#gamma value

Work Heat Distinction and Energy Transfer Modes

A gas undergoes an adiabatic expansion from 28 L to 84 L , reducing its pressure from 15 atm to 3 atm . What is the valu

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

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Heat Transfer Work Distinction and Internal Energy Change

0.3 moles of an ideal gas at 400 K expand adiabatically from 6 atm to 2 atm. If gamma = 1.5 , what is the final temperat

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. Adiabatic: T₁ V₁^γ-1 = T₂ V₂^γ-1 , P V = μ R T ⇒ V₁ = (μ R T₁)/(P₁) = (0.3 × 8.3 × 400)/(6) = 166 L , V₂ = (0.3 × 8.3 × T₂)/(2) = 1.245 T₂ . 400 × 166⁰.5 = T₂ × (1.245 T₂)⁰.5 .

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

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Specific Heat Capacity and Latent Heat

A gas expands adiabatically from 8 atm and 16 L to 2 atm . What is the final volume? ( gamma = 1.5 )

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. P₁ V₁^γ = P₂ V₂^γ . 8 × 16¹.5 = 2 × V₂¹.5 . V₂¹.5 = (8)/(2) × 16¹.5 = 4 × 16¹.5 . 16¹.5 = 16 × 16⁰.5 = 64 , V₂¹.5 = 4 × 64 = 256 . V₂ = 256¹/1.5 = 256²/3 ≈ 40.3 L .

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Second Law Heat Engines and Kelvin-Planck

0.4 moles of an ideal gas at 340 K expand adiabatically from 7 atm to 1 atm. If gamma = 1.4 , what is the final temperat

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Adiabatic: T₁ V₁^γ-1 = T₂ V₂^γ-1 , V₁ = (μ R T₁)/(P₁) = (0.4 × 8.3 × 340)/(7) ≈ 161.37 L , V₂ = (0.4 × 8.3 × T₂)/(1) = 3.32 T₂ . 340 × 161.37⁰.4 = T₂ × (3.32 T₂)⁰.4 . Approximate: T₂ ≈ 245 K . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

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Isobaric and Isothermal Processes Work Calculation

A gas is compressed adiabatically from 8 L to 2 L . If the initial pressure is 1 atm and gamma = 1.4 , what is the final

**Isobaric process** constant pressure, work W = P ΔV = P(V₂ - V₁) = n R ΔT, for expansion ΔV positive W positive, for compression negative, heat Q = n C_p ΔT, ΔU = n C_v ΔT. Isothermal process constant temperature ΔU=0, work W = n R T ln(V₂/V₁) = n R T ln(P₁/P₂), Q = W, heat absorbed equals work done. P₁ V₁^γ = P₂ V₂^γ . P₁ = 1 atm , V₁ = 8 L , V₂ = 2 L , γ = 1.4 . 1 × 8¹.4 = P₂ × 2¹.4 . P₂ = 8¹.42¹.4 = ((8)/(2))¹.4 = 4¹.4 . 4¹.4 =

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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