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#pressure change

39 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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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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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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Isobaric and Isothermal Processes Work Calculation

A gas undergoes an adiabatic expansion from 22 L to 66 L , reducing its pressure from 12 atm to 2 atm . What is the valu

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 12 × 22^γ = 2 × 66^γ . 12 / 2 = ((66)/(22))^γ ⇒ 6 = 3^γ . 3^γ = 3¹.63 , γ ≈ 1.63 ≈ 1.67 (standard value from context). Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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RMS Speed and Temperature Dependence

A gas at 3 atm and 600 K has a volume of 15 litres. If the pressure decreases to 1.5 atm at constant temperature, what i

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 3 atm, V₁ = 15 litres, P₂ = 1.5 atm.V₂ = (P₁ V₁)/(P₂) = (3 × 15)/(1.5) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

The mean free path of a gas molecule is 1.0 × 10⁻⁶ m at 0.5 atm. What will it be at 2 atm if temperature remains constan

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.5 to 2), n increases 4 times, l reduces to (1)/(4).New l = 1.0 × 10⁻⁶/4 = 2.5 × 10⁻⁷ m. Substituting values gives 2.5 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

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Degrees of Freedom and Molar Specific Heat

A gas at 2 atm and 400 K has a volume of 6 litres. If the pressure decreases to 1 atm at constant temperature, what is t

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 2 atm, V₁ = 6 litres, P₂ = 1 atm.V₂ = (P₁ V₁)/(P₂) = (2 × 6)/(1) = 12 litres. Substituting values gives 12 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Gas Laws and Volume-Temperature Relations

A gas at 3 atm and 400 K has a volume of 15 litres. If the pressure drops to 1.5 atm at constant temperature, what is th

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 3 atm, V₁ = 15 litres, P₂ = 1.5 atm.V₂ = (P₁ V₁)/(P₂) = (3 × 15)/(1.5) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ =

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Fluid Pressure, Pascal's Law and Manometers

What happens to the boiling point of a liquid when external pressure decreases?

Decreasing external pressure reduces the energy needed for molecules to escape into the vapor phase, lowering the boiling point, as seen in high-altitude cooking. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It decreases. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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