The critical temperature and pressure of a van der Waal's gas is -177°C and 20 atm, respectively. At — States of Matter and Gaseous State Chemistry Question
Question
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: (A) -177°C, 20 atm (B) 51°C, 6.48 atm (C) 77°C, 7.0 atm (D) 27°C, 6.0 atm (E) 51°C, 64.8 atm Column II: (P) 821 ml (Q) 28.8 ml (R) 760 ml (S) 800 ml (T) 85 ml
💡 Solution & Explanation
$T_c=96$ K, $P_c=20$ atm. $b=RT_c/8P_c=0.048$ L/mol; $V_c=3b=0.144$ L/mol. $T_b=3.375T_c=324$ K ($51^\circ$C). (A) Critical state: $V=0.2\times0.144=0.0288$ L $=28.8$ mL → Q. (B) 51°C (=$T_b$), 6.48 atm: ideal behaviour, $V=nRT/P=0.2\times0.08\times324/6.48=0.800$ L $=800$ mL → S. (C) 77°C ($>T_b$, repulsive): $V>V_{ideal}=800$ mL; $V\approx800+(0.2\times48)=821$ mL → P. (D) 27°C ($<T_b$, attractive): $V<V_{ideal}=800$ mL; $V\approx760$ mL → R. (E) 51°C, 64.8 atm (high P): $V=nRT/P+nb=80+9.6\approx85$ mL → T.