Applications to Homogeneous Systems

Excerpts

• Applications to Homogeneous Systems

  The results of the experiment show that when the flow has become steady there is, for air, a very small change of temperature, and, for hydrogen, a still smaller, hardly appreciable change.

• Applications to Homogeneous Systems

  Strictly speaking, this expression is vague, since a process presupposes changes, and, therefore, disturbances of equilibrium.

• Applications to Homogeneous Systems

  Wherever external pressure enters---as, for instance, in the calculation of the work of compression---a very small error will then be committed, if the pressure of the gas be substituted for the external pressure.

• Applications to Homogeneous Systems

  Only for infinitesimal changes, *i.e.* when $1$ and $2$ are infinitely near one another and $\alpha$ shrinks to a curve element, is $W$ determined by the initial and final points of the curve alone.

• Applications to Homogeneous Systems

  Carnot’s cycle, performed with a perfect gas, thus affords a means of drawing heat from a body and of gaining work in its stead, without introducing any changes in nature except the transference of a certain quantity of heat from a body of higher temperature to one of lower temperature.

• Applications to Homogeneous Systems

  The term *homogeneous* is used here in the sense of *physically homogeneous*, and is applied to any system which appears of completely uniform structure throughout.

• Applications to Homogeneous Systems

  He put the two communicating vessels, one filled with air at high pressure, the other exhausted, into a common water-bath at the same temperature, and found that, after the air had expanded and equilibrium had been established, the change of temperature of the water-bath was inappreciable.

• Applications to Homogeneous Systems

  In other words, *the internal energy of a perfect gas depends only on the temperature, and not on the volume*.

• Applications to Homogeneous Systems

  Thus, a gas may be compressed very slowly to any fraction of its original volume, by making the external pressure, at each moment, just a trifle greater than the internal pressure of the gas.

• Applications to Homogeneous Systems

  In fact, it assumes an entirely different value along a different curve, $\beta$, joining $1$ and $2$. Therefore $p\, dV$ is not a perfect differential.

• Applications to Homogeneous Systems

  In isothermal changes $C$ is evidently $= \pm\infty$, because $d\theta = 0$, and the heat added or withdrawn is a finite quantity. In adiabatic changes $C = 0$, for here the temperature may change in any way, while no heat is added or withdrawn.

• Applications to Homogeneous Systems

  It is usual to follow the example of Clausius, and denote this quantity by $dQ$, to indicate that it is infinitely small. This notation, however, has frequently given rise to misunderstanding, for $dQ$ has been repeatedly regarded as the differential of a known finite quantity $Q$.

• Applications to Homogeneous Systems

  By reversing Carnot’s cycle, we have, then, a means of transferring heat from a colder to a hotter body without introducing any other changes in nature than the transformation of a certain amount of mechanical work into heat.

Equations

• Applications to Homogeneous Systems

  Q = U_{2} - U_{1} + \int_{1}^{2} p\, dV

  The heat absorbed along a reversible path equals the change of internal energy plus the integral of pressure over volume, so Q depends on the path.

• Applications to Homogeneous Systems

  U_{2} - U_{1} = Q + W

  The change in internal energy equals the heat added plus the work done, as stated for the homogeneous system (equation 17 applied).

• This equation is in General Exposition (General Exposition)

• Applications to Homogeneous Systems

  \theta_{2} = \theta_{1}

  Joule's experiment found that for perfect gases the temperature is unchanged by free expansion.

• Applications to Homogeneous Systems

  \left(\frac{\dd U}{\dd V}\right)_{\theta} = 0\Add{.}

  For a perfect gas the internal energy does not change with volume at constant temperature, so it depends only on temperature.

• Applications to Homogeneous Systems

  Q = 0

  No heat passes through the non-conducting porous plug tube, so the heat term vanishes.

• Applications to Homogeneous Systems

  W = p_{1}V_{1} - p_{2}V_{2},

  The external work done on a nearly perfect gas pushed through the porous plug is the difference of the pressure-volume products at the two ends.

• Applications to Homogeneous Systems

  W = -\int_{1}^{2} p\, dV

  The external work of a reversible process equals minus the integral of pressure over volume along the path from state 1 to state 2.

• Applications to Homogeneous Systems

  Q = -W

  For a complete cycle returning to the initial state, the heat absorbed equals minus the external work.

• Applications to Homogeneous Systems

  W = -\int_{1}^{1} p\, dV

  The external work over a complete cycle is minus the integral of pressure around the closed curve, which equals the area enclosed.

• Applications to Homogeneous Systems

  Q = dU + p\, dV

  For an infinitesimal reversible change, the heat absorbed equals the increment of internal energy plus pressure times the increment of volume.

• Applications to Homogeneous Systems

  U_{2} - U_{1} = W + Q

  Along a reversible curve from state 1 to state 2, the energy increase equals the work expended plus the heat absorbed.

• Applications to Homogeneous Systems

  q = du + p\, dv

  Per unit mass, the heat absorbed equals the increment of specific internal energy plus pressure times the increment of specific volume.

