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#temperature increase

8 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 0.5 moles of gas expand from 300 K to 450 K . What is the heat supplied if C_p = 29.1 J mol⁻¹ K⁻

**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. Δ Q = μ C_p Δ T . μ = 0.5 , C_p = 29.1 , Δ T = 450 - 300 = 150 . Δ Q = 0.5 × 29.1 × 150 = 2182.5 J . Using first law ΔU = Q - W, W = ∫ P dV,

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

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First Law of Thermodynamics Applications

How much heat is required to raise the temperature of 0.45 kg of aluminium from 30^circ C to 60^circ C ? (Specific heat

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Δ Q = m s Δ T . m = 0.45 , s = 900 , Δ T = 60 - 30 = 30 . Δ Q = 0.45 × 900 × 30 = 12150 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

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Partial Pressures and Gas Mixtures

A gas at 2 atm and 300 K has a volume of 5 litres. If the temperature rises to 600 K at constant pressure, what is the n

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 5 litres, T₁ = 300 K, T₂ = 600 K.V₂ = V₁ × (T₂)/(T₁) = 5 × (600)/(300) = 10 litres. Substituting values gives 10.0 litres, 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 > Partial Pressures and Gas Mixtures

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Molecular Mass Density and Ideal Gas Equation

A gas at 1 atm and 273 K has a volume of 15 litres. If the temperature increases to 819 K at constant pressure, what is

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 15 litres, T₁ = 273 K, T₂ = 819 K.V₂ = V₁ × (T₂)/(T₁) = 15 × (819)/(273) = 45 litres. Substituting values gives 45 litres, 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 > Molecular Mass Density and Ideal Gas Equation

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Internal Energy of Ideal Gases

How much heat is required to raise the temperature of 0.5 moles of neon by 25 K at constant volume? (R = 8.31 J mol⁻¹ K⁻

**Equipartition theorem** energy ½ k_B T per degree of freedom per molecule, f degrees give U = f/2 k_B T per molecule, f/2 R T per mole, internal energy function of T only for ideal gas, no intermolecular potential. For monatomic gas, C_v = (3)/(2) R.Q = μ C_v Δ T = 0.5 × (3)/(2) × 8.31 × 25 = 155.8125 J ≈ 155.8 J. Substituting values gives 155.8 J, 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 > Internal Energy of Ideal Gases

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Degrees of Freedom and Molar Specific Heat

How much heat is required to raise the temperature of 0.1 moles of a monatomic gas by 30 K at constant volume? (R = 8.31

**Molar specific heat** from equipartition C_v = f/2 R, C_p = f/2 R + R, γ = C_p/C_v =1+2/f, for f=3 γ=1.67, f=5 γ=1.4, f=6 γ=1.33, explaining specific heat variation with molecular structure, degrees of freedom determine heat capacity. Monatomic gas: C_v = (3)/(2) R.Q = μ C_v Δ T = 0.1 × (3)/(2) × 8.31 × 30 = 37.395 J ≈ 37.4 J. Substituting values gives 37.4 J, 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 > Degrees of Freedom and Molar Specific Heat

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Mean Free Path and Molecular Diameter

A gas at 1 atm and 273 K has a volume of 11.2 litres. If the temperature increases to 546 K at constant pressure, what i

**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. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 11.2 litres, T₁ = 273 K, T₂ = 546 K.V₂ = V₁ × (T₂)/(T₁) = 11.2 × (546)/(273) = 22.4 litres. Substituting values gives 22.4 litres, 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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