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Next: Physical Constants Up: Multi-Phase Systems Previous: Phase Transformations in Van


  1. Consider a relatively small region of a homogeneous substance, containing $ N\gg 1$ particles, that is in equilibrium with the remainder of the substance. The region is characterized by a volume, $ V$ , and a temperature, $ T$ .
    1. Show that the properly normalized probability that the volume lies between $ V$ and $ V + dV$ , and the temperature lies between $ T$ and $ T+dT$ , is

      $\displaystyle P(V, T) dV dT = \frac{1}{2\pi {\mit\Delta}^\ast V {\mit\Delta...
...t[-\frac{(T-\overline{T})^{ 2}}{2 ({\mit\Delta}^\ast T)^{ 2}}\right]dV dT,


      $\displaystyle {\mit\Delta}^{\ast} V$ $\displaystyle =\left(k \overline{T} \kappa_T \overline{V}\right)^{1/2},$    
      $\displaystyle {\mit\Delta}^{\ast} T$ $\displaystyle =\left(\frac{k \overline{T}^{ 2}}{C_V}\right)^{1/2}.$    

      Here, the mean temperature, $ \overline{T}$ , is the same as that of the remainder of the substance, whereas the mean volume, $ \overline{V}$ , is such that when $ T=\overline{T}$ the pressure of the region matches that of the surrounding substance.

    2. Hence, deduce that

      $\displaystyle \langle (V-\overline{V}) (T-\overline{T})\rangle = 0.

      In other words, the volume and temperature fluctuations are uncorrelated.

    3. Show that for the case of a monatomic ideal gas

      $\displaystyle \frac{{\mit\Delta}^\ast V}{\overline{V}}$ $\displaystyle = \left(\frac{1}{N}\right)^{1/2},$    
      $\displaystyle \frac{{\mit\Delta}^\ast T}{\overline{T}}$ $\displaystyle = \left(\frac{2}{3 N}\right)^{1/2},$    
      $\displaystyle \frac{{\mit\Delta}^\ast p}{\bar{p}}$ $\displaystyle = \left(\frac{5}{3 N}\right)^{1/2},$    

      where $ p$ is the pressure of the region. Furthermore, demonstrate that

      $\displaystyle \frac{\langle (p-\bar{p}) (V-\overline{V})\rangle}{\bar{p} \overline{V}}$ $\displaystyle = -\frac{1}{N},$    
      $\displaystyle \frac{\langle (p-\bar{p}) (T-\overline{T})\rangle}{\bar{p} \overline{T}}$ $\displaystyle = \frac{2}{3 N}.$    

  2. A substance of molecular weight $ \mu$ has its triple point at the absolute temperature $ T_0$ and pressure $ p_0$ . At this point, the mass densities of the solid and liquid phases are $ \rho_s$ and $ \rho_l$ , respectively, while the vapor phase can be approximated as a dilute ideal gas. If, at the triple point, the slope of the melting curve is $ (dp/dT)_m$ , and that of the liquid vaporization curve is $ (dp/dT)_v$ , show that the slope of the sublimation curve can be written

    $\displaystyle \left(\frac{dp}{dT}\right)_s = \frac{\mu p_0}{R T_0}\left(\frac...
..._s \rho_l}\right)\left(\frac{dp}{dT}\right)_m + \left(\frac{dp}{dT}\right)_v.

  3. The vapor pressure of solid ammonia (in millimeters of mercury) is given by $ \ln p = 23.02-3754/T$ , and that of liquid ammonia by $ \ln p = 19.48-3063/T$ . Here, $ T$ is in degrees kelvin.
    1. Deduce that the triple point of ammonia occurs at $ 195.2$ K.
    2. Show that, at the triple point, the latent heats of sublimation, vaporization, and melting of ammonia are $ 31.22 {\rm kJ/mol}$ , $ 25.47 {\rm kJ/mol}$ , and $ 5.75 {\rm kJ/mol}$ , respectively.

