Skip to content
Login Register

#molar heat capacity

15 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 1.2 moles of gas expand from 350 K to 420 K . What is the heat supplied if C_p = 25.5 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 = 1.2 , C_p = 25.5 , Δ T = 420 - 350 = 70 . Δ Q = 1.2 × 25.5 × 70 = 2142 J . Using first law ΔU = Q - W, W = ∫ P dV,

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

View Details
Cyclic Processes and Reversibility Concepts

What is the molar specific heat capacity at constant pressure for a monatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. For monatomic gas: C_v = (3)/(2) R , C_p = C_v + R . C_v = (3)/(2) × 8.3 = 12.45 . C_p = 12.45 + 8.3 = 20.75 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 20.75 J mol⁻¹ K⁻¹, consistent

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

View Details
Cyclic Processes and Reversibility Concepts

In an isobaric process, 0.9 moles of gas expand from 360 K to 450 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻

**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. Δ Q = μ C_p Δ T . μ = 0.9 , C_p = 25.5 , Δ T = 450 - 360 = 90 . Δ Q = 0.9 × 25.5 × 90 = 2065.5 J ≈ 2066 J . 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

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

View Details
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
Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant volume for a diatomic gas if R = 8.3 J mol⁻¹ K⁻¹ ?

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. For diatomic gas: C_v = (5)/(2) R . C_v = (5)/(2) × 8.3 = 20.75 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 20.75 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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
Partial Pressures and Gas Mixtures

A gas has a C_p of 33.24 J mol⁻¹ K⁻¹. What is its C_v? (R = 8.31 J mol⁻¹ K⁻¹)

**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_p - C_v = R, C_v = C_p - R.C_v = 33.24 - 8.31 = 24.93 J mol⁻¹ K⁻¹. Substituting values gives 24.93 J mol⁻¹ K⁻¹, 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
Partial Pressures and Gas Mixtures

How much heat is required to raise the temperature of 1 mole of argon gas by 20 K at constant volume? (R = 8.31 J mol⁻¹

**Dalton's law of partial pressures** total pressure P_total = Σ P_i, P_i = X_i P_total, X_i mole tion, each gas exerts pressure as if alone, ideal gas mixture P_i V = n_i R T, partial pressure proportional to mole tion, e.g., air 79% N₂ 21% O₂ P_N₂=0.79 atm P_O₂=0.21 atm at 1 atm total. For monatomic gas, C_v = (3)/(2) R.Heat Q = μ C_v Δ T = 1 × (3)/(2) × 8.31 × 20 = 249.3 J ≈ 249 J . Substituting values gives 249 J, 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 > Partial Pressures and Gas Mixtures

View Details
Collision Frequency and Mean Free Path Variation

A solid has a molar specific heat capacity of 24.9 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. C = f × (R)/(2), 24.9 = f × (8.31)/(2).f = (24.9 × 2)/(8.31) ≈ 5.99 ≈ 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
Mean Free Path and Molecular Diameter

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** 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. 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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

View Details