One mole of oxygen at 273 K and atmospheric pressure is adiabatically compressed to 5 atm. Calculate the final temperature. Also ?

Adiabatically compression of O₂:  P₁V₁¹·⁴ = P₂V₂¹·⁴ …… [1]
Combined gas law:  P₁V₁/T₁ = P₂V₂/T₂
Then, P₁¹·⁴V₁¹·⁴/T₁¹·⁴ = P₂¹·⁴V₂¹·⁴/T₂¹·⁴ …… [2]

[2]/[1]:  P₁⁰·⁴/T₁¹·⁴ = P₂⁰·⁴/T₂¹·⁴
Then, T₂ = ¹·⁴√[T₁¹·⁴ × (P₂/P₁)⁰·⁴]
Final temperature, T₂ = ¹·⁴√[273¹·⁴ × (5/1)⁰·⁴] K = 432 K

Work done one the gas = ΔPV = nRΔT = 1 × 8.31 × (432 - 273) W = 573 J

PV=nRT <<<memorize!!
n is moles, R is gas constant (use correct one!), T is temp in K (C+273.15)
Since you have a before and after with the same n &R simplifies to
PV/T initial  = PV/T final
Sub in your values and solve for T final