UPSC Combined Geo-Scientist, Mains · 2025 · Mains Paper II
Q1(a). Why Z > 1 for H₂ and He at all pressures
Question
Z (compressibility factor) versus P plots of H₂ and He gas are always above the plot of ideal gases. Explain why.
Answer in brief
In H₂ and He the attraction term a is very small, so repulsion (molecular size, b) dominates at every pressure and Z > 1.
Key points
- Weak attraction: only two electrons, low polarisability ⇒ very small a.
- So a/RT ≪ b at ordinary T ⇒ slope (b − a/RT)/RT > 0 ⇒ Z > 1 from P = 0.
- Same statement: room temperature is far above their Boyle temperature (He 23 K, H₂ 110 K).
- Limitation: below TB even H₂ and He show the dip. “Always” holds only for T > TB.
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.
Coverscompressibility factorvan der Waals equationBoyle temperaturereal gaseshydrogen and helium
All parts of Question 1
- 1(a) Why Z > 1 for H₂ and He at all pressures5 marks
- 1(b) Corresponding state of H₂S for N₂ at 25 °C and 1 atm5 marks
- 1(c) Band structure of n-type and p-type semiconductors5 marks
- 1(d) Symmetry elements of orthorhombic Mg₂SiO₄5 marks
- 1(e) Fall of G with temperature at constant pressure5 marks
- 1(f) Z > 1 and f > P for the gas P(Vm − b) = RT5 marks
- 1(g) Overall equation and the role of each species5 marks
- 1(h) log k against pH for specific H⁺ catalysis5 marks
- 1(i) KCl salt bridge for a silver–lead cell5 marks
- 1(j) Equilibrium constant from standard potentials5 marks
- 1(k) Conditions for an acceptable wave function5 marks
- 1(l) No solution with E = 0 for the particle in a box5 marks
- 1(m) Discrete levels and the resonance condition5 marks
- 1(n) Rotational activity of CO and CO₂5 marks
- 1(o) Measuring light intensity with a chemical actinometer5 marks
- 1(p) Why watch hands glow in the dark5 marks