Chemistry - Real Gas Question with Solution | TestHub
The critical temperature and pressure of a van der Waal's gas is -145°C and 32 atm, respectively. At the Boyle temperature, the gas behaves ideally up to 80 atm. If 1 mole 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) -145°C, 32 atm | P) 120 ml |
B) 159°C, 10.8 atm | Q) 2.0 L |
C) 159°C, 108 atm | R) 3.2 L |
D) 227°C, 10 atm | S) 360 ml |
E) 127°C, 16 atm | T) 4.0 L |
Options:
Answer:
Solution:
Tc = -145°C = 128 K, Pc = 32 atm, so b = RTc/8Pc = (0.08 × 128)/(8 × 32) = 0.04 l/mol and Vc = 3b = 0.12 l/mol; TB = 27Tc/8 = 432 K = 159°C. (A) is the critical point: V = nVc = 1 × 120 = 120 ml (P). (B) 159°C = TB, 10.8 atm < 80 atm, ideal: V = 0.08 × 432/10.8 = 3.2 L (R). (C) 159°C = TB but 108 atm > 80 atm, outside the ideal range: with Z = 1 + bP/RT, V = nRT/P + nb = 0.08 × 432/108 + 0.04 = 0.32 + 0.04 = 0.36 L = 360 ml (S). (D) 227°C = 500 K, 10 atm: V = 0.08 × 500/10 = 4.0 L (T). (E) 127°C = 400 K, 16 atm: V = 0.08 × 400/16 = 2.0 L (Q). Hence AP; BR; CS; DT; EQ.
