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#adiabatic compression

11 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

A gas is compressed adiabatically, doing 300 J of work on the system. What is the change in internal energy?

**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. For adiabatic ( Δ Q = 0 ), First Law: Δ U = -Δ W . Work on system: Δ W = -300 J (negative by convention). Δ U = -(-300) = 300 J . Using first law ΔU = Q - W,

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

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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

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 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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Specific Heat Capacity and Latent Heat

A gas is compressed adiabatically from 24 L to 6 L , increasing its pressure from 5 atm to 20 atm . What is gamma ?

**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. P₁ V₁^γ = P₂ V₂^γ . 5 × 24^γ = 20 × 6^γ . (24^γ)/(6^γ) = (20)/(5) ⇒ ((24)/(6))^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,

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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Adiabatic Processes and Gamma Determination

A gas undergoes an adiabatic compression from 16 L to 4 L , increasing its pressure from 1 atm to 8 atm . What is the va

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 1 × 16^γ = 8 × 4^γ . 16^γ = 8 × 4^γ . ((16)/(4))^γ = 8 ⇒ 4^γ = 8 . 4^γ = 2³ ⇒ 2²γ = 2³ ⇒ 2γ = 3 ⇒ γ = 1.5 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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Isobaric and Isothermal Processes Work Calculation

A gas is compressed adiabatically, increasing its pressure from 1 atm to 4 atm in a 10 L container. What is the final vo

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. P₁ V₁^γ = P₂ V₂^γ . 1 × 10¹.6 = 4 × V₂¹.6 . V₂¹.6 = 10¹.64 . V₂ = (10¹.64)¹/1.6 = 10 × 4⁻¹/1.6 . 4⁻⁰.625 ≈ 0.315 , V₂ ≈ 10 × 0.315 ≈ 3.15 L . 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 > Isobaric and Isothermal Processes Work Calculation

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First Law of Thermodynamics and Enthalpy and Internal Energy

For an ideal gas ( γ = 1.4 ) compressed adiabatically from 20 L to 5 L at 400 K, calculate Δ U if Cv = 20.785 J/mol·K .

For adiabatic compression, T₂ = 400 × (20/5)⁰.⁴ ≈ 400 × 1.741 = 696.4 K . Then, Δ T = 696.4 - 400 = 296.4 K . Internal energy change: Δ U = nCvΔ T = 1 × 20.785 × 296.4 ≈ 6160 J .

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

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