Temperature

Excerpts

• Temperature

  Two bodies of equal temperature are, therefore, in thermal equilibrium, and *vice versâ*.

• Temperature

  If a body, $A$, be in thermal equilibrium with two other bodies, $B$ and $C$, then $B$ and $C$ are in thermal equilibrium with one another.

• Temperature

  The definition of temperature remains arbitrary in cases where the requirements of accuracy cannot be satisfied by the agreement between the readings of the different gas thermometers, for there is no sufficient reason for the preference of any one of these gases.

• Temperature

  *Coefficient of elasticity* is the ratio of an infinitely small increase of pressure to the resulting contraction of unit volume of the substance.

• Temperature

  For lower pressures (*i.e.* to the left of the minimum), the volume decreases at a more rapid rate, with increasing pressure, than in the case of perfect gases; for higher pressures (to the right of the minimum), at a slower rate.

• Temperature

  Above the critical temperature and critical pressure, condensation does not exist, as the diagram plainly shows.

• Temperature

  The earlier fundamental distinction between liquids, vapours, and gases should therefore be dropped as no longer tenable.

• Temperature

  Only for gases and vapours does Dalton’s law hold, at least with great approximation, that the total pressure of a mixture is the sum of the partial pressures which each gas would exert if it alone filled the total volume at the given temperature.

• Temperature

  This direct sensation, however, furnishes no quantitative scientific measure of a body’s state with regard to heat; it yields only qualitative results, which vary according to external circumstances. For quantitative purposes we utilize the change of volume which takes place in all bodies when heated under constant pressure, for this admits of exact measurement.

• Temperature

  From this follows the important proposition: *If a body, $A$, be in thermal equilibrium with two other bodies, $B$ and $C$, then $B$ and $C$ are in thermal equilibrium with one another.*

• Temperature

  The definition of temperature is therefore somewhat arbitrary. This we may remedy to a certain extent by taking gases, in particular those hard to condense, such as hydrogen, oxygen, nitrogen, and carbon monoxide, as thermometric substances.

• Temperature

  The pressure of an atmosphere is the weight of a column of mercury at $0°$ C., $76~\Unit{cm.}$ high, and $1~\Unit{sq.}\ \Unit{cm.}$ in cross-section, when placed in mean geographical latitude. This latter condition must be added, because the weight, *i.e.* the force of the earth’s attraction, varies with the locality.

• Temperature

  Of these three values (indicated on the figure by $\alpha$, $\beta$, $\gamma$, for instance) only the smallest ($\alpha$) and the largest ($\gamma$) represent practically realizable states, for at the middle point ($\beta$) the pressure along the isotherm would increase with increasing volume, and the compressibility would accordingly be negative. Such a state has, therefore, only a theoretical signification.

• Temperature

  Condensation nowhere occurs in this process, which leads, nevertheless, to a region of purely liquid states. The earlier fundamental distinction between liquids, vapours, and gases should therefore be dropped as no longer tenable.

• Temperature

  Here the water vapour cannot be supposed to be subject to a pressure of $1~\Unit{atm.}$, since at $0°$ C. no water vapour exists at this pressure. The only choice remaining is to assign to the air and water vapour a common volume (that of the mixture) and different pressures (partial pressures).

Equations

• Temperature

  \frac{V}{M} = v

  The specific volume v is the volume V divided by the mass M of the substance.

• Temperature

  p = f(v, t)

  Every substance has a characteristic relation giving its pressure as a function of specific volume and temperature.

• Temperature

  pv = T

  At constant temperature the product of pressure and specific volume of a perfect gas is constant; T depends only on the temperature for a given gas.

• Temperature

  t = (v - v_{0})P

  At constant pressure the temperature is proportional to the difference between the present specific volume and the specific volume at 0 degrees C.

• Temperature

  pv_{0} = T_{0}

  The product of pressure and the specific volume at 0 degrees C equals the value of the temperature function T at t = 0 degrees C.

• Temperature

  v - v_{0} = \alpha v_{0}

  Heating a permanent gas from 0 to 1 degree C expands it by the same fraction alpha of its volume at 0 degrees C.

• Temperature

  1 = \alpha v_{0} P\Add{.}

  Substituting t = 1 into the temperature relation gives a condition on the constant P, the coefficient alpha and the specific volume at 0 degrees C.

• Temperature

  T = T_{0} (1 + \alpha t)

  The temperature function of a perfect gas is a linear function of the temperature t in degrees C.

• Temperature

  p = \frac{T_{0}}{v} (1 + \alpha t)

  The characteristic equation of a perfect gas at constant volume-pressure relation in degrees C, with pressure as a linear function of t at fixed specific volume.

• Temperature

  t + \dfrac{1}{\alpha} = \theta

  The absolute temperature theta is the Centigrade temperature shifted by 1/alpha degrees, so that the zero of theta lies about 273 degrees below the melting point of ice.

• Temperature

  \alpha T_{0} = C

  The constant C is defined as alpha times T_0, the characteristic constant for the perfect gas under consideration.

• Temperature

  p = \frac{C}{v} \theta = \frac{CM}{V} \theta\Add{.}

  In absolute temperature the characteristic equation of a perfect gas is pressure equal to C times theta over specific volume, or C M theta over V.

• Temperature

  \frac{1}{273} = \alpha

  The coefficient of expansion of a perfect gas equals 1/273, the fraction of its volume at 0 degrees C gained per degree.

• Temperature

  dV = \frac{CM\theta}{p^{2}}\, dp = \frac{V}{p}\, dp

  At constant temperature an infinitely small increase of pressure dp contracts the volume of a perfect gas by dV, which equals V dp over p.

