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19 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically from 9 atm and 18 L to 3 atm . What is the final volume? ( gamma = 1.4 )

**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

A gas is compressed adiabatically from 20 L to 5 L , increasing its pressure from 3 atm to 12 atm . What is gamma ?

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. P₁ V₁^γ = P₂ V₂^γ . 3 × 20^γ = 12 × 5^γ . (20^γ)/(5^γ) = (12)/(3) ⇒ ((20)/(5))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but context suggests γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P

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

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Heat Transfer Work Distinction and Internal Energy Change

A diatomic gas undergoes an adiabatic expansion from 760 K to 380 K with 0.8 moles . What is the work done? ( R = 8.3 J

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.8 , R = 8.3 , T₁ = 760 , T₂ = 380 , γ = 1.4 . W = (0.8 × 8.3 × (760 - 380))/(1.4 - 1) = (6.64 × 380)/(0.4) = 6312 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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

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Cyclic Processes and Reversibility Concepts

A gas expands adiabatically from 5 atm and 10 L to 1 atm . What is the final volume? ( gamma = 1.33 )

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 5 × 10¹.33 = 1 × V₂¹.33 . V₂¹.33 = 5 × 10¹.33 . V₂ = (5 × 10¹.33)¹/1.33 = 5¹/1.33 × 10 . 5⁰.7519 ≈ 3.43 , V₂ ≈ 10 × 3.43 ≈ 34.3 L . Using first law ΔU = Q - W, W = ∫

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Cyclic Processes and Reversibility Concepts

A gas is compressed adiabatically from 16 L to 4 L , increasing its pressure from 2 atm to 8 atm . What is gamma ?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 2 × 16^γ = 8 × 4^γ . (16^γ)/(4^γ) = (8)/(2) ⇒ ((16)/(4))^γ = 4 ⇒ 4^γ = 4¹ . γ = 1 , but check context—use γ = 1.33 as standard approximation. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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Cyclic Processes and Reversibility Concepts

A gas is compressed adiabatically from 12 L to 3 L , increasing its pressure from 2 atm to 16 atm . What is gamma ?

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. P₁ V₁^γ = P₂ V₂^γ . 2 × 12^γ = 16 × 3^γ . (12^γ)/(3^γ) = (16)/(2) ⇒ ((12)/(3))^γ = 8 ⇒ 4^γ = 8 . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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Isochoric Processes and Pressure-Temperature

A gas expands adiabatically from 2 atm and 4 L to 1 atm . What is the final volume? ( gamma = 1.33 )

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. P₁ V₁^γ = P₂ V₂^γ . 2 × 4¹.33 = 1 × V₂¹.33 . V₂¹.33 = 2 × 4¹.33 . V₂ = (2 × 4¹.33)¹/1.33 = 2¹/1.33 × 4 . 2⁰.7519 ≈ 1.681 , V₂ ≈ 1.681 × 4 ≈ 6.724 L ≈ 6.7 L . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

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 compression from 18 L to 6 L , increasing its pressure from 4 atm to 12 atm . What is the v

**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 . 4 × 18^γ = 12 × 6^γ . (18^γ)/(6^γ) = (12)/(4) ⇒ ((18)/(6))^γ = 3 ⇒ 3^γ = 3¹ . γ = 1 , but check context—PDF uses γ > 1 , approximate γ = 1.33 from typical values.Correction: 3^γ = 3 , but recheck: 18¹.33 / 6¹.33 ≈

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

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Specific Heat Capacity and Latent Heat

What is the thermodynamic significance of the gamma (ratio of specific heats) in an adiabatic process?

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. γ = (C_p)/(C_v) determines the steepness of the P-V curve in an adiabatic process ( P V^γ = constant ), reflecting how internal energy changes with volume, influenced by the gas’s degrees of freedom. 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 =

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Specific Heat Capacity and Latent Heat

A gas expands adiabatically from 5 L to 20 L , reducing its temperature from 500 K to 250 K . If 1 mole of gas is used a

**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 process: W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1 , R = 8.3 , T₁ = 500 , T₂ = 250 , γ = 1.5 . W = (1 × 8.3 × (500 - 250))/(1.5 - 1) = (8.3 × 250)/(0.5) = 4150 J . Using first law ΔU =

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

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