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#final volume

8 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

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

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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 10 atm and 5 L to 2 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 . 10 × 5¹.33 = 2 × V₂¹.33 . V₂¹.33 = (10)/(2) × 5¹.33 = 5 × 5¹.33 . 5¹.33 ≈ 9.62 , V₂¹.33 = 5 × 9.62 ≈ 48.1 . V₂ = (48.1)¹/1.33 ≈ 14.5 L . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

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

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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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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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