Pressure and volume of an ideal gas undergoing a revisable adiabatic change of state are related as[1]:
P∙V^γ = C ( a constant )
When you substitute pressure using ideal gas law you can derive a relation for temperature and volume for such process:
P = n∙R∙T/V
=>
(n∙R∙T/V)∙V^γ = C
<=>
T∙V^(γ - 1) = C/(n∙R∙) (also a constant)
So volume and temperature in initial (1) and final (2) state are related as:
T₁∙V₁^(γ - 1) = T₂∙V₂^(γ - 1)
<=>
T₁^(1/(γ - 1))∙V₁ = T₂^(1/(^(γ - 1))∙V₂
So final volume is given by:
V₂ = V₁ ∙ (T₁/T₂∙)^(1/(γ - 1))
The temperatures are given. The initial volume can be found from ideal gas law:
V₁ = n∙R∙T₁∙/P₁
= 20mol ∙ 8.314 J∙mol⁻¹∙K⁻¹ ∙ 450K / 400×10³ Pa
= 0.187 m³
= 187 L
Hence,
V₂ = 187L ∙ (450K / 320∙)^(1/(1.67 - 1)) = 311 L ≈ 310 L
Hope this helps :)