8 Phase transitions questions

Michael Mombourquette

8 Phase transitions questions

Michael Mombourquette

Pentane, C₅H₁₂, boils at 309 K. Estimate the critical temperature using information from table 4. [475 K]
Which straight-chain hydrocarbon has a critical temperature closest to that of carbon dioxide?
Which of the substances in table 3, besides carbon dioxide, will not be found in the liquid state at normal atmospheric pressure?
Use the data from table 4 to calculate the van der Waals parameters for oxygen gas and Chloromethane. are they close to those found in table 2?
Use the data from the tables in this chapter to sketch the phase diagram for carbon dioxide. Label the triple point and the critical point. Explain why there is no melting point or boiling point data in the tables?
Use data from table 1 to calculate the temperature at which the vapour pressure of ethanol is 33.0 kPa?

Answers:

1. we can use the boiling temperature to estimate the critical temperature, using $\frac{T_b}{T_c}$ ≅ 0.6 + .04.
  actually, if you look at the alkanes in the table, you’ll see that they tend to be closer to .65, so let’s use that number.
  $\frac{T_b}{T_c}=.65.$ so $T_c = \frac{T_b}{0.65} = \frac{309\;K}{0.65} = 475\textrm{ K}$ . that’s in between the values given in table 4 for propane and hexane so this seems about right.
2. Looking at the data, we see ethane has T_c = 305.32 K which seems closest to the value for carbon dioxide of 304.14 K.
3. In Table 3, we see the triple point pressures. Any such pressure (Pt) that is above 1 bar (100 kPa) will not have a liquid phase under standard pressures. In table 3, the only two materials with this feature are:
  SiF₄ (P_t = 220.8 kPa) and SF₆ (P_t = 232 kPa).
4. ₂: $T_c = 154.59\textrm{ K } P_c = 5.043\textrm{ MPa}$
  Van der Waals coefficients can be calculated using the equations
  
  $a = \frac{27R^2T_c^2}{64P_c}$
  
  and
  
  $b = \frac{RT_c}{8P_c}$
  
  I’ll use basic SI units here (so convert things like MPa into Pa in the background)
  
  $a = \frac{27 \left(8.3145\mathrm{\frac{J}{K\;mol}}+ 154.59\mathrm{K}\right)^2}{64\times5.043\times 10^6\textrm{ kPa}} = 0.13282\mathrm{ \;Pa \;m^6 \;mol^{-2}}$
  
  and
  
  $b = \frac{8.3145\mathrm{\frac{J}{K\;mol}}\;154.59\mathrm{K}}{8\times5.043\times 10^6\textrm{ kPa}} = 3.186\times 10^{-5}\mathrm{m^3} = 31.86\mathrm{cm^3}$
  
  CH3Cl: $T_c = 416.25\textrm{ K } P_c = 6.679\textrm{ MPa}$
  
  $a = \frac{27 \left(8.3145\mathrm{\frac{J}{K\;mol}}+ 416.25\mathrm{K}\right)^2}{64\times6.679\times 10^6\textrm{ kPa}} = 0.7566\mathrm{ \;Pa \;m^6 \;mol^{-2}}$
  
  and
  
  $b = \frac{8.3145\mathrm{\frac{J}{K\;mol}}\;416.25\mathrm{K}}{8\times6.679\times 10^6\textrm{ kPa}} = 6.477\times 10^{-5}\mathrm{m^3} = 64.77\mathrm{cm^3}$
6. The boiling point of ethanol at 101.325 kPa is 351.3 K from table 1 and the enthalpy of vaporization at the boiling point is 38.56 kJ/mol (from the thermodynamics tables). We have one temperature/pressure points on the curve for ethanol and the enthalpy change. We can use the Clausius-Clapeyron equation to get the second temperature at 33.0 kPa..
  $ln \left(\frac{101.325}{33.0}\right) = - \,\frac{38560}{8.3145}\;\left(\frac{1}{T} - \frac{1}{351.3} \right)$
  
  solve for T = 325 K = 51°C.