• Applications to Homogeneous Systems

  c = \frac{q}{d\theta} = \frac{du}{d\theta} + p\, \frac{dv}{d\theta}

  The specific heat for any heating process equals heat per unit mass divided by the temperature increment, expressed through internal energy and volume changes.

• Applications to Homogeneous Systems

  c_{v} = \left(\frac{\dd u}{\dd \theta}\right)_{v}\Add{,}

  The specific heat at constant volume is the rate of change of specific internal energy with temperature at fixed volume.

• Applications to Homogeneous Systems

  c_{v} = \left(\frac{\dd u}{\dd p}\right)_{v} \left(\frac{\dd p}{\dd \theta}\right)_{v}\Add{.}

  The specific heat at constant volume can be written as the product of internal energy change with pressure and pressure change with temperature, both at constant volume.

• Applications to Homogeneous Systems

  c_{p} = \left(\frac{\dd u}{\dd \theta}\right)_{p} + p\left(\frac{\dd v}{\dd \theta}\right)_{p}\Add{,}

  The specific heat at constant pressure equals the internal energy rate plus pressure times the volume rate, both at constant pressure.

• Applications to Homogeneous Systems

  c_{p} = \left[\left(\frac{\dd u}{\dd v}\right)_{p} + p\right]\left(\frac{\dd v}{\dd \theta}\right)_{p}\Add{.}

  The specific heat at constant pressure equals the bracket of internal energy change with volume plus pressure, times the volume change with temperature at constant pressure.

• Applications to Homogeneous Systems

  c_{p} = c_{v} + \left[\left(\frac{\dd u}{\dd v}\right)_{\theta} + p\right]\left(\frac{\dd v}{\dd \theta}\right)_{p}

  The difference between the specific heats at constant pressure and constant volume is given by the internal energy dependence on volume plus pressure, times the expansion coefficient at constant pressure.

• Applications to Homogeneous Systems

  (c_{p} - c_{v})\, \frac{\dd^{2} \theta}{\dd p\, \dd v} + \frac{\dd c_{p}}{\dd p} · \frac{\dd \theta}{\dd v} - \frac{\dd c_{v}}{\dd v} · \frac{\dd \theta}{\dd p} = 1\Add{.}

  A relation among measurable quantities for any homogeneous substance, which tests the first law of thermodynamics by experiment.

• This equation is in Molecular Weight (Molecular Weight)

• Applications to Homogeneous Systems

  \theta = \frac{m}{R}\, pv

  Rearranging the characteristic equation, temperature equals molecular weight over the gas constant times pressure times volume.

• Applications to Homogeneous Systems

  c_{p} - c_{v} + p\, \frac{\dd c_{p}}{\dd p} - v\, \frac{\dd c_{v}}{\dd v} = \frac{R}{m}

  For a perfect gas, the first-law test relation reduces to this condition involving the two specific heats and the gas constant per molecular weight.

• Applications to Homogeneous Systems

  \left(\frac{\dd u}{\dd v}\right)_{\theta} = 0

  For a perfect gas the internal energy per unit mass is independent of volume at constant temperature.

• Applications to Homogeneous Systems

  du = \left(\frac{\dd u}{\dd \theta}\right)_{v} d\theta + \left(\frac{\dd u}{\dd v}\right)_{\theta} dv

  The general differential of specific internal energy as a function of temperature and volume.

• Applications to Homogeneous Systems

  du = \left(\frac{\dd u}{\dd \theta}\right)_{v} d\theta

  For a perfect gas the increment of specific internal energy depends on temperature alone.

• Applications to Homogeneous Systems

  du = c_{v} · d\theta

  For a perfect gas the increment of specific internal energy equals the specific heat at constant volume times the temperature increment.

• Applications to Homogeneous Systems

  c_{p} = c_{v} + p \left(\frac{\dd v}{\dd \theta}\right)_{p}

  For a perfect gas the difference of specific heats equals pressure times the volume change with temperature at constant pressure.

• Applications to Homogeneous Systems

  c_{p} = c_{v} + \frac{R}{m}

  For a perfect gas the specific heat at constant pressure exceeds that at constant volume by the gas constant divided by molecular weight.

• Applications to Homogeneous Systems

  mc_{p} - mc_{v} = R

  The molecular heats at constant pressure and at constant volume differ by the gas constant, independently of the nature of the gas.

• Applications to Homogeneous Systems

  mc_{p} - mc_{v} = \frac{R}{J} = \frac{826 · 10^{5}}{419 · 10^{5}} = 1.971\Add{.}

  With molecular heats in calories, the difference of molecular heats equals the gas constant divided by Joule's equivalent, about 1.971.

• Applications to Homogeneous Systems

  u = c_{v} \theta + \const

  Over a range where specific heat is constant, specific internal energy is proportional to temperature plus a constant set by the zero of energy.

• Applications to Homogeneous Systems

  0 = du + p\, dv

  In an adiabatic process no heat is absorbed, so the increment of internal energy plus pressure times volume increment is zero.

