This MCQ module is based on: First Law Internal Energy
TOPIC 20 OF 28
First Law Internal Energy
🎓 Class 11
Chemistry
CBSE
Theory
Ch 5 – Thermodynamics
⏱ ~14 min
🌐 Language: [gtranslate]
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First Law of Thermodynamics, Internal Energy and Enthalpy
5.6 The First Law of Thermodynamics
The First Law of Thermodynamics is the law of conservation of energy applied to thermodynamic processes:
First Law: Energy can neither be created nor destroyed — only transformed from one form to another. Mathematically:
\[\Delta U = q + w\]
where ΔU = change in internal energy, q = heat supplied to the system, and w = work done on the system.
Consequence: for an isolated system q = 0 and w = 0, so ΔU = 0 — the energy of an isolated system is constant.
5.7 Work in Expansion / Compression of Gases
Consider a gas confined in a cylinder by a piston of cross-sectional area A. If the gas pushes the piston outward by distance dx against external pressure P_ext, the work done by the gas is P_ext · A · dx = P_ext · dV. By IUPAC convention (work done on the system positive):
\[w = -P_\text{ext}\,\Delta V\]5.7.1 Free Expansion (against vacuum)
If P_ext = 0 (expansion into vacuum), w = 0. No work is done by or on the system.
5.7.2 Reversible Isothermal Expansion of an Ideal Gas
For a reversible expansion at constant T, P_ext = P_gas at every instant. Integrating:
\[w_\text{rev} = -\int_{V_1}^{V_2} P\,dV = -nRT\ln\frac{V_2}{V_1} = -2.303\,nRT\log\frac{V_2}{V_1}\]For an isothermal ideal gas, ΔU = 0 (U depends only on T). Therefore q = −w = +nRT ln(V₂/V₁).
5.7.3 Isobaric Process
If P_ext = P (constant), w = −P·ΔV.
5.7.4 Isochoric Process
ΔV = 0 → w = 0 → ΔU = q_v (the entire heat raises U).
5.8 Enthalpy (H) — A Useful State Function
Most laboratory chemistry happens in open vessels at constant atmospheric pressure, not constant volume. For such conditions, internal energy isn't directly the most useful quantity. We define a new state function called enthalpy:
Enthalpy (H):
\[H = U + PV\]
At constant pressure, the change in enthalpy equals the heat absorbed:
\[\Delta H = q_p\]
Derivation (constant P): ΔU = q_p + w = q_p − PΔV ⇒ q_p = ΔU + PΔV. But ΔH = Δ(U + PV) = ΔU + Δ(PV) = ΔU + PΔV (since P is constant). Therefore q_p = ΔH. ✓
5.8.1 Relationship between ΔH and ΔU
For reactions involving gases, the change in moles of gaseous species (Δn_g) matters:
\[\Delta H = \Delta U + \Delta n_g RT\]where Δn_g = (moles of gaseous products) − (moles of gaseous reactants), and R = 8.314 J K⁻¹ mol⁻¹.
| Reaction type | Δn_g | Relation |
|---|---|---|
| No gas, only solids/liquids | 0 | ΔH = ΔU |
| Same number of gas moles on both sides | 0 | ΔH = ΔU |
| More gas moles produced (e.g., CaCO₃ → CaO + CO₂) | +ve | ΔH > ΔU |
| Fewer gas moles produced (e.g., 2H₂ + O₂ → 2H₂O(l)) | −ve | ΔH < ΔU |
5.8.2 Exothermic vs Endothermic Reactions
An exothermic reaction releases heat: ΔH < 0. An endothermic reaction absorbs heat: ΔH > 0.
5.9 Extensive vs Intensive Properties
| Type | Depends on amount? | Examples |
|---|---|---|
| Extensive | Yes (proportional to size) | U, H, V, mass, n, heat capacity (C) |
| Intensive | No (independent of size) | P, T, density, molar volume, specific heat (c) |
🎯 Interactive: ΔH ↔ ΔU Converter
Convert between ΔU and ΔH using ΔH = ΔU + Δn_g RT. Adjust the temperature and Δn_g to see the conversion live.
ΔH = ΔU + Δn_g RT = −100.00 kJ
No change in moles of gas → ΔH = ΔU.
🧪 Activity 5.2 — Endothermic Bag Demonstration
Setup: Place 10 g of solid Ba(OH)₂·8H₂O and 5 g of solid NH₄SCN in a small plastic bag. Seal it and shake. Place it on a wet wooden block.
Predict: Will the bag warm up or cool down? Will the wooden block stick to the bag?
