For an adiabatic process (ΔQ=0) with an ideal gas: TVγ−1=constant
Given: Vf=8Vi and Tf=41Ti
Applying the adiabatic relation:
TiViγ−1=TfVfγ−1
TiViγ−1=4Ti(8Vi)γ−1
1=41⋅8γ−1
4=8γ−1
22=23(γ−1)
2=3(γ−1)
γ=35=1.67
This value corresponds to a monatomic gas. Among the options, only He is monatomic (γ=5/3 for monatomic ideal gases). O₂ and N₂ are diatomic (γ=7/5), CO₂ is polyatomic (γ≈1.30), and NH₃ is polyatomic.