#heat engine

Cyclic Processes and Reversibility Concepts

Which of the following statements is correct regarding the Kelvin-Planck statement?

**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. The Kelvin-Planck statement asserts that no engine can absorb heat from a single reservoir and convert it entirely to work, requiring heat rejection to a colder reservoir. Option D is correct. 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

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

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Adiabatic Processes and Gamma Determination

Why does the Second Law impose a limit on the efficiency of a heat engine?

**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. The Second Law (e.g., Kelvin-Planck) requires some heat to be rejected to a cold reservoir, preventing complete conversion of heat to work. This inherent loss sets a maximum efficiency below 100%, dependent on temperature difference. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁),

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

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Isobaric and Isothermal Processes Work Calculation

Why can’t a heat engine operate with a single reservoir according to the Second Law?

**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. The Kelvin-Planck statement of the Second Law prohibits a heat engine from converting all heat from a single reservoir into work without rejecting some to a colder reservoir, as this would violate the natural tendency toward equilibrium. 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 > Isobaric and Isothermal Processes Work Calculation

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Isobaric and Isothermal Processes Work Calculation

According to the Kelvin-Planck statement, what is impossible for a heat engine?

**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. The Kelvin-Planck statement of the Second Law states that no process can absorb heat from a single reservoir and convert it entirely into work without rejecting some heat to a colder reservoir. This limits efficiency to less than 100%. Using first law ΔU

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

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