5.1 First Law of Thermodynamics: Work, Heat, Enthalpy, Adiabatic Changes, Joule–Thomson Effect and Thermochemistry
★★★ 2021–2025: 9 Mains parts, 73 marks (Civil Services 7 parts / 55 m · IFoS 2 / 18 m) and 35 Geo-Scientist Prelims questions (2021–2026), about six every year. Short numericals (bomb calorimeter, tyre, porous plug, ice) recur in Civil Services Paper I Q1 and Q3.
Syllabus lines. Civil Services: “Work, heat and internal energy; first law of thermodynamics … J-T effect and inversion temperature”. IFoS: “Thermodynamic systems, states and processes, work, heat and internal energy; first law of thermodynamics, work done on the systems and heat absorbed in different types of processes; calorimetry, energy and enthalpy changes in various processes and their temperature dependence”. Geo-Scientist: Mains Paper II starts thermodynamics at chemical potential, so the first law is examined in the Prelims chemistry paper.
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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes
5.1.1 Basic terms: system, state functions, work and heat ★★
- System: open (matter and energy cross), closed (energy only), isolated (neither). State functions (U, H, S, p, V, T) depend only on the present state: their change between two states is the same for every path, and round a cycle it is zero. Work and heat are path functions (inexact differentials đw, đq) Geo Pre 2024 Q50 Geo Pre 2024 Q51.SN
- Intensive properties do not depend on the amount (T, p, concentration, density, surface tension, dielectric constant, e.m.f., any molar quantity such as Hm, J mol⁻¹, or Cp,m, J K⁻¹ mol⁻¹); extensive ones are additive (V, U, H, S, G, Cp in J K⁻¹). The ratio of two extensive properties is intensive Geo Pre 2021 Q52 Geo Pre 2022 Q49 Geo Pre 2024 Q49 Geo Pre 2026 Q51.SN
- For a state function z(x, y) the cross derivatives are equal (Euler test of exactness) and the cyclic rule holds: ; with U(p, V) the product is −1 Geo Pre 2022 Q50.SN
- Sign convention (IUPAC): q and w are energy received by the system; work done by an expanding gas is negative. Older books write ΔU = q − w with w done by the system: same physics, opposite sign of w Geo Pre 2025 Q52.
5.1.2 First law and expansion work ★★★
- First law: energy is conserved; the internal energy of an isolated system is constant. For a closed system the change in U equals the heat supplied plus the work done on it.SN
- Free expansion (pext = 0): w = 0. In an insulated box q = 0 as well, so ΔU = 0; for a perfect gas U depends on T only, so ΔT = 0 and ΔH = 0. The process is still irreversible: ΔSuniv = nR ln[(V1 + V2)/V1] > 0 (R ln 2 = 5.76 J K⁻¹ mol⁻¹ when the volume doubles) CSE 2021 Q3(c)(ii) 10m Geo Pre 2022 Q54 Geo Pre 2024 Q53.SN
- Constant external pressure: w = −pextΔV. An ideal gas going from 3.00 to 6.00 L against 2.50 atm while absorbing 1000 J: w = −7.50 L atm = −760 J, ΔU = 1000 − 760 = +240 J Geo Pre 2026 Q50. If đw = 0 and đq < 0, dU < 0 and the temperature of an ideal gas falls Geo Pre 2025 Q54.SN
- Reversible = maximum work: the external pressure is kept infinitesimally below the gas pressure, so the gas does the most work possible (largest area); compression the other way needs the least work. Any real (one-step, fast) expansion does less Geo Pre 2023 Q48.SN
- Quick prelims values: 0.001 mol, 25 → 250 mL at 300 K: w = −nRT ln 10 = −5.74 J Geo Pre 2025 Q51; 5.00 mol, 0.500 → 1.500 dm³ at 300 K: w = −13.7 kJ, q = +13.7 kJ, ΔU = ΔH = 0 Geo Pre 2026 Q52; 2.0 mol compressed 1 → 100 bar at 300 K: w = +23 kJ, ΔU = ΔH = 0 Geo Pre 2023 Q49; 2.0 mol He, 1 → 2 bar at 100 K: w = +1145 J, q = −1145 J, ΔU = 0 Geo Pre 2025 Q55.SN
Solved PYQ CSE 2023 Q7(e) 10m
An ideal gas expands reversibly and isothermally at 25 °C from 0.05 L, 8 atm to 0.40 L, 1 atm. Find w, ΔU, ΔH, ΔS. Is q equal to ΔH?
- nRT = p1V1 = 0.40 L atm; w = −0.40 ln 8 = −0.832 L atm = −84.3 J.
- ΔU = ΔH = 0 (ideal gas, constant T); q = −w = +84.3 J; ΔS = qrev/T = 84.3/298.15 = +0.283 J K⁻¹.
- No: q = ΔH only at constant pressure with expansion work alone; here p falls from 8 to 1 atm. ΔSsurr = −0.283 J K⁻¹, total zero, as for any reversible change.SN
5.1.3 Enthalpy, heat capacities and calorimetry ★★★
- At constant volume no expansion work is done, so heat = ΔU; at constant pressure (only pV work) heat = ΔH. A bomb calorimeter measures ΔU; an open calorimeter or a dish in air measures ΔH Geo Pre 2026 Q54.SN
- Joule’s law: for a perfect gas (∂U/∂V)T = 0 (Joule’s free-expansion experiment showed no temperature change), so U and H depend on T only and ΔU = nCV,mΔT for any process of a perfect gas.SN
- Monatomic perfect gas: CV,m = (3/2)R, Cp,m = (5/2)R, γ = 5/3; diatomic near room temperature: (5/2)R, (7/2)R, γ = 1.40. Cp > CV because at constant pressure part of the heat goes into expansion work.SN
- Δng counts gas molecules only. Benzoic acid, C₇H₆O₂(s) + 15/2 O₂ → 7 CO₂ + 3 H₂O(l): Δng = −1/2 per mole. 1.22 g (0.0100 mol) burnt in an open dish at 27 °C releasing 31 700 J: ΔH = −31 700 J, ΔU = ΔH − ΔngRT = −31 688 J Geo Pre 2026 Q54. Glucose combustion has Δng = 0, so ΔcH = ΔcU Geo Pre 2026 Q55.SN
Solved PYQ CSE 2025 Q1(g) 5m
0.500 g benzoic acid (ΔcU = −3227 kJ mol⁻¹) raises the temperature of a bomb calorimeter by 1.236 K; 0.300 g naphthalene in the same set-up gives 1.128 K. Find ΔcU of naphthalene.
- n(benzoic acid) = 0.500/122.12 = 4.094 × 10−3 mol; heat = 13.21 kJ; calorimeter constant C = 13.21/1.236 = 10.69 kJ K⁻¹.
- Naphthalene: heat = 10.69 × 1.128 = 12.06 kJ; n = 0.300/128.17 = 2.341 × 10−3 mol.
- ΔcU = −12.06/2.341 × 10−3 = −5.15 × 103 kJ mol⁻¹; C₁₀H₈ + 12 O₂ → 10 CO₂ + 4 H₂O(l), Δng = −2, ΔcH = −5157 kJ mol⁻¹.SN
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