For an adiabatic process of an ideal gas, TV^(γ-1) = constant. If T₁ = 300 K and V₂ = 2V₁ for γ = 5/3, then T₂:
A150 K
B100 K
C225 K
D189 K
Answer & Solution
Correct answer: D. 189 K
T₁V₁^(γ-1) = T₂V₂^(γ-1). T₂ = T₁(V₁/V₂)^(γ-1) = 300 × (1/2)^(2/3) = 300 × 0.63 ≈ 189 K. Expansion cools the gas adiabatically.
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