \log v + \const\}

  The free energy of a perfect gas, which contains an arbitrary linear function of temperature.

• General Deductions

  dF = -\frac{M\theta R}{m} · \frac{dv}{v} = -p\, dV \leq W

  For isothermal changes of a perfect gas the free energy change equals minus p dV and is bounded by the work done on the gas.

• This equation is in Proof (Proof)

• General Deductions

  d\left(\Phi - \frac{U + pV}{\theta}\right) \geq 0

  At constant temperature and pressure the quantity Phi minus (U+pV)/theta cannot decrease in a natural change.

• General Deductions

  \Phi - \frac{U + pV}{\theta} = \Psi

  Defines the function Psi as entropy minus (internal energy plus pV) over temperature.

• General Deductions

  \delta\left(\Phi - \frac{U}{\theta}\right) + \frac{W}{\theta} = 0

  The equilibrium condition at constant temperature, in virtual variations.

• General Deductions

  - \delta F = -W

  At constant temperature the virtual change of free energy equals the virtual work done on the system.

• General Deductions

  \delta F = 0

  Equilibrium at constant temperature with no external work: the free energy is at a minimum.

• General Deductions

  W = -p\, \delta V

  Virtual external work at constant pressure equals minus p times the virtual volume change.

• General Deductions

  \delta\Psi = 0

  At constant temperature and pressure, equilibrium is an absolute maximum of the Psi function.

• General Deductions

  \delta\Phi - \frac{\delta U - W}{\theta} \leq 0

  The general condition that equilibrium is maintained when only virtual changes satisfying it are permitted.

• General Deductions

  \delta\Phi - \frac{\delta U - W}{\theta} = 0

  The condition for equilibrium (76): for every permitted virtual change the variation vanishes.

• General Deductions

  \delta U = W

  In the first case (no heat exchange) the virtual energy change equals the virtual work.

• General Deductions

  \delta \Phi = 0

  Adiabatic equilibrium: the entropy is at a maximum among states reachable by adiabatic processes.

Problems

No exercises in this chapter.