#volume expansion

Zeroth Law Thermal Equilibrium and Quasi-static

An ideal gas expands isothermally at 540 K from 10 L to 30 L with 0.3 moles . What is the work done by the gas? ( R = 8.

**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. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.3 , R = 8.3 , T = 540 , V₂ = 30 , V₁ = 10 . W = 0.3 × 8.3 × 540 × ln((30)/(10)) = 1344.6 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1344.6 × 1.0986 ≈ 1477 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

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

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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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Fluid Pressure, Pascal's Law and Manometers

Which of the following is true about the coefficient of volume expansion compared to the coefficient of linear expansion

The coefficient of volume expansion (αv) is three times the coefficient of linear expansion (αl), as αv = 3αl (Section 10.5). As per NCERT, applying relevant law/formula with correct units and sign convention leads to αv\=3αl. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

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

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Specific Heat and Calorimetry

A metal cube has a volume of 1000cm3 at 0∘C. What will be its volume at 100∘C if the coefficient of volume expansion is

Given: V0 = 1000cm3, ΔT = 100∘C, αv = 5.4×10−5K−1. ΔV = V0αvΔT = 1000×5.4×10−5×100 = 5.4cm3. New volume: V = V0+ΔV = 1000+5.4 = 1005.4cm3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1005.4 cm³. 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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