Chemistry - Real Gas Question with Solution | TestHub
The critical temperature and pressure of a van der Waal's gas is -177°C and 20 atm, respectively. At the Boyle temperature, the gas behaves ideally up to 50 atm. If 0.2 moles of this gas is taken at the temperature and pressure given in Column I, then match with the expected volume of system in Column II (R = 0.08 l-atm/K-mol)
Column I | Column II |
|---|---|
A) -177°C, 20 atm | P) 801 ml |
B) 51°C, 6.48 atm | Q) 28.8 ml |
C) 77°C, 7.0 atm | R) 799 ml |
D) 27°C, 6.0 atm | S) 800 ml |
E) 51°C, 64.8 atm | T) 90 ml |
Options:
Answer:
Solution:
Tc = -177°C = 96 K, Pc = 20 atm, so b = RTc/8Pc = (0.08 × 96)/(8 × 20) = 0.048 l/mol, a = 27R2Tc2/64Pc ≈ 1.244 l2-atm/mol2, and Vc = 3b = 0.144 l/mol; TB = 27Tc/8 = 324 K = 51°C. (A) is the critical point: V = nVc = 0.2 × 144 = 28.8 ml (Q). (B) 51°C = TB, 6.48 atm < 50 atm, ideal: V = 0.2 × 0.08 × 324/6.48 = 800 ml (S). (C) 77°C = 350 K > TB: Z = 1 + (b - a/RT) P/RT = 1.0009, so V = 801 ml (P). (D) 27°C = 300 K < TB: Z = 0.9990, so V = 799 ml (R). (E) 51°C = TB but 64.8 atm > 50 atm: Z ≈ 1 + bP/RT = 1.12, so V = 90 ml (T). Hence A→Q; B→S; C→P; D→R; E→T.
