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#isothermal expansion

13 public questions tagged with this topic.

First Law of Thermodynamics Applications

0.2 moles of an ideal gas expand isothermally at 350 K from 4 L to 10 L. What is the heat absorbed? ( R = 8.3 J mol⁻¹ K⁻

**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. Isothermal: Δ U = 0 , Δ Q = Δ W = μ R T ln((V₂)/(V₁)) . μ = 0.2 , T = 350 , V₂ = 10 , V₁ = 4 . Δ Q = 0.2 × 8.3 × 350 × ln((10)/(4)) = 581 × 0.916 ≈ 532 J . Using first law ΔU

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

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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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First Law of Thermodynamics Applications

An ideal gas expands isothermally at 570 K from 12 L to 36 L with 0.4 moles . What is the work done by the gas? ( R = 8.

**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. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.4 , R = 8.3 , T = 570 , V₂ = 36 , V₁ = 12 . W = 0.4 × 8.3 × 570 × ln((36)/(12)) = 1892.4 × ln(3) . ln(3) ≈ 1.0986 , W

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

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Isobaric and Isothermal Processes Work Calculation

An ideal gas expands isothermally at 480 K from 6 L to 18 L with 0.6 moles . What is the work done by the gas? ( R = 8.3

**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. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.6 , R = 8.3 , T = 480 , V₂ = 18 , V₁ = 6 . W = 0.6 × 8.3 × 480 × ln((18)/(6)) = 2390.4 × ln(3) .

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

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