#physics homework

Superposition Principle and Equilibrium of Charges

Two point charges \( 6 \times 10^{-7} \, \text{C} \) and \( 9 \times 10^{-7} \, \text{C} \) are 45 cm apart in air. What

**Independent action of charges** allows total force or field as vector sum. Geometry dictates distances to evaluation point, and resultant follows Σ k q_i/r_i², explaining zero field at symmetric centres for equal charges. Using Coulomb’s law: F = k (|q₁ q₂|/r²) . k = 9 × 10⁹ N·m²/C² , q₁ = 6 × 10⁻⁷ C , q₂ = 9 × 10⁻⁷ C , r = 0.45 m . |q₁ q₂| = 6 × 9 × 10⁻¹⁴ = 54 × 10⁻¹⁴ C² . r² = (0.45)² = 0.2025 m² . F = 9 × 10⁹ × (54 × 10⁻¹⁴/0.2025) = 9 × 10⁹ × 2.667 ×

Ref: NCERT > Physics Book > Electric Charges and Fields > Superposition Principle and Equilibrium of Charges

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First Law of Thermodynamics Applications

An ideal gas expands isothermally at 570 K from 12 L to 36 L with 0.4 moles . What is the work done by the gas? ( R = 8.

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.4 , R = 8.3 , T = 570 , V₂ = 36 , V₁ = 12 . W = 0.4 × 8.3 × 570 × ln((36)/(12)) = 1892.4 × ln(3) . ln(3) ≈ 1.0986 , W

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

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Internal Energy of Ideal Gases

How much heat is required to raise the temperature of 0.15 moles of a triatomic gas by 25 K at constant volume, with no

**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. Triatomic gas: 6 degrees of freedom, C_v = 3 R.Q = μ C_v Δ T = 0.15 × 3 × 8.31 × 25 = 93.4875 J ≈ 93.5 J. Substituting values gives 93.5 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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Internal Energy of Ideal Gases

A gas at 1.5 atm and 300 K has a volume of 12 litres. If the temperature increases to 450 K at constant pressure, what i

**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. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 12 litres, T₁ = 300 K, T₂ = 450 K.V₂ = V₁ × (T₂)/(T₁) = 12 × (450)/(300) = 18 litres. Substituting values gives 18 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 > Internal Energy of Ideal Gases

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