NTA Abhyas JEE Main2020ChemistryStates of MatterPractice
The reduced temperature = θ = T T C The reduced pressure = π = P P C The reduced volume = ϕ = V V C Hence, it can be said that the reduced equation of state may be given as
Options
- Aπ 3 + 1 ϕ 2 3 ϕ - 1 = 8 3 θ
- Bπ 3 + 1 ϕ ϕ - 1 = 8 3 θ
- Cπ 4 + 1 ϕ 3 θ - 1 = 8 3 ϕ
- Dπ 3 + 1 ϕ 3 ϕ - 1 = 8 3 θ
Correct answer
A. π 3 + 1 ϕ 2 3 ϕ - 1 = 8 3 θ
Step-by-step solution
P = P c π = a 27 b 2 π V = V c ϕ = 3 b ϕ T = T c θ = 8 a 27 R b θ Hence substituting in the van der Waal's equation P + a V 2 V - b = R T ⇒ π + 3 ϕ 2 3 ϕ - 1 = 8 θ This is reduced state equation π + 3 ϕ 2 3 ϕ 8 θ - 1 8 θ = 1 π 3 + 1 ϕ 2 3 ϕ - 1 = 8 3 θ