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#gas work

3 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 1.1 moles of an ideal gas expand from 7 L to 14 L at 390 K . What is the work done by the gas? (

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

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Second Law Heat Engines and Kelvin-Planck

In an isobaric process, 0.9 moles of an ideal gas expand from 4 L to 10 L at 340 K . What is the work done by the gas? (

**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

A gas undergoes an isothermal expansion at 300 K from a volume of 2 L to 6 L . If the number of moles of the gas is 0.1

**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 an isothermal process, W = μ R T ln((V₂)/(V₁)) .Substitute: μ = 0.1 , R = 8.3 , T = 300 , V₂ = 6 , V₁ = 2 . W = 0.1 × 8.3 × 300 × ln((6)/(2)) = 249 × ln(3) . ln(3) ≈ 1.0986 , so W ≈ 249 × 1.0986

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

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