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#adiabatic expansion

30 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A gas expands adiabatically, doing 360 J of work. What is the change in its internal energy?

**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

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

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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

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

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Work Heat Distinction and Energy Transfer Modes

A monatomic gas undergoes an adiabatic expansion from 700 K to 350 K with 2 moles . What is the work done? ( R = 8.3 J m

**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

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

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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

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

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Heat Transfer Work Distinction and Internal Energy Change

A monatomic gas undergoes an adiabatic expansion from 720 K to 360 K with 1.5 moles . What is the work done? ( R = 8.3 J

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1.5 , R = 8.3 , T₁ = 720 , T₂ = 360 , γ = 1.67 . W = (1.5 × 8.3 × (720 - 360))/(1.67 - 1) = (12.45

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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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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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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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 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 .

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

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Specific Heat Capacity and Latent Heat

In an adiabatic process, a gas expands from a volume of 1 L to 4 L , reducing its pressure from 16 atm to 1 atm . What i

**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. For an adiabatic process, P₁ V₁^γ = P₂ V₂^γ .Substitute: 16 × 1^γ = 1 × 4^γ . 16 = 4^γ .Taking log: log(16) = γ log(4) . log(16) = log(2⁴) = 4 log(2) , log(4) = log(2²) = 2 log(2) . 4 log(2) = γ × 2 log(2) ⇒ γ = (4)/(2) = 2 . Using first law ΔU = Q -

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

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Specific Heat Capacity and Latent Heat

A gas undergoes an adiabatic expansion, doing 450 J of work. What is the change in its internal energy?

**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. Adiabatic: Δ Q = 0 , Δ U = -Δ W . Work by gas: Δ W = 450 J . Δ U = -450 J (internal energy decreases). 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 = 1

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

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