• Applications to Homogeneous Systems

  0 = c_{v}\, d\theta + \frac{R}{m} · \frac{\theta}{v}\, dv

  For a perfect gas undergoing an adiabatic change, the specific heat times the temperature increment plus the gas-law term times the volume increment vanishes.

• Applications to Homogeneous Systems

  \log \theta + (\gamma - 1) \log v = \const

  During an adiabatic change of a perfect gas, the log of temperature plus (gamma minus one) times the log of volume is constant.

• Applications to Homogeneous Systems

  -\gamma \log \theta + (\gamma - 1) \log p = \const

  During adiabatic compression of a perfect gas the temperature rises, as expressed in terms of temperature and pressure.

• Applications to Homogeneous Systems

  \log p + \gamma \log v = \const

  For an adiabatic change of a perfect gas, log pressure plus gamma times log volume is constant.

• Applications to Homogeneous Systems

  pv^{\gamma} = \const

  For a perfect gas in an adiabatic change, pressure times volume raised to gamma is constant, so adiabatic curves are steeper than isotherms.

• Applications to Homogeneous Systems

  \frac{p}{\rho^{\gamma}} = \const

  In the adiabatic compressions of sound waves in a perfect gas, pressure over density raised to gamma is constant.

• Applications to Homogeneous Systems

  \frac{dp}{d\rho} = \frac{\gamma p}{\rho} = \gamma pv

  Differentiating the adiabatic relation gives the rate of change of pressure with density as gamma times pressure over density, equal to gamma times pressure times volume.

• Applications to Homogeneous Systems

  \frac{dp}{d\rho} = \gamma \frac{R}{m} \theta

  For a perfect gas the rate of change of pressure with density equals gamma times the gas constant over molecular weight times temperature.

• Applications to Homogeneous Systems

  \gamma = \frac{m}{R\theta} · \frac{dp}{d\rho}

  The ratio of specific heats can be determined from the velocity of sound, using molecular weight, gas constant, temperature and the pressure-density derivative.

• Applications to Homogeneous Systems

  \sqrt{\dfrac{dp}{d\rho}}

  The velocity of sound in a fluid equals the square root of the rate of change of pressure with density.

• Applications to Homogeneous Systems

  \gamma = \frac{28.8}{826 · 10^{5}} · \frac{33280^{2}}{273} = 1.41

  From the measured velocity of sound in air at 0°, the ratio of specific heats for air comes out at 1.41, agreeing with the calorimetric value.

• Applications to Homogeneous Systems

  W = -\frac{R}{m} \left(\theta_{2} \log \frac{v_{2}'}{v_{2}} + \theta_{1} \log \frac{v_{1}'}{v_{1}}\right)

  For the Carnot cycle with a perfect gas, the external work is the sum of the isothermal contributions at the two reservoir temperatures.

• Applications to Homogeneous Systems

  \frac{v_{2}'}{v_{2}} = \frac{v_{1}'}{v_{1}}

  The two adiabatic branches of the Carnot cycle give equal volume ratios for the isothermal expansions.

• Applications to Homogeneous Systems

  W = -\frac{R}{m} (\theta_{2} - \theta_{1}) \log \frac{v_{1}'}{v_{1}}

  The external work of the Carnot cycle with a perfect gas equals minus the gas constant term times the temperature difference times the log of the volume ratio.

• Applications to Homogeneous Systems

  Q = Q_{1} + Q_{2}= -W

  Over the Carnot cycle the net heat absorbed equals minus the external work, so the net heat is positive when the cycle gains work.

• Applications to Homogeneous Systems

  Q = Q_{1} + Q_{2} = \frac{R}{m} (\theta_{2} - \theta_{1}) \log \frac{v_{1}'}{v_{1}}

  The net heat taken in during the Carnot cycle of a perfect gas equals gas constant over molecular weight times the temperature difference times the log volume ratio.

• Applications to Homogeneous Systems

  Q_{2} = \frac{R}{m} \theta_{2} \log \frac{v_{2}'}{v_{2}} = \frac{R}{m} \theta_{2} \log \frac{v_{1}'}{v_{1}}

  The heat absorbed in the isothermal expansion at the upper reservoir equals the gas-law work of that expansion.

• Applications to Homogeneous Systems

  Q_{1} = \frac{R}{m} \theta_{1} \log \frac{v_{1}}{v_{1}'} = -\frac{R}{m} \theta_{1} \log \frac{v_{1}'}{v_{1}}

  The heat absorbed in the isothermal compression at the lower reservoir is negative and equals the gas-law work of that compression.

• Applications to Homogeneous Systems

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

  For a reversible Carnot cycle of a perfect gas, the heats and work are in the ratio of the reservoir temperatures.

• Applications to Homogeneous Systems

  Q + W = 0

  Over a complete cycle the sum of heat absorbed and work done on the system is zero.

• Applications to Homogeneous Systems

  Q = Q_{1} + Q_{2}

  The total heat absorbed over the Carnot cycle is the sum of the heats taken from the two reservoirs, with Q_1 negative.

Problems

No exercises in this chapter.