#thermodynamics problem

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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Heat Transfer Work Distinction and Internal Energy Change

A gas undergoes an isothermal compression from 8 L to 2 L at 350 K with 0.2 moles . What is the heat released? ( R = 8.3

**Heat transfer** occurs via conduction, convection, radiation, work via volume change W=∫ P dV, electrical work, etc., first law distinguishes, internal energy includes kinetic and potential of molecules, for ideal gas only kinetic, U = f/2 n R T. For isothermal: W = μ R T ln((V₂)/(V₁)) , Δ U = 0 , Q = W . W = 0.2 × 8.3 × 350 × ln((2)/(8)) = 581 × ln(0.25) . ln(0.25) = -ln(4) ≈ -1.386 . W = 581 × (-1.386) ≈ -805 J . Q = -805 J (negative implies heat released). Using first law ΔU = Q - W, W = ∫

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

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Cyclic Processes and Reversibility Concepts

In an isobaric process, 0.9 moles of gas expand from 360 K to 450 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻

**Reversibility** reversible process can be reversed by infinitesimal change, no entropy production, quasi-static without friction, e.g., Carnot cycle reversible, irreversible processes involve friction, free expansion, heat transfer across finite temperature difference, entropy increases, most real processes irreversible. Δ Q = μ C_p Δ T . μ = 0.9 , C_p = 25.5 , Δ T = 450 - 360 = 90 . Δ Q = 0.9 × 25.5 × 90 = 2065.5 J ≈ 2066 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ

Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts

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First Law of Thermodynamics Applications

A system absorbs 450 J of heat and performs 150 J of work. What is the change in internal energy?

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. First Law: Δ Q = Δ U + Δ W . Δ Q = 450 , Δ W = 150 . 450 = Δ U + 150 ⇒ Δ U = 450 - 150 = 300 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

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

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First Law of Thermodynamics Applications

What is the change in internal energy for 0.5 moles of an ideal gas heated from 260 K to 310 K at constant volume? ( C_v

**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. Δ U = μ C_v Δ T . μ = 0.5 , C_v = 20.8 , Δ T = 310 - 260 = 50 . Δ U = 0.5 × 20.8 × 50 = 520 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV,

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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Specific Heat and Calorimetry

A 0.05kg gold block at 1200∘C is placed in 0.5kg water at 20∘C. What is the final temperature? (Specific heat of gold =

0.05×134×(1200−T) = 0.5×4186×(T−20). 8040−6.7T = 2093T−41860. 8040+41860 = 2093T+6.7T. 49900 = 2099.7T⇒T≈23.77∘C≈23.8∘C. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 23.8°C. This satisfies dimensional consistency and physical conditions given, so option D is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Specific Heat and Calorimetry

How much heat is required to convert 0.3kg of ice at −18∘C to water at 40∘C in a 0.1kg copper calorimeter initially at 2

Q1 = 0.3×2100×18 = 11340J (ice to 0°C). Q2 = 0.3×3.35×105 = 100500J (melting). Heat gained by water and calorimeter: (0.3×4186+0.1×386)×(40−0) = (1255.8+38.6)×40 = 1294.4×40 = 51776J. Heat lost by calorimeter: 0.1×386×(25−0) = 965J (assume it cools to 0°C first). Total heat supplied: Q = 11340+100500+51776−965 = 162651J = 162.65kJ.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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