• Temperature

  -\frac{dV}{V} = \frac{dp}{p}

  The fractional contraction of unit volume of a perfect gas equals the fractional increase of pressure.

• Temperature

  \frac{\;\;dp\;\;}{\dfrac{dp}{p}} = p

  The coefficient of elasticity of a perfect gas equals its pressure.

• Temperature

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

  The total differential of the characteristic equation: a change of pressure is the sum of its changes with temperature at constant volume and with volume at constant temperature.

• Temperature

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

  The change of volume with temperature at constant pressure equals minus the ratio of the pressure coefficient to the derivative of pressure with volume.

• Temperature

  \left(\frac{\dd p}{\dd \theta}\right)_{v} = -\left(\frac{\dd p}{\dd v}\right)_{\theta} · \left(\frac{\dd v}{\dd \theta}\right)_{p}

  The pressure coefficient equals minus the product of the compressibility-type derivative and the expansion-type derivative, so the three coefficients are linked.

• Temperature

  \left(\frac{\dd v}{\dd \theta}\right)_{p} · \frac{1}{v_{0}} = 0.00018

  For mercury at 0 degrees C and atmospheric pressure the coefficient of expansion is 0.00018.

• Temperature

  -\left(\frac{\dd v}{\dd p}\right)_{\theta} · \frac{1}{v_{0}} = 0.000003

  For mercury at 0 degrees C and atmospheric pressure the coefficient of compressibility, in atmospheres, is 0.000003.

• Temperature

  p = \frac{C_{1}M_{1} \theta}{V_{1}}

  Before diffusion, the first gas alone in volume V_1 at absolute temperature theta has pressure C_1 M_1 theta over V_1.

• Temperature

  V = V_{1} + V_{2} + \dots

  The total volume of a gas mixture before diffusion is the sum of the volumes of its constituents and remains constant during diffusion.

• Temperature

  p_{1} = \frac{C_{1}M_{1} \theta}{V} = \frac{V_{1}}{V} p

  After diffusion each gas fills the total volume, so its partial pressure p_1 is the fraction V_1/V of the total pressure.

• Temperature

  p_{1} + p_{2} + \dots = \frac{V_{1} + V_{2} + \dots}{V} p = p\Add{.}

  In a homogeneous mixture of gases the total pressure equals the sum of the partial pressures of the gases.

• Temperature

  p_{1} : p_{2} : \dots = V_{1} : V_{2} : \dots = C_{1}M_{1} : C_{2}M_{2} : \dots\Add{,}

  The partial pressures of the gases are proportional to their volumes before diffusion and to C_i M_i.

• Temperature

  p = (C_{1}M_{1} + C_{2}M_{2} + \dots) \frac{\theta}{V}

  The characteristic equation of a mixture of perfect gases has the form of that of a single perfect gas with summed constants.

• Temperature

  C = \frac{C_{1}M_{1} + C_{2}M_{2} + \dots}{M_{1} + M_{2} + \dots}

  The characteristic constant of a gas mixture is the mass-weighted average of the constants of its constituents.

• Temperature

  0.0014291 : 0.0012571 : 0.0012930 = \frac{1}{C_{1}} : \frac{1}{C_{2}} : \frac{1}{C_{3}}

  The ratio of the densities of oxygen, atmospheric nitrogen and air equals the ratio of the reciprocals of their characteristic constants.

• Temperature

  C = \frac{C_{1}M_{1} + C_{2}M_{2}}{M_{1} + M_{2}}

  The characteristic constant of a two-gas mixture is the mass-weighted average of the constants of the two gases.

• Temperature

  M_{1} : M_{2} = 0.2998

  The mass ratio of oxygen to atmospheric nitrogen in air is 0.2998, i.e. 23.1 per cent oxygen by weight.

• Temperature

  C_{1}M_{1} : C_{2}M_{2} = p_{1} : p_{2} = V_{1} : V_{2} = 0.2637

  The volume (and partial pressure) ratio of oxygen to nitrogen in air is 0.2637, i.e. 20.9 per cent oxygen by volume.

• Temperature

  pv = \const

  At constant temperature the product of pressure and volume of a perfect gas is constant, so its isotherms are equilateral hyperbolae.

• Temperature

  p = \frac{R\theta}{v - b} - \frac{a}{v^{2}}

  An approximate characteristic equation for gases and liquids, which reduces to that of a perfect gas for large specific volumes.

• Temperature

  p = \frac{R\theta}{v - a} - \frac{c}{\theta(v + b)^{2}}

  Clausius' characteristic equation, an improvement on van der Waals' equation with an extra constant c, approaches the perfect-gas form for large specific volumes.

• Temperature

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

  At the critical point the tangent to the isotherm is parallel to the axis of abscissae, so the slope of pressure with volume vanishes.

• Temperature

  \left(\frac{\dd^{2} p}{\dd v^{2}}\right)_{\theta} = 0

  At the critical point the isotherm has a point of inflection, so the second derivative of pressure with volume vanishes.

• Temperature

  \theta^{2} = \frac{8c}{27(a + b)R}

  From Clausius' equation the critical absolute temperature squared equals 8c over 27(a+b)R.

• Temperature

  p^{2} = \frac{cR}{216(a + b)^{3}}

  From Clausius' equation the square of the critical pressure equals cR over 216(a+b)^3.

• Temperature

  v = 3a + 2b

  From Clausius' equation the critical specific volume equals 3a + 2b.

Problems

No exercises in this chapter.