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#isobaric

2 public questions tagged with this topic.

First Law of Thermodynamics Applications

In an isobaric process, 0.7 moles of gas expand from 320 K to 400 K . What is the heat supplied if C_p = 29.1 J mol⁻¹ K⁻

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Δ Q = μ C_p Δ T . μ = 0.7 , C_p = 29.1 , Δ T = 400 - 320 = 80 . Δ Q = 0.7 × 29.1 × 80 = 1632 J . Using first law ΔU = Q - W, W = ∫

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

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Isobaric and Isothermal Processes Work Calculation

In an isobaric process, 0.8 moles of an ideal gas expand from 6 L to 12 L at 350 K . What is the work done by the gas? (

**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. W = P Δ V , P V = μ R T . Δ V = 12 - 6 = 6 L .Initial P = (μ R T)/(V₁) , but directly: W = μ R T ((V₂ - V₁)/(V₁)) , adjust via P

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

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