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MHT-CET Chemistry · Chemical Thermodynamics and Energetics

First Law of Thermodynamics, Internal Energy and Work

Energy is conserved: ΔU = q + w, with heat absorbed and work done ON the system positive; pressure–volume work against a constant external pressure is −P_ext ΔV, and reversible isothermal work is −2.303 nRT log(V₂/V₁).

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Why this matters

52 PYQs, 2 HARD — the biggest subtopic in the chapter and the one every paper draws from. Five shapes: add q and w with the right signs; compute −P_ext ΔV and convert dm³ bar or L atm to joules; combine the two; work for a gas reaction from Δn_g RT; and the reversible-isothermal logarithm. Get the sign convention and the 100 J per dm³ bar right and every one of these is arithmetic.

Concept 1 of 5

ΔU = q + w and the Sign Convention

Intuition

Think from the system's point of view. Energy coming IN — heat absorbed, work done ON it — is positive; energy going OUT — heat released, work done BY it — is negative. Add the two and you have the change in internal energy, a state function that rises with temperature.

Definition

  • ΔU=q+w\Delta U = q + w. Heat absorbed q>0q > 0; heat released q<0q < 0. Work done ON the system w>0w > 0; work done BY the system w<0w < 0.
  • 40 J absorbed, 8 J done by the system: ΔU=40−8=32\Delta U = 40 - 8 = 32 J. 605 J absorbed, 380 J done by: +225 J.
  • 300 J released, 150 J done by: −300−150=−450-300 - 150 = -450 J. 8 kJ released, 660 J done by: −8660 J. 15 kJ done by, 2 kJ lost: −17 kJ.
  • 20 kJ done ON, 10 kJ released: −10+20=+10-10 + 20 = +10 kJ. x kJ released, y kJ done on: y−xy - x.
  • At constant volume w=0w = 0 so qV=ΔUq_V = \Delta U — ΔU is the heat of reaction at constant volume. The first law says the energy of the UNIVERSE is constant, not of the system.

First law

ΔU=q+w,qabsorbed>0,won system>0\Delta U = q + w,\qquad q_{\text{absorbed}} > 0,\quad w_{\text{on system}} > 0

Worked example

A system does 250 J of work on the surroundings while absorbing 90 J of heat. Find ΔU. Then, for a different step, 400 J of work is done on the system and it loses 120 J of heat.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Chemical Thermodynamics and EnergeticsEASY
In a process 605 J heat is absorbed by the system and 380 J work is done by the system on surrounding. What is the value of ΔU\Delta U?

[Q76 · 2nd May Shift 1 · 2023]

'Work done BY the system' entered as positive

That is the old convention and it flips every answer's sign. In the current one, work done BY the system LEAVES it: w is negative. 'Work done ON the system' is the positive case.
Drill 15 more on δu = q + w and the sign convention

Concept 2 of 5

Work Against a Constant External Pressure: w = −P_ext ΔV

Intuition

Push the piston out against a fixed opposing pressure and the system pays P_ext × ΔV of energy, so its work is negative. Compress it and w is positive. The product comes out in dm³ bar or L atm and must be converted: 1 dm³ bar = 100 J, 1 L atm = 101.3 J, 1 m³ Pa = 1 J.

Definition

  • w=−Pext(V2−V1)w = -P_{\text{ext}}(V_2 - V_1). Expansion (V₂ > V₁): w negative. Compression: w positive. Free expansion (P_ext = 0): w = 0.
  • 1.5 bar, 5 → 10 dm³: −7.5-7.5 dm³ bar =−750= -750 J. 1 bar, 3 → 15 L: −1200 J. 1.9 bar, 0.3 → 2.5 dm³: −418 J.
  • SI: 2×1052 \times 10^5 N m⁻², 200 cm³ =2×10−4= 2 \times 10^{-4} m³: −40-40 J. 2.02×1052.02 \times 10^5 Pa, 5 → 7 dm³: −404 J. 2 atm, 1.5 L: −303.9 J; 1 atm, 4.5 dm³: −456 J.
  • Compression: 3 bar, 24 → 13 dm³: +33+33 dm³ bar =+3300= +3300 J; 4 bar, 25 → 13: +4800 J.
  • Solve for the pressure: Pext=−w/ΔVP_{\text{ext}} = -w/\Delta V. 500 J = 5 dm³ bar over 2 L: 2.5 bar; −600 J over 5 dm³: 1.2 bar.

PV work

w=−Pext ΔV,1 dm3 bar=100 J,1 L atm=101.3 Jw = -P_{\text{ext}}\,\Delta V,\qquad 1\ \text{dm}^3\,\text{bar} = 100\ \text{J},\quad 1\ \text{L atm} = 101.3\ \text{J}

Worked example

A gas expands from 2 dm³ to 6.5 dm³ against 2.5 bar. Find the work in joules. What external pressure would make the same expansion cost 900 J?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Chemical Thermