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