One mole of ar initially at 305 k undergoes an adiabatic expansion against a pressure pexternal = 0 from a volume of 8.5 l to a volume of 82.0 l. part a

An adiabatic process is when the system is insulated that no heat is released to the surroundings. For this type of process, we have a derived formula written below:

(T₂/T₁)^C = (V₁/V₂)
where C = Cv/nR

From the complete problem shown in the attached picture, Cv = (3/2)R. Thus,
C= (3/2)/1 mol = 3/2

(T₂/305 K)^(3/2) = (8.5 L/82 L)
Solving for T₂,
T₂ = 67.3 K