  4. Water boils when its vapor pressure is equal to that of the atmosphere. The boiling point of pure water at ground level is $ 100^\circ$ C. Moreover, the latent heat of vaporization at this temperature is $ 40.66 {\rm kJ/mol}$ . Show that the boiling point of water decreases approximately linearly with increasing altitude such that

    $\displaystyle T_b(z) \simeq 100 - \frac{z}{295.0},

    where $ z$ measures altitude above ground level (in meters), and $ T_b(z)$ is the boiling point in degrees centigrade at this altitude. In other words, the boiling point of pure water is depressed by about $ 3.4^\circ$ C per kilometer increase in altitude.

  5. The relative humidity of air is defined as the ratio of the partial pressure of water vapor to the equilibrium vapor pressure at the ambient temperature. The dew point is defined as the temperature at which the relative humidity becomes $ 100\%$ . Show that if the relative humidity of air at (absolute) temperature $ T$ is $ h(T)$ then the dew point is given by

    $\displaystyle T_d \simeq T-\frac{R T^{ 2}}{l} \ln\left[\frac{1}{h(T)}\right].

    Here, $ R$ is the ideal gas constant, and $ l$ the molar latent heat of vaporization of water. The molar latent heat of water at $ 25^\circ$ C is $ 43.99 {\rm kJ/mol}$ . Suppose that air at $ 25^\circ$ C has a relative humidity of $ 90\%$ . Estimate the dew point. Suppose that the relative humidity is $ 40\%$ . Estimate the dew point.

  6. When a rising air mass in the atmosphere becomes saturated (i.e., attains $ 100\%$ relative humidity), condensing water droplets give up energy, thereby slowing the adiabatic cooling process.
    1. Use the first law of thermodynamics to show that, as condensation forms during adiabatic expansion, the temperature of the air mass changes by

      $\displaystyle dT = \frac{2}{7} \frac{T}{p} dp -\frac{2}{7} \frac{l}{R} \frac{d\nu_w}{d\nu},

      where $ \nu_w$ is the number of moles of water vapor, $ \nu$ the total number of moles, and $ l$ the molar latent heat of vaporization of water. Here, we have assumed that $ \gamma=7/5$ for air.
    2. Assuming that the air is always saturated during this process, show that

      $\displaystyle d\left(\frac{\nu_w}{\nu}\right) = \frac{p_v(T)}{p} \frac{l}{R T} \frac{dT}{T} - \frac{p_v}{p} \frac{dp}{p},

      where $ p_v(T)$ is the vapor pressure of water at temperature $ T$ , and $ p$ is the pressure of the atmosphere.
    3. Use the equation of hydrostatic equilibrium of the atmosphere,

      $\displaystyle \frac{dp}{p} =-\frac{\mu g}{R T} dz,

      where $ z$ represents altitude, $ \mu$ is the molecular weight of air, and $ g$ the acceleration due to gravity, to obtain the following expression so-called wet adiabatic lapse-rate of the atmosphere:

      $\displaystyle \frac{dT}{dz} =-\left(\frac{2}{7} \frac{\mu g}{R}\right) \frac{1+(p_v/p) (l/R T)}{1+(2/7) (p_v/p) (l/R T)^{ 2}}.

      Of course, $ dT/dz = -(2/7) (\mu g/R)$ is the dry adiabatic lapse-rate derived in Section 6.8.
    4. At $ 25^\circ$ C, the vapor pressure of water is $ 0.0317$ bar, and the molar latent heat of vaporization is $ 43.99 {\rm kJ/mol}$ . At $ 0^\circ$ C, the vapor pressure of water is $ 0.00611$ bar, and the molar latent heat of vaporization is $ 51.07 {\rm kJ/mol}$ . What is the ratio of the wet adiabatic lapse-rate to the dry adiabatic lapse-rate at these two temperatures?

  7. Consider a phase transition in a van der Waals fluid whose reduced equation of state is

    $\displaystyle p'=\frac{8}{3} \frac{T'}{v'-1/3} - \frac{3}{v'^{ 2}}.

