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#monatomic gas

15 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A monatomic gas undergoes an adiabatic expansion from 700 K to 350 K with 2 moles . What is the work done? ( R = 8.3 J m

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

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

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Heat Transfer Work Distinction and Internal Energy Change

A monatomic gas undergoes an adiabatic expansion from 720 K to 360 K with 1.5 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1.5 , R = 8.3 , T₁ = 720 , T₂ = 360 , γ = 1.67 . W = (1.5 × 8.3 × (720 - 360))/(1.67 - 1) = (12.45

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 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

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Cyclic Processes and Reversibility Concepts

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

**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 monatomic gas: C_v = (3)/(2) R . C_v = (3)/(2) × 8.3 = 12.45 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 12.45 J mol⁻¹ K⁻¹, consistent with thermodynamic laws and energy conservation.

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

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Isochoric Processes and Pressure-Temperature

A monatomic gas undergoes an adiabatic expansion from 820 K to 410 K with 0.9 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.9 , R = 8.3 , T₁ = 820 , T₂ = 410 , γ = 1.67 . W = (0.9 × 8.3 × (820 - 410))/(1.67 - 1) = (7.47 × 410)/(0.67) ≈ 4570.15 J ≈ 4570 J . Using first law ΔU = Q - W, W = ∫ P

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

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Specific Heat Capacity and Latent Heat

What is the molar specific heat capacity at constant pressure for a monatomic gas if C_v = 12.45 J mol⁻¹ K⁻¹ and R = 8.3

**Specific heat capacity** c = Q/(m ΔT) (J/kg·K), molar C = Q/(n ΔT), heat required to raise temperature, Q = m c ΔT, for water c=4186 J/kg·K, latent heat L = Q/m for phase change at constant temperature, fusion L_f and vaporization L_v, Q = m L, e.g., ice melting L_f=3.34×10⁵ J/kg, water vaporization 2.26×10⁶ J/kg. C_p - C_v = R . C_p = C_v + R = 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

Ref: NCERT > Physics Book > Thermodynamics > Specific Heat Capacity and Latent Heat

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the molar specific heat capacity at constant volume for a monatomic ideal gas? (Take R = 8.3 J mol⁻¹ K⁻¹ )

**Quasi-static process** infinitely slow, system always near equilibrium, reversible, can be represented as continuous path on P-V diagram, non-quasi-static rapid process non-equilibrium, work W = ∫ P_ext dV, for quasi-static P_ext = P_system, work = ∫ P dV, zeroth law ensures temperature defined throughout quasi-static. For a monatomic gas, C_v = (3)/(2) R . C_v = (3)/(2) × 8.3 = 12.45 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 12.45 J

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What is the molar specific heat capacity at constant pressure for a monatomic gas if C_v = 12.45 J mol⁻¹ K⁻¹ and R = 8.3

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. C_p - C_v = R . C_p = C_v + R = 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

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

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Isobaric and Isothermal Processes Work Calculation

A monatomic gas undergoes an adiabatic expansion from 740 K to 370 K with 0.7 moles . What is the work done? ( R = 8.3 J

**Isobaric process** constant pressure, work W = P ΔV = P(V₂ - V₁) = n R ΔT, for expansion ΔV positive W positive, for compression negative, heat Q = n C_p ΔT, ΔU = n C_v ΔT. Isothermal process constant temperature ΔU=0, work W = n R T ln(V₂/V₁) = n R T ln(P₁/P₂), Q = W, heat absorbed equals work done. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.7 , R = 8.3 , T₁ = 740 , T₂ = 370 , γ = 1.67 . W = (0.7 × 8.3 × (740 - 370))/(1.67 - 1) = (5.81

Ref: NCERT > Physics Book > Thermodynamics > Isobaric and Isothermal Processes Work Calculation

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Molecular Mass Density and Ideal Gas Equation

What is the total internal energy of 2 moles of a monatomic gas at 400 K? (R = 8.31 J mol⁻¹ K⁻¹)

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. For monatomic gas, U = (3)/(2) μ R T.U = (3)/(2) × 2 × 8.31 × 400 = 9972 J ≈ 9.97 kJ . Substituting values gives 9.97 kJ, 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 > Molecular Mass Density and Ideal Gas Equation

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RMS Speed and Temperature Dependence

What is the total internal energy of 1.5 moles of a monatomic gas at 250 K? (R = 8.31 J mol⁻¹ K⁻¹)

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. Monatomic gas: U = (3)/(2) μ R T.U = (3)/(2) × 1.5 × 8.31 × 250 = 4674.375 J ≈ 4.67 kJ. Substituting values gives 4.67 kJ, 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 > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

A monatomic gas has a molar specific heat at constant pressure of 20.8 J mol⁻¹ K⁻¹. What is the value of C_v?

**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. For an ideal gas: C_p - C_v = R, where R = 8.31 J mol⁻¹ K⁻¹.C_v = C_p - R = 20.8 - 8.31 = 12.49 J mol⁻¹ K⁻¹ ≈ 12.5 J mol⁻¹ K⁻¹ . Substituting values gives 12.5 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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

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