#physics problems

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First Law of Thermodynamics and Enthalpy and Internal Energy

For an ideal gas ( γ = 1.67 ) compressed adiabatically from V₁ = 12 L to V₂ = 3 L, starting at T₁ = 300 K, calculate the

For an adiabatic ideal-gas process, T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × 4⁰.⁶⁷ ≈ 300 × 2.5315 = 759.45 K .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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First Law of Thermodynamics and Enthalpy and Internal Energy

Calculate the work done when 1 mol of an ideal gas ( γ = 1.67 ) is compressed adiabatically from 20 L to 4 L at 300 K. (

For adiabatic compression, T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (20/4)⁰.⁶⁷ ≈ 300 × 2.63 = 789.0 K . Then, Cv = R/(γ-1) = 8.314 / 0.67 ≈ 12.41 J/mol·K . Work done: w = nCvΔ T = 1 × 12.41 × (789.0 - 300) ≈ 6068.3 J .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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#physics problems

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For an ideal gas ( γ = 1.67 ) compressed adiabatically from V₁ = 12 L to V₂ = 3 L, starting at T₁ = 300 K, calculate the

Calculate the work done when 1 mol of an ideal gas ( γ = 1.67 ) is compressed adiabatically from 20 L to 4 L at 300 K. (

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