UPSC Combined Geo-Scientist, Mains · 2019 · Mains Paper II
Q1(b). Units of attraction parameter
Question
The pressure correction in the van der Waals equation is n²a/V². Find the SI unit of a.
Answer in brief
a has SI unit Pa m⁶ mol⁻²=kg m⁵ s⁻² mol⁻².
Key points
- Since the correction is added to pressure, an²/V² must have units of pascal.
- Therefore [a]=[p][V]²/[n]²=Pa(m³)²/mol².
- Using 1 Pa=1 kg m⁻¹ s⁻² gives the equivalent base-unit expression. The mole factor must be retained because n is amount of substance.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversUnits of attraction parameterReal gases
All parts of Question 1
- 1(a) Atomicity from heat capacities5 marks
- 1(b) Units of attraction parameter5 marks
- 1(c) Surface energy of cube and sphere5 marks
- 1(d) Atoms in a BCC sample5 marks
- 1(e) Volume independence of ideal-gas heat capacity5 marks
- 1(f) Entropy on doubling volume5 marks
- 1(g) Single-ion activity coefficients5 marks
- 1(h) Ammonia buffer5 marks
- 1(i) Pure-component vapour pressures5 marks
- 1(j) Zero-order half-life5 marks
- 1(k) Physical and chemical adsorption5 marks
- 1(l) Iron redox-electrode potential5 marks
- 1(m) Energy of one einstein5 marks
- 1(n) Bohr quantisation from matter waves5 marks
- 1(o) Spectral wavelength shifts5 marks
- 1(p) Bromobutane with one NMR signal5 marks