#constant temperature

Heat Transfer Work Distinction and Internal Energy Change

Which thermodynamic process involves a constant temperature?

**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. An isothermal process is characterized by constant temperature ( T = constant ). For an ideal gas, this implies P V = constant , with heat exchange balancing work done. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

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

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Molecular Mass Density and Ideal Gas Equation

A gas at 1.5 atm and 300 K has a volume of 18 litres. If the pressure decreases to 0.75 atm at constant temperature, wha

**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. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 1.5 atm, V₁ = 18 litres, P₂ = 0.75 atm.V₂ = (P₁ V₁)/(P₂) = (1.5 × 18)/(0.75) = 36 litres. Substituting values gives 36 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas

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

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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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Degrees of Freedom and Molar Specific Heat

A gas at 4 atm and 500 K has a volume of 20 litres. If the pressure increases to 8 atm at constant temperature, what is

**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 = 4 atm, V₁ = 20 litres, P₂ = 8 atm.V₂ = (P₁ V₁)/(P₂) = (4 × 20)/(8) = 10 litres. Substituting values gives 10 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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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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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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Mean Free Path and Molecular Diameter

The mean free path of a gas molecule is 4 × 10⁻⁷ m at 2 atm. What will it be at 4 atm if temperature remains constant?

**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 doubles, n doubles, l halves.New l = 4 × 10⁻⁷/2 = 2 × 10⁻⁷ m. Substituting values gives 2 × 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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Specific Heat and Calorimetry

Which gas law relates pressure and volume when temperature is held constant?

Boyle’s Law describes the inverse relationship between pressure and volume of an ideal gas at constant temperature (PV = constant). As per NCERT, applying relevant law/formula with correct units and sign convention leads to Boyle’s Law. 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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Thermodynamics

Which thermodynamic process occurs at constant temperature?

Isothermal is the scientifically accurate answer to this question. Within the study of Thermodynamics, this concept is well-established through extensive research and is documented in standard scientific literature. The specific properties, mechanisms, or characteristics of Isothermal directly address what is being asked. Among the other options, Isochoric, Adiabatic, and Isobaric do not correctly answer this question because they either refer to different concepts, describe properties of other molecules or processes, or represent common misconceptions about this topic.

Ref: Lehninger Principles of Biochemistry, Nelson & Cox, 8th Ed., Ch. 1

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