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
1 answer:
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,
<em>T₂ = 67.3 K</em>
(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,
<em>T₂ = 67.3 K</em>
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