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#isothermal process

15 public questions tagged with this topic.

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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Isochoric Processes and Pressure-Temperature

An ideal gas expands isothermally at 460 K from 3 L to 9 L with 0.5 moles . What is the work done by the gas? ( R = 8.3

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.5 , R = 8.3 , T = 460 , V₂ = 9 , V₁ = 3 . W = 0.5 × 8.3 × 460 × ln((9)/(3)) = 1909 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1909 × 1.0986 ≈ 2097 J . Using first law ΔU = Q - W,

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Isochoric Processes and Pressure-Temperature

0.15 moles of an ideal gas at 320 K expand isothermally from 3 L to 9 L. What is the work done by the gas? ( R = 8.3 J m

**Pressure-temperature relation** at constant volume Gay-Lussac law P ∝ T, for V constant, P₁/T₁ = P₂/T₂, if T doubles from 300 K to 600 K P doubles, e.g., P₁=1 atm at 300 K P₂=2 atm at 600 K, no work done, ΔU = n C_v ΔT = Q. Isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.15 , T = 320 , V₂ = 9 , V₁ = 3 . W = 0.15 × 8.3 × 320 × ln((9)/(3)) = 398.4 × 1.0986 ≈ 438 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W =

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

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Zeroth Law Thermal Equilibrium and Quasi-static

An ideal gas expands isothermally at 540 K from 10 L to 30 L with 0.3 moles . What is the work done by the gas? ( R = 8.

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.3 , R = 8.3 , T = 540 , V₂ = 30 , V₁ = 10 . W = 0.3 × 8.3 × 540 × ln((30)/(10)) = 1344.6 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1344.6 × 1.0986 ≈ 1477 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Zeroth Law Thermal Equilibrium and Quasi-static

A gas at 4 atm and 300 K in a 5 L container is compressed isothermally to 2 L. What is the work done on the gas? ( R = 8

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. Isothermal: W = μ R T ln((V₂)/(V₁)) . P₁ V₁ = μ R T ⇒ 4 × 5 = μ × 8.3 × 300 ⇒ μ = (20)/(2490) ≈ 0.008 mol . W = 0.008 × 8.3 × 300 × ln((2)/(5)) = 19.92 × (-0.916) ≈ -18.25 J (work by gas negative). Work

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

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Adiabatic Processes and Gamma Determination

Which of the following statements is correct about isothermal processes for an ideal gas?

**Gamma determination** γ = C_p/C_v, C_p - C_v = R, for monatomic f=3 C_v=3/2 R C_p=5/2 R γ=1.67, diatomic f=5 C_v=5/2 R C_p=7/2 R γ=1.4, adiabatic relation P V^γ = const allows γ determination from P-V measurements, slope of log P vs log V = -γ. In an isothermal process ( T = constant ), Δ U = 0 for an ideal gas, and P V = constant . Option C is correct. 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 = const and η

Ref: NCERT > Physics Book > Thermodynamics > Adiabatic Processes and Gamma Determination

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Isobaric and Isothermal Processes Work Calculation

A gas expands isothermally at 400 K absorbing 800 J of heat. What is the change in its internal energy?

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For an ideal gas in an isothermal process, Δ U = 0 (since U depends only on temperature). 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 = const and η = 1 - T_c/T_h, evaluation yields 0 J, consistent with thermodynamic laws and energy conservation.

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

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Isobaric and Isothermal Processes Work Calculation

An ideal gas expands isothermally at 420 K from 7 L to 21 L with 0.4 moles . What is the work done by the gas? ( R = 8.3

**Work calculation** isobaric W = P(V₂ - V₁), e.g., P=1 atm=1.013×10⁵ Pa ΔV=0.01 m³ W=1013 J. Isothermal W = n R T ln(V₂/V₁), for n=1 mol T=300 K V₂/V₁=2 W=1×8.314×300×ln2=1729 J, work done by gas during expansion, area under P-V curve. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.4 , R = 8.3 , T = 420 , V₂ = 21 , V₁ = 7 . W = 0.4 × 8.3 × 420 × ln((21)/(7)) = 1394.4 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1394.4 × 1.0986 ≈ 1532 J . Using first law ΔU = Q - W,

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

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Isobaric and Isothermal Processes Work Calculation

An ideal gas expands isothermally at 450 K from 4 L to 12 L with 0.3 moles . What is the work done by the gas? ( R = 8.3

**Isobaric and isothermal** are fundamental thermodynamic processes, isobaric P constant horizontal line on P-V diagram, isothermal hyperbolic P = n R T/V, work equals area under curve, isothermal work larger than adiabatic for same volume change because pressure higher. For isothermal: W = μ R T ln((V₂)/(V₁)) . μ = 0.3 , R = 8.3 , T = 450 , V₂ = 12 , V₁ = 4 . W = 0.3 × 8.3 × 450 × ln((12)/(4)) = 1120.5 × ln(3) . ln(3) ≈ 1.0986 , W ≈ 1120.5 × 1.0986 ≈ 1231 J . Using first law ΔU = Q - W, W =

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

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CHEMISTRY

What is the entropy change when a gas expands isothermally?

Isothermal expansion increases volume, increasing disorder, so Δ S > 0. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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