The reaction Ba(OH)₂·8H₂O + 2NH₄SCN → Ba(SCN)₂ + 2NH₃ + 10 H₂O is strongly endothermic (ΔH ≈ +60 kJ/mol).
The bag becomes ICY COLD (~−20 °C). Water on the wooden block freezes solid, gluing the block to the bag — you can lift the block by the bag!
Why? Heat needed for reaction is drawn from surroundings, lowering their temperature. This proves that endothermic reactions absorb heat from the environment.
Worked Example 5.3: Reversible Isothermal Expansion
2 moles of an ideal gas expand isothermally and reversibly at 300 K from 5 L to 50 L. Calculate q, w, ΔU and ΔH. (R = 8.314 J K⁻¹ mol⁻¹)
For an isothermal process on an ideal gas: ΔU = 0, ΔH = 0.
w_rev = −2.303 × nRT × log(V₂/V₁) = −2.303 × 2 × 8.314 × 300 × log(50/5) = −2.303 × 4988.4 × 1 = −11489 J ≈ −11.49 kJ.
q = ΔU − w = 0 − (−11489) = +11489 J ≈ +11.49 kJ.
The gas absorbs 11.49 kJ from the surroundings and uses it entirely to do 11.49 kJ of expansion work.
w_rev = −2.303 × nRT × log(V₂/V₁) = −2.303 × 2 × 8.314 × 300 × log(50/5) = −2.303 × 4988.4 × 1 = −11489 J ≈ −11.49 kJ.
q = ΔU − w = 0 − (−11489) = +11489 J ≈ +11.49 kJ.
The gas absorbs 11.49 kJ from the surroundings and uses it entirely to do 11.49 kJ of expansion work.
Worked Example 5.4: ΔH from ΔU
For the reaction C(graphite) + ½ O₂(g) → CO(g), ΔU = −110.5 kJ at 300 K. Calculate ΔH. (R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)
Δn_g = (moles of gaseous products) − (moles of gaseous reactants) = 1 − 0.5 = +0.5
ΔH = ΔU + Δn_g RT = (−110.5) + (0.5)(8.314 × 10⁻³)(300)
= −110.5 + 1.247 = −109.25 kJ.
Slightly less negative than ΔU because more moles of gas means PV-work is done on the surroundings.
ΔH = ΔU + Δn_g RT = (−110.5) + (0.5)(8.314 × 10⁻³)(300)
= −110.5 + 1.247 = −109.25 kJ.
Slightly less negative than ΔU because more moles of gas means PV-work is done on the surroundings.
🎯 Competency-Based Questions
Q1. The mathematical statement of the first law of thermodynamics for a closed system is: L1 Remember
Answer: (b) ΔU = q + w with IUPAC sign convention (work done ON system positive).
Q2. For which of the following processes is q_p = ΔH valid? L2 Understand
Answer: q_p = ΔH is valid only for processes occurring at constant pressure. This is why open-vessel reactions (atmospheric P) directly give ΔH from heat measurements, while bomb-calorimeter reactions (constant V) give ΔU.
Q3. For the reaction 2H₂(g) + O₂(g) → 2H₂O(l), what is Δn_g and how does ΔH compare with ΔU? L3 Apply
Answer: Δn_g = (0) − (2 + 1) = −3 (water is liquid, so 0 gas moles in product).
ΔH = ΔU + Δn_g RT = ΔU − 3RT, so ΔH is MORE NEGATIVE than ΔU. The system loses extra energy to the atmosphere as the surroundings push in (compression work received reduces |ΔU| but heat released exceeds it).
ΔH = ΔU + Δn_g RT = ΔU − 3RT, so ΔH is MORE NEGATIVE than ΔU. The system loses extra energy to the atmosphere as the surroundings push in (compression work received reduces |ΔU| but heat released exceeds it).
Q4. Fill in the blank: For free expansion of an ideal gas into vacuum, w = ____ and q = ____. L3 Apply
Answer: w = 0 (P_ext = 0) and q = 0 (Joule's experiment showed no temperature change for ideal gas; isothermal). Therefore ΔU = 0 — consistent with the fact that for an ideal gas, U depends only on temperature, which is unchanged.
Q5. HOT (Analyse): The combustion of glucose in a bomb calorimeter gives q_v = −2802 kJ/mol at 298 K. Calculate ΔH for the reaction C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l). L4 Analyse
Answer: ΔU = q_v = −2802 kJ.
Δn_g = 6 (CO₂ products) − 6 (O₂ reactants) = 0.
So ΔH = ΔU + 0 = −2802 kJ/mol.
This is the energy your body extracts per mole of glucose (~180 g) you metabolise — the basis of calorie counting in nutrition.
Δn_g = 6 (CO₂ products) − 6 (O₂ reactants) = 0.
So ΔH = ΔU + 0 = −2802 kJ/mol.
This is the energy your body extracts per mole of glucose (~180 g) you metabolise — the basis of calorie counting in nutrition.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: For an isothermal expansion of an ideal gas, q = −w.
R: Internal energy of an ideal gas depends only on temperature.
Answer: (A). Both true; R explains A. ΔT = 0 → ΔU = 0 → q + w = 0 → q = −w.
A: Enthalpy is more useful than internal energy for chemists working at atmospheric pressure.
R: The change in enthalpy at constant pressure equals the heat exchanged.
Answer: (A). Both true; R explains A. Most laboratory chemistry happens at constant atmospheric pressure, so q_p = ΔH directly.
A: ΔH = ΔU + Δn_g RT applies to all reactions.
R: Only gaseous species contribute appreciably to PΔV at moderate temperatures.
Answer: (A). Both true; R explains A. The contribution of solids/liquids to the volume change is negligible compared to that of gases (Vmolar of gas ~24 L vs ~25 mL for liquid water).
Frequently Asked Questions — First Law of Thermodynamics, Internal Energy and Enthalpy
What is the first law of thermodynamics?
The first law of thermodynamics is the principle of conservation of energy applied to a thermodynamic system. It states that the energy of an isolated system is constant; energy can only be converted from one form to another, never created or destroyed. Mathematically: ΔU = q + w, where ΔU is the change in internal energy of the system, q is the heat absorbed by the system from the surroundings, and w is the work done on the system. NCERT Class 11 Chemistry Chapter 5 uses this law to analyse all energy exchanges in chemical reactions, gas expansions and phase changes.
What is internal energy of a system?
Internal energy (U) is the total energy contained within a thermodynamic system, including the kinetic energy of molecular motion (translational, rotational, vibrational), potential energy of intermolecular interactions, electronic energy and nuclear binding energy. It is a state function, so ΔU depends only on initial and final states. Internal energy cannot be measured directly — only ΔU can be measured during a process. In NCERT Class 11 Chemistry, ΔU equals heat exchanged at constant volume: ΔU = q_v. Bomb calorimeters measure ΔU directly because the rigid container prevents volume work.
How is work calculated in thermodynamics?
Work in thermodynamics is energy transferred when a system pushes against or is pushed by its surroundings. In NCERT Class 11 Chemistry Chapter 5, the most common type is pressure-volume work: w = −p_ext ΔV (sign convention: work done on the system is positive). For free expansion (p_ext = 0), w = 0; for expansion against constant external pressure, w = −p_ext (V₂ − V₁); for reversible isothermal expansion of an ideal gas, w = −nRT ln(V₂/V₁). The negative sign indicates that when the gas expands (ΔV > 0), the system does work on surroundings (w < 0).
What is enthalpy and why is it useful?
Enthalpy (H) is a state function defined as H = U + PV. The change in enthalpy at constant pressure equals the heat exchanged: ΔH = q_p. Since most chemical reactions occur at constant atmospheric pressure (in open beakers), ΔH is the most commonly used heat quantity in chemistry. If ΔH < 0, the reaction is exothermic (releases heat); if ΔH > 0, endothermic (absorbs heat). NCERT Class 11 Chemistry Chapter 5 uses enthalpy throughout — standard enthalpies of formation, combustion, neutralisation, atomisation and bond dissociation are all examples.
What is heat capacity and how is it defined?
Heat capacity is the amount of heat required to raise the temperature of a substance by 1°C (or 1 K). Specific heat capacity (c) is heat per unit mass per degree (J g⁻¹ K⁻¹), while molar heat capacity (C) is per mole per degree (J mol⁻¹ K⁻¹). In NCERT Class 11 Chemistry, two molar heat capacities are defined for gases: at constant volume (C_v) and at constant pressure (C_p). For an ideal gas, C_p − C_v = R (Mayer's relation). The ratio γ = C_p/C_v is used in adiabatic process equations and determines gas behaviour.
What is the difference between isothermal and adiabatic processes?
In an isothermal process, the temperature of the system remains constant (ΔT = 0), so for an ideal gas ΔU = 0 and q = −w. Heat must flow between system and surroundings to maintain constant temperature. In an adiabatic process, no heat is exchanged between the system and surroundings (q = 0), so ΔU = w; the temperature changes as a result. Isothermal processes follow PV = constant, while reversible adiabatic processes follow PV^γ = constant. NCERT Class 11 Chemistry Chapter 5 derives work and heat for both processes — a frequent CBSE board exam topic.
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