An ideal gas expands isothermally at 380 K from 5 L to 15 L with 0.25 moles . What is the work done by the gas? ( R = 8.
**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.25 , R = 8.3 , T = 380 , V₂ = 15 , V₁ = 5 . W = 0.25 × 8.3 × 380 × ln((15)/(5)) = 788.5 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 788.5 × 1.0986 ≈ 866 J
Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications