For an adiabatic process of an ideal gas, TV^(γ-1) = constant. If T₁ =…

Question

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:

Accepted 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.

Answer

150 K

Answer

100 K

Answer

225 K

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: