#gas expansion

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,

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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

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Isochoric Processes and Pressure-Temperature

A gas undergoes an adiabatic expansion from 25 L to 100 L , reducing its pressure from 16 atm to 1 atm . What is the val

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. For adiabatic: P₁ V₁^γ = P₂ V₂^γ . 16 × 25^γ = 1 × 100^γ . 16 = ((100)/(25))^γ ⇒ 16 = 4^γ . 4^γ = 2⁴ ⇒ 2²γ = 2⁴ ⇒ 2γ = 4 ⇒ γ = 2 . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P

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

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Isochoric Processes and Pressure-Temperature

A gas expands adiabatically from 10 atm and 5 L to 2 atm . What is the final volume? ( gamma = 1.33 )

**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. P₁ V₁^γ = P₂ V₂^γ . 10 × 5¹.33 = 2 × V₂¹.33 . V₂¹.33 = (10)/(2) × 5¹.33 = 5 × 5¹.33 . 5¹.33 ≈ 9.62 , V₂¹.33 = 5 × 9.62 ≈ 48.1 . V₂ = (48.1)¹/1.33 ≈ 14.5 L . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

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

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Specific Heat Capacity and Latent Heat

In an isobaric process, 0.6 moles of gas expand from 400 K to 480 K . What is the heat supplied if C_p = 25.0 J mol⁻¹ K⁻

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. Δ Q = μ C_p Δ T . μ = 0.6 , C_p = 25.0 , Δ T = 480 - 400 = 80 . Δ Q = 0.6 × 25.0 × 80 = 1200 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic

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Specific Heat Capacity and Latent Heat

In an isobaric process, 1.5 moles of an ideal gas expand from 5 L to 15 L at 300 K . What is the work done by the gas? (

**Latent heat** energy needed for phase change without temperature change, overcomes intermolecular forces, e.g., heating ice at 0°C to water at 0°C requires 334 kJ/kg, then heating water to 100°C requires c ΔT, then vaporization 2260 kJ/kg, illustrating two types of heat. W = P Δ V , P V = μ R T .Initial P = (μ R T)/(V₁) = (1.5 × 8.3 × 300)/(5) = 747 atm (unit adjustment needed).Correctly: W = μ R T ((V₂ - V₁)/(V₁)) , but simply W = P Δ V . Δ V = 15 - 5 = 10 L , adjust units: W = μ R

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

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Specific Heat Capacity and Latent Heat

A gas expands adiabatically from 5 L to 20 L , reducing its temperature from 500 K to 250 K . If 1 mole of gas is used a

**Heat capacity** at constant pressure C_p and volume C_v, C_p = C_v + R per mole, for solids Dulong-Petit C_v≈3R≈25 J/mol·K. Specific heat and latent heat govern temperature changes and phase transitions, Q = m c ΔT for heating, Q = m L for melting/boiling at constant T. For adiabatic process: W = (μ R (T₁ - T₂))/(γ - 1) . μ = 1 , R = 8.3 , T₁ = 500 , T₂ = 250 , γ = 1.5 . W = (1 × 8.3 × (500 - 250))/(1.5 - 1) = (8.3 × 250)/(0.5) = 4150 J . Using first law ΔU =

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

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Zeroth Law Thermal Equilibrium and Quasi-static

In an isobaric process, 2 moles of an ideal gas expand from 10 L to 20 L at 400 K . What is the work done by the gas? (T

**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. Work done: W = P Δ V = μ R T ((Δ V)/(V₁)) , but directly, W = μ R Δ T .Here, Δ V = 20 - 10 = 10 L , use W = P Δ V = μ R T (Δ V)/(V) , but since P V = μ R

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

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Zeroth Law Thermal Equilibrium and Quasi-static

What happens to the temperature of an ideal gas during an adiabatic expansion?

**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. In an adiabatic expansion ( Δ Q = 0 ), the gas does work on the surroundings ( W > 0 ), reducing its internal energy ( Δ U = -W ). For an ideal gas, U depends only on temperature, so temperature decreases. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

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

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Second Law Heat Engines and Kelvin-Planck

In an isobaric process, 1 mole of an ideal gas expands from 8 L to 16 L at 360 K . What is the work done by the gas? ( R

**Second law Kelvin-Planck statement** no process possible whose sole result is absorption of heat from reservoir and complete conversion to work, heat engine must have at least two reservoirs hot and cold, efficiency η = W/Q_h =1 - Q_c/Q_h

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

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First Law of Thermodynamics Applications

In an isobaric process, 0.7 moles of gas expand from 320 K to 400 K . What is the heat supplied if C_p = 29.1 J mol⁻¹ K⁻

**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. Δ Q = μ C_p Δ T . μ = 0.7 , C_p = 29.1 , Δ T = 400 - 320 = 80 . Δ Q = 0.7 × 29.1 × 80 = 1632 J . Using first law ΔU = Q - W, W = ∫

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

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Adiabatic Processes and Gamma Determination

In an isobaric process, 1.6 moles of an ideal gas expand from 6 L to 12 L at 400 K . What is the work done by the gas? (

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. W = P Δ V , P V = μ R T . Δ V = 12 - 6 = 6 L . P = (μ R T)/(V₁) = (1.6 × 8.3 × 400)/(6) = 885.33 atm (unit correction needed).Directly: W = μ R T ((V₂ - V₁)/(V₁)) , but W = P Δ V . W = 1.6

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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