Question

5.(20pts) () For each of the following pairs, state which has the greater Van der Waals "a", which has the greater van der Waals "b", and which has the greater T: (a) He or Ne, (b) ethane (8) or propane (g),(c) H2O(g) or H,S(g) (ii) Read Sections 8,3 and 8.4 in Levine's Textbook. The normal boiling point of benzene is 80°C. and the density of liquid benzene at 80 °C is 0.81 g/cm? Estimate Po Te amd Vme.

Answer 1

Answer (i) (a) Helium has highen values of 'a and b. (6) Propane has higher values of 'a' and 's. (c) H0 has higher a - H₂S has higher 'b The Vander waal's equation - (Pt an) (v-nb) = nRT where a - Correction for intermolecular forces b- Correction for Volume of O molecules. 16 In case of ethane and Propane - Propane is larger so more London forces attractive) holding the molecule together compared to ethane. (6) Among H₂O and H₂S - D' is more electronegative so Hydrogen - Oxygen bond is more polar and hence strong intermolecular dipol-dipole interactions so value of à is more for H₂O. But since this is larger because of larger Sulphur compared to oxygen that is why 'b value is more for H₂S. Te is directly related to the attraction between individual atoms. Therefore (a) He or Ne – It is Helium having high to (6) propane are than – Propane with high to (c) H₂0 or H₂S - H₂O with high Te. (1) Normal Boiling point To = 80c = 353k. To Te - 3 To = 3 x 353=529.5K From the Vander waal's equation and critical state: (Pt 2) (Vm- b) = RT we get the following equations- [Vmx = 36 and P= 216 Knowing the values of a and 'b for liquid Benzene we get values of Vmic and Pc.