#gamma calculation

Isochoric Processes and Pressure-Temperature

A gas undergoes an adiabatic expansion from 25 L to 100 L , reducing its pressure from 16 atm to 1 atm . What is the val

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 16 × 25^γ = 1 × 100^γ . 16 = ((100)/(25))^γ ⇒ 16 = 4^γ . 4^γ = 2⁴ ⇒ 2²γ = 2⁴ ⇒ 2γ = 4 ⇒ γ = 2 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Specific Heat Capacity and Latent Heat

A gas undergoes an adiabatic expansion from 10 L to 40 L , reducing its pressure from 8 atm to 1 atm . What is the value

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 8 × 10^γ = 1 × 40^γ . 8 = ((40)/(10))^γ ⇒ 8 = 4^γ . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Second Law Heat Engines and Kelvin-Planck

A gas is compressed adiabatically from 15 L to 5 L , increasing its pressure from 3 atm to 12 atm . What is gamma ?

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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Second Law Heat Engines and Kelvin-Planck

A gas undergoes an adiabatic expansion from 30 L to 90 L , reducing its pressure from 9 atm to 1 atm . What is the value

**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. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 9 × 30^γ = 1 × 90^γ . 9 = ((90)/(30))^γ ⇒ 9 = 3^γ . 3^γ = 3² ⇒ γ = 2 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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Molecular Mass Density and Ideal Gas Equation

A gas has a C_p of 29.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. C_v = C_p - R = 29.8 - 8.31 = 21.49 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (29.8)/(21.49) ≈ 1.39 ≈ 1.40. Substituting values gives 1.40, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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