A monoatomic ideal gas initially has pressure P and volume V. It expands adiabatically until its volume becomes 27V. What is the change in internal energy of the gas?
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A monoatomic ideal gas initially has pressure P and volume V. It expands adiabatically until its volume becomes 27V. What is the change in internal energy of the gas?

G1 Physics · Accepted Answer

For a monoatomic ideal gas, γ = 5 ÷ 3. For an adiabatic process, P × (V raised to power γ) = constant. Therefore, P₁ × (V₁ raised to power γ) = P₂ × (V₂ raised to power γ). Given, P₁ = P, V₁ = V, V₂ = 27V. Hence, P₂ = P ÷ [27 raised to power (5 ÷ 3)]. Now, 27 raised to power (5 ÷ 3) = (cube root of 27)⁵ = 3⁵ = 243. Thus, P₂ = P ÷ 243. Therefore, P₂V₂ = (P ÷ 243)(27V) = PV ÷ 9. For an ideal gas, U = PV ÷ (γ-1). Hence, ΔU = [P₂V₂-P₁V₁] ÷ (γ-1). Substitute: ΔU = [PV ÷ 9-PV] ÷ [(5 ÷ 3)-1] = [-8PV ÷…