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#gas pressure

14 public questions tagged with this topic.

Superposition and Interference of Waves

What happens to the speed of a longitudinal wave in a gas if the pressure is increased while temperature remains constan

**Superposition principle** states resultant displacement equals algebraic sum of individual waves, y = y₁ + y₂. For coherent waves with phase difference φ, resultant amplitude A = √(a₁² + a₂² + 2a₁a₂ cosφ), equal amplitudes give A = 2a cos(φ/2), constructive when φ = 2nπ, destructive when φ = (2n+1)π. For an ideal gas, v = √((gamma P/rho)) , and P/rho = RT/M (constant at constant temperature). Thus, speed depends only on temperature and gamma , not pressure alone, so it remains unchanged. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Remain

Ref: NCERT > Physics Book > Waves > Superposition and Interference of Waves

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Work Heat Distinction and Energy Transfer Modes

A gas at 10 atm and 80^circ C in a 8 L container is heated isochorically to 140^circ C . What is the final pressure?

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 10 atm , T₁ = 80 + 273 = 353 K , T₂ = 140 + 273 = 413 K . (10)/(353) = (P₂)/(413) ⇒ P₂ = (10 × 413)/(353) ≈ 11.7 atm . Using first law ΔU = Q - W,

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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Cyclic Processes and Reversibility Concepts

A gas at 8 atm and 60^circ C in a 7 L container is heated isochorically to 120^circ C . What is the final pressure?

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 8 atm , T₁ = 60 + 273 = 333 K , T₂ = 120 + 273 = 393 K . (8)/(333) = (P₂)/(393) ⇒ P₂ = (8 × 393)/(333) ≈ 9.44 atm . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic

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

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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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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 5 atm in a 6 L container is heated from 20°C to 60°C at constant volume. What is the final pressure?

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Isochoric: (P₁)/(T₁) = (P₂)/(T₂) . P₁ = 5 atm , T₁ = 20 + 273 = 293 K , T₂ = 60 + 273 = 333 K . (5)/(293) = (P₂)/(333) ⇒ P₂ = (5 × 333)/(293) ≈ 5.68 atm . 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 =

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Molecular Mass Density and Ideal Gas Equation

The mean free path of a gas molecule is 3.0 × 10⁻⁶ m at 0.25 atm. What will it be at 1 atm if temperature remains consta

**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. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.25 to 1), n increases 4 times, l reduces to (1)/(4).New l = 3.0 × 10⁻⁶/4 = 7.5 × 10⁻⁷ m. Substituting values gives 7.5 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ

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

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Mean Free Path and Molecular Diameter

The mean free path of a gas molecule is 2.0 × 10⁻⁶ m at 0.1 atm. What will it be at 0.4 atm if temperature remains const

**Kinetic theory mean free path** λ = 1/(√2 π d² n) quantifies collision frequency. With n =1.0×10²⁵ m⁻³, λ=9×10⁻⁷ m, d² =1/(1.414×10²⁵×3.14×9×10⁻⁷)=2.5×10⁻²⁰ m², d≈1.58×10⁻¹⁰ m, typical molecular size ~10⁻¹⁰ m, consistent with gas kinetic theory. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.1 to 0.4), n increases 4 times, l reduces to (1)/(4).New l = 2.0 × 10⁻⁶/4 = 5.0 × 10⁻⁷ m. Substituting values gives 5.0 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Mean Free Path and Molecular Diameter

The mean free path of a gas molecule is 8 × 10⁻⁷ m at 1 atm. What will it be at 0.25 atm if temperature remains constant

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. l ∝ (1)/(n), n ∝ P. If P reduces to (1)/(4), n reduces to (1)/(4), l increases 4 times.New l = 8 × 10⁻⁷ × 4 = 3.2 × 10⁻⁶ m. Substituting values gives 3.2 × 10⁻⁶ m, which matches expected kinetic theory result, confirming mean

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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