Skip to content
Login Register

#solid state

6 public questions tagged with this topic.

Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant volume for a solid predicted by the law of equipartition? ( R = 8.3

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. For a solid: C = 3R (from equipartition, 3 degrees of freedom). C = 3 × 8.3 = 24.9 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 24.9 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

View Details
Zeroth Law Thermal Equilibrium and Quasi-static

A solid has a molar specific heat capacity of 24.9 J mol⁻¹ K⁻¹ . Which element could it be?

**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. Tungsten has C = 24.9 J mol⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields Tungsten, consistent with thermodynamic laws and energy conservation.

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

View Details
Partial Pressures and Gas Mixtures

A solid has a molar specific heat capacity of 26.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. C = f × (R)/(2), 26.5 = f × (8.31)/(2).f = (26.5 × 2)/(8.31) ≈ 6.38 ≈ 6. Substituting values gives 6, 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

View Details
Collision Frequency and Mean Free Path Variation

A solid has a molar specific heat capacity of 24.4 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. C = f × (R)/(2), 24.4 = f × (8.31)/(2).f = (24.4 × 2)/(8.31) ≈ 5.87 ≈ 6. Substituting values gives 6, 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 > Collision Frequency and Mean Free Path Variation

View Details
Gas Laws and Volume-Temperature Relations

A solid has a molar specific heat capacity of 25.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Gas laws** Boyle, Charles, Gay-Lussac are special cases of ideal gas equation, for constant pressure volume-temperature relation V ∝ T, for constant temperature pressure-volume inverse, for constant volume pressure-temperature direct, enabling calculation of new volume from temperature ratio. C = f × (R)/(2), 25.5 = f × (8.31)/(2).f = (25.5 × 2)/(8.31) ≈ 6.14 ≈ 6. Substituting values gives 6, 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 > Gas Laws and Volume-Temperature Relations

View Details
Mean Free Path and Molecular Diameter

A solid has a molar specific heat capacity of 25.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**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. C = f × (R)/(2), 25.5 = f × (8.31)/(2).f = (25.5 × 2)/(8.31) ≈ 6.14 ≈ 6. Substituting values gives 6, 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

View Details