#work done

Third Law of Thermodynamics and Applications

Calculate the work done when 2 mol of an ideal gas expands isothermally and reversibly from 8 L to 24 L at 400 K. (R = 8

w = -nRT ln(V₂/V₁) = -2 × 8.314 × 400 × ln(24/8) = -6651.2 × ln(3) = -6651.2 × 1.099 = -7309.7 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

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Thermodynamic Terms - System Surroundings and Types of Systems

A system at constant pressure releases 800 J of heat and contracts by 0.3 L against 2 atm. What is Δ U ? (1 atm·L = 101.

Work done: w = -PΔ V = -2 × (-0.3) × 101.325 = 60.795 J . Given q = -800 J , Δ U = q + w = -800 + 60.795 = -739.21 ≈ -739.2 J .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Thermodynamic Terms - System Surroundings and Types of Systems

A system releases 200 J of heat and has 150 J of work done on it. What is Δ U?

Δ U = q + w. Here, q = -200 J (heat released), w = +150 J (work done on system), so Δ U = -200 + 150 = -50 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Thermodynamic Terms - System Surroundings and Types of Systems

Calculate the work done when 2 mol of an ideal gas expands adiabatically and reversibly from 10 L to 20 L at 600 K (γ =

T₂ = 600 × (10/20)⁰.⁴ = 600 × 0.7579 = 454.74 K, Cv = R/(γ-1) = 8.314 / 0.4 = 20.785 J/mol·K, w = nCvΔ T = 2 × 20.785 × (454.74 - 600) = -6042 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Thermodynamic Terms - System Surroundings and Types of Systems

The work done in the reversible adiabatic expansion of 1 mol of an ideal gas from 10 L to 20 L is ( γ = 1.4 ) at 300 K.

For adiabatic reversible expansion, T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (10/20)⁰.⁴ ≈ 300 × 0.7579 = 227.4 K . Then, Cv = R/(γ-1) = 8.314 / 0.4 = 20.785 J/mol·K . Work done: w = nCvΔ T = 1 × 20.785 × (227.4 - 300) ≈ -1509.2 J .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Thermodynamic Terms - System Surroundings and Types of Systems

What is the work done when 1 mol of an ideal gas expands isothermally and reversibly from 1 atm to 0.5 atm at 300 K? (R

w = -nRT ln(P₁/P₂) = -1 × 8.314 × 300 × ln(1/0.5) = -2494.2 × 0.693 = -1728 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Thermodynamic Terms - System Surroundings and Types of Systems

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Heat Capacity and Calorimetry and Measurement of Enthalpy

For an ideal gas ( γ = 1.4 ) expanding adiabatically from 5 L to 10 L at 500 K, calculate the work done if Cv = 20.785 J

For adiabatic expansion, T₂ = 500 × (5/10)⁰.⁴ ≈ 500 × 0.7579 = 378.9 K . Then, Δ T = 378.9 - 500 = -121.1 K . Work done: w = nCvΔ T = 1 × 20.785 × (-121.1) ≈ -2516 J .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy

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Heat Capacity and Calorimetry and Measurement of Enthalpy

For a reversible isothermal expansion of an ideal gas, the work done is given by:

For reversible isothermal expansion, w = -nRT ln(V₂/V₁), where V₂ > V₁.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Heat Capacity and Calorimetry and Measurement of Enthalpy

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Gibbs Energy and Gibbs Energy Change and Equilibrium

Calculate the work done when 1 mol of an ideal gas expands reversibly and isothermally from 10 atm to 2 atm at 300 K. (

For reversible isothermal expansion, w = -nRT ln(P₁/P₂) = -1 × 8.314 × 300 × ln(10/2) = -2494.2 × 1.6094 ≈ -4013 J .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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First Law of Thermodynamics and Enthalpy and Internal Energy

A system undergoes a process where 500 J of heat is released, and 300 J of work is done by the system. What is the chang

Δ U = q + w. Here, q = -500 J (heat released), w = -300 J (work done by system), so Δ U = -500 - 300 = -800 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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First Law of Thermodynamics and Enthalpy and Internal Energy

For an ideal gas compressed reversibly and isothermally from 15 L to 3 L at 350 K, calculate the work done. ( R = 8.314

For reversible isothermal compression, w = -nRT ln(V₂/V₁) = -1 × 8.314 × 350 × ln(3/15) = -2909.9 × (-1.6094) ≈ 4685 J .

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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First Law of Thermodynamics and Enthalpy and Internal Energy

A system absorbs 600 J of heat at constant volume, and its internal energy increases by 600 J. What is the work done?

At constant volume, w = -PΔ V = 0. First Law: Δ U = q + w = 600 + 0 = 600 J, so w = 0 J.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: First Law of Thermodynamics and Enthalpy and Internal Energy

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