    Let $ T'$ and $ p_v'$ be the constant temperature and reduced vapor pressure at the transition, respectively, and let $ v_l'$ and $ v_g'$ be the reduced molar volumes of the liquid and gas phases, respectively. The fact that the phase transition takes place at constant temperature and pressure implies that

    $\displaystyle p_v' =\frac{8}{3} \frac{T'}{v_l'-1/3}-\frac{3}{v_l'^{ 2}} = \frac{8}{3} \frac{T'}{v_g'-1/3}-\frac{3}{v_g'^{ 2}}.

    Moreover, the Maxwell construction yields

    $\displaystyle \int_l^g v' dp' = 0.

    1. Demonstrate that the Maxwell construction implies that

      $\displaystyle p_v' (v_g'-v_l')= \frac{8}{3} T' \ln\left(\frac{v_g'-1/3}{v_l'-1/3}\right) + 3\left(\frac{1}{v_g'}-\frac{1}{v_l'}\right).

    2. Eliminate $ T'$ and $ p_v'$ from the previous equations to obtain the transcendental equation

      $\displaystyle \ln\left(\frac{v_g'-1/3}{v_l'-1/3}\right) =\left(\frac{v_g'-v_l'}{v_g'+v_l'}\right)\left(\frac{v_g'}{v_g'-1/3} + \frac{v_l'}{v_l'-1/3}\right)

      that relates the molar volumes of the liquid and gas phases.
    3. Writing $ x_+ = 1/(3 v_l'-1)$ and $ x_-=1/(3 v_g'-1)$ , show that the previous equation reduces to

      $\displaystyle \ln\left(\frac{x_+}{x_-}\right) = \frac{(x_+-x_-) (x_++x_-+2)}{(x_++x_-+2 x_+ x_-)}.

    4. By setting both sides of the previous equation equal to $ y$ , show that it can be solved parametrically to give

      $\displaystyle x_+$ $\displaystyle = {\rm e}^{+y/2} f(y/2),$    
      $\displaystyle x_-$ $\displaystyle = {\rm e}^{-y/2} f(y/2),$    


      $\displaystyle f(x) = \frac{x \cosh x -\sinh x}{\sinh x \cosh x - x}.

    5. Furthermore, demonstrate that

      $\displaystyle T'$ $\displaystyle = \frac{27}{4} \frac{f (c+f)}{g^{ 2}},$    
      $\displaystyle p_v'$ $\displaystyle = \frac{27 f^{ 2} (1-f^{ 2})}{g^{ 2}},$    
      $\displaystyle v_g'-v_l'$ $\displaystyle = \frac{2 s}{3 f},$    
      $\displaystyle \frac{s_g-s_l}{R}$ $\displaystyle = y,$    


      $\displaystyle f$ $\displaystyle = f(y/2),$    
      $\displaystyle c$ $\displaystyle = \cosh(y/2),$    
      $\displaystyle s$ $\displaystyle = \sinh(y/2),$    
      $\displaystyle g$ $\displaystyle = 1+ 2 c f+f^{ 2}.$    

      Here, $ s_g$ and $ s_l$ denote the molar entropies of the gas and liquid phases, respectively.

    6. Finally, by considering the limits $ y\gg 1$ and $ y\ll 1$ , show that

      $\displaystyle p_v'$ $\displaystyle = 27 \exp\left(-\frac{27}{8 T'}\right),$    
      $\displaystyle \frac{l_{lg}}{R T_c}$ $\displaystyle = \frac{27}{4},$    

      in the limit $ T'\ll 1$ , and

      $\displaystyle v_g'-v_l'$ $\displaystyle =4 (1-T')^{1/2},$    
      $\displaystyle \frac{l_{lg}}{R T_c}$ $\displaystyle = 6 (1-T')^{1/2},$    

      in the limit $ T'\rightarrow 1$ . Here, $ l_{lg}$ is the molar latent heat of vaporization.

next up previous
Next: Physical Constants Up: Multi-Phase Systems Previous: Phase Transformations in Van
Richard Fitzpatrick 2016-01-25