First Law of Thermodynamics and Enthalpy and Internal Energy

First Law of Thermodynamics and Enthalpy and Internal Energy

For an ideal gas ( γ = 1.4 ) compressed adiabatically from 20 L to 5 L at 400 K, calculate Δ U if Cv = 20.785 J/mol·K .

For adiabatic compression, T₂ = 400 × (20/5)⁰.⁴ ≈ 400 × 1.741 = 696.4 K . Then, Δ T = 696.4 - 400 = 296.4 K . Internal energy change: Δ U = nCvΔ T = 1 × 20.785 × 296.4 ≈ 6160 J .

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

View Details

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

A gas absorbs 1200 J of heat at constant pressure, and its volume increases by 0.5 L against 2 atm. What is Δ U ? (1 atm

Work done: w = -PΔ V = -2 × 0.5 × 101.325 = -101.325 J . Given q = 1200 J , Δ U = q + w = 1200 - 101.325 = 1098.68 ≈ 1098.7 J .

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

Which of the following is a state function?

A state function depends only on the initial and final states, not the path taken. Internal energy (U) is a state function, while work (w) and heat (q) are path-dependent.

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

Using Hess’s law, calculate Δ H for C(graphite) + 2H₂(g) → CH₄(g), given: C(graphite) + O₂(g) → CO₂(g), Δ H = -393.5 kJ/

Reverse the third equation and add the first two: Δ H = -393.5 + (-2 × 285.8) + 890.3 = -74.8 kJ/mol.

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

The enthalpy of combustion of benzene is -3267.6 kJ/mol. What is the heat released when 39 g of benzene is burnt? (Molar

Moles = 39 / 78 = 0.5 mol, heat released = 0.5 × 3267.6 = 1633.8 kJ.

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

View Details

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

Calculate Δ H for C₂H₂(g) + 2H₂(g) → C₂H₆(g) using: Δ Hf° (C₂H₂,g) = 226.7 kJ/mol, Δ Hf° (C₂H₆,g) = -84.7 kJ/mol.

Δ H = Δ Hf° (C₂H₆) - [Δ Hf° (C₂H₂) + 2 × Δ Hf° (H₂)] = -84.7 - [226.7 + 0] = -84.7 - 226.7 = -311.4 kJ/mol.

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

For a reaction at constant volume, the heat absorbed is equal to:

At constant volume, w = -pΔ V = 0, so from the first law, Δ U = qv, the heat absorbed at constant volume.

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

View Details

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

For C₂H₄(g) + H₂(g) → C₂H₆(g) , Δ H = -137.00 kJ/mol at 298 K. What is Δ U ? ( R = 8.314 J/mol·K )

For Δ ng = 1 - 2 = -1 , RT = 8.314 × 298 × 10⁻³ = 2.4776 kJ . Using Δ H = Δ U + Δ ng RT , Δ U = -137.00 - (-1 × 2.4776) = -137.00 + 2.4776 = -134.52 kJ/mol .

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

View Details

First Law of Thermodynamics and Enthalpy and Internal Energy

For a chemical reaction carried out at constant temperature (ideal-gas approximation) with Δ U = 250 J and no change in

For reactions at constant temperature (ideal gas), Δ H = Δ U + Δ ng RT . With Δ ng = 0 , Δ H = Δ U = 250 J .

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

View Details