Skip to content
Login Register

#isobaric process

13 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

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
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

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

View Details
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

View Details
Specific Heat Capacity and Latent Heat

A gas at 1 atm and 300 K in a 10 L container is heated isobarically to 600 K . What is the work done by the gas? ( R = 8

**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. For isobaric: W = P Δ V .Initial: P V₁ = μ R T₁ ⇒ 1 × 10 = μ × 8.3 × 300 ⇒ μ = (10)/(2490) ≈ 0.004 moles .Final: V₂ = (μ R T₂)/(P) = (0.004 × 8.3 × 600)/(1) = 19.92 L . Δ V = 19.92 - 10 ≈ 9.92 L . W = P Δ V =

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

View Details
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

View Details
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

View Details
Adiabatic Processes and Gamma Determination

Why does an isobaric process involve both internal energy change and work?

**Adiabatic process** no heat exchange Q=0, first law ΔU = -W, for ideal gas P V^γ = constant, T V^{γ-1}= constant, P^{1-γ} T^{γ}= constant, γ = C_p/C_v = (f+2)/f, monatomic γ=5/3, diatomic γ=7/5. Work done W = (P₁V₁ - P₂V₂)/(γ-1), temperature changes due to work. In an isobaric process ( P = constant ), heat added ( Δ Q ) increases internal energy ( Δ U ) and does work ( W = P Δ V ) due to volume expansion, as per the First Law: Δ Q = Δ U + P Δ V . Using first law ΔU = Q - W, W

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

View Details
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

View Details
Isobaric and Isothermal Processes Work Calculation

In an isobaric process, 1.5 moles of an ideal gas expand from 9 L to 15 L at 380 K . What is the work done by the gas? (

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. W = P Δ V , P V = μ R T . Δ V = 15 - 9 = 6 L . P = (μ R T)/(V₁) = (1.5 × 8.3 × 380)/(9) = 526 atm (unit correction needed).Directly: W = μ R T ((V₂ - V₁)/(V₁)) , but W = P Δ V . W = 1.5 × 8.3 × 380 =

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

View Details