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#diatomic gas

15 public questions tagged with this topic.

Heat Transfer Work Distinction and Internal Energy Change

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

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. 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

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

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Heat Transfer Work Distinction and Internal Energy Change

A diatomic gas undergoes an adiabatic expansion from 760 K to 380 K with 0.8 moles . What is the work done? ( R = 8.3 J

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.8 , R = 8.3 , T₁ = 760 , T₂ = 380 , γ = 1.4 . W = (0.8 × 8.3 × (760 - 380))/(1.4 - 1) = (6.64 × 380)/(0.4) = 6312 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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

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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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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

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Zeroth Law Thermal Equilibrium and Quasi-static

A diatomic gas undergoes an adiabatic expansion from 860 K to 430 K with 0.5 moles . What is the work done? ( R = 8.3 J

**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. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 0.5 , R = 8.3 , T₁ = 860 , T₂ = 430 , γ = 1.4 . W = (0.5 × 8.3 × (860 - 430))/(1.4 - 1) = (4.15 × 430)/(0.4) = 4467.5 J ≈ 4468 J . Using first law ΔU = Q - W,

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

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Second Law Heat Engines and Kelvin-Planck

A diatomic gas undergoes an adiabatic expansion from 780 K to 390 K with 1.3 moles . What is the work done? ( R = 8.3 J

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1.3 , R = 8.3 , T₁ = 780 , T₂ = 390 , γ = 1.4 . W = (1.3 × 8.3 × (780 - 390))/(1.4 - 1) = (10.79 × 390)/(0.4) = 10520.25 J ≈ 10520 J . Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

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First Law of Thermodynamics Applications

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

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. For diatomic gas: C_v = (5)/(2) R , C_p = C_v + R = (7)/(2) R . C_p = (7)/(2) × 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

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

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Collision Frequency and Mean Free Path Variation

What is the total internal energy of 0.5 moles of a diatomic gas at 200 K with no vibrational modes? (R = 8.31 J mol⁻¹ K

**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. Diatomic gas: 5 degrees of freedom (3 translational + 2 rotational).U = (5)/(2) μ R T = (5)/(2) × 0.5 × 8.31 × 200 = 2077.5 J ≈ 2.08 kJ. Substituting values gives 2.08 kJ, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²),

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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Molecular Mass Density and Ideal Gas Equation

What is the total internal energy of 0.7 moles of a diatomic gas at 200 K with no vibrational modes? (R = 8.31 J mol⁻¹ K

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. Diatomic gas: 5 degrees of freedom, U = (5)/(2) μ R T.U = (5)/(2) × 0.7 × 8.31 × 200 = 2908.5 J ≈ 2.91 kJ. Substituting values gives 2.91 kJ, 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 > Molecular Mass Density and Ideal Gas Equation

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Molecular Mass Density and Ideal Gas Equation

How much heat is required to raise the temperature of 0.25 moles of a diatomic gas by 20 K at constant volume, with no v

**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. Diatomic gas: 5 degrees of freedom, C_v = (5)/(2) R.Q = μ C_v Δ T = 0.25 × (5)/(2) × 8.31 × 20 = 207.75 J ≈ 207.8 J. Substituting values gives 207.8 J, 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 molar specific heat at constant volume (C_v) for a diatomic gas with no vibrational modes? (R = 8.31 J mol⁻¹

**RMS speed** v_rms = √(3 R T/M) = √(3 k_B T/m) where M molar mass (kg/mol), m molecular mass (kg), k_B=1.38×10⁻²/³ J/K, R=8.314 J/mol·K, T absolute temperature (K). Proportional to √T and 1/√M, lighter gases faster at same T, e.g., H₂ faster than O₂, temperature increase raises v_rms as √T. For a diatomic gas (rigid rotator), 5 degrees of freedom (3 translational + 2 rotational).C_v = (5)/(2) R = (5)/(2) × 8.31 = 20.775 J mol⁻¹ K⁻¹ ≈ 20.8 J mol⁻¹ K⁻¹ . Substituting values gives 20.8 J mol⁻¹ K⁻¹, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

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 vessel contains 1 mole of a diatomic gas at 300 K. What is its total internal energy if vibrational modes are not exci

**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 diatomic gas (rigid rotator): U = (5)/(2) μ R T.U = (5)/(2) × 1 × 8.31 × 300 = 6232.5 J ≈ 6.23 kJ . Substituting values gives 6.23 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 > Internal Energy of Ideal Gases

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