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13 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

Which of the following statements is correct about C_p and C_v for an ideal gas?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For an ideal gas, C_p > C_v because at constant pressure, heat supplies both internal energy increase and work ( C_p = C_v + R ), while at constant volume, heat only increases internal energy. Option A is correct. Using first law ΔU

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

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Cyclic Processes and Reversibility Concepts

What is the change in internal energy for 0.5 moles of an ideal gas heated from 250 K to 300 K at constant volume? ( C_v

**Cyclic process** system returns to initial state, ΔU=0 over cycle, net work W_net = area enclosed in P-V diagram, Q_net = W_net from first law ΔU= Q - W =0 => Q_net = W_net, clockwise cycle work done by system positive, counterclockwise work done on system negative, efficiency η = W_net/Q_in. Δ U = μ C_v Δ T . μ = 0.5 , C_v = 20.8 , Δ T = 300 - 250 = 50 . Δ U = 0.5 × 20.8 × 50 = 520 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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

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Isochoric Processes and Pressure-Temperature

What is the molar specific heat capacity at constant pressure for a diatomic gas if C_v = 20.75 J mol⁻¹ K⁻¹ and R = 8.3

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. C_p - C_v = R . C_p = C_v + R = 20.75 + 8.3 = 29.05 J mol⁻¹ K⁻¹ ≈ 29.1 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

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

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First Law of Thermodynamics Applications

What is the change in internal energy for 0.4 moles of an ideal gas heated from 290 K to 330 K at constant volume? ( C_v

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Δ U = μ C_v Δ T . μ = 0.4 , C_v = 20.8 , Δ T = 330 - 290 = 40 . Δ U = 0.4 × 20.8 × 40 = 332.8 J ≈ 333 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

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

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Collision Frequency and Mean Free Path Variation

A gas has a C_v of 24.93 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**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_p = C_v + R = 24.93 + 8.31 = 33.24 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (33.24)/(24.93) ≈ 1.33. Substituting values gives 1.33, 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

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Degrees of Freedom and Molar Specific Heat

A gas has a C_v of 20.8 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**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. C_p = C_v + R = 20.8 + 8.31 = 29.11 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (29.11)/(20.8) ≈ 1.40. Substituting values gives 1.40, 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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Degrees of Freedom and Molar Specific Heat

What is the ratio of C_p to C_v for a gas with 3 translational and 2 rotational degrees of freedom?

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. Degrees of freedom = 3 + 2 = 5.C_v = (5)/(2) R, C_p = C_v + R = (5)/(2) R + R = (7)/(2) R.γ = (C_p)/(C_v) = (7)/(2) R(5)/(2) R = (7)/(5) = 1.4. Substituting values gives 1.4, 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 has a C_v of 12.7 J mol⁻¹ K⁻¹. What is the ratio of specific heats (gamma)? (R = 8.31 J mol⁻¹ K⁻¹)

**Molecular diameter from mean free path** uses λ =1/(√2 n π d²), solving d = √(1/(√2 n π λ)). At higher pressure n ∝ P, λ ∝1/P, so doubling P halves λ, illustrating pressure dependence of collision distance. C_p = C_v + R = 12.7 + 8.31 = 21.01 J mol⁻¹ K⁻¹.γ = (C_p)/(C_v) = (21.01)/(12.7) ≈ 1.65. Substituting values gives 1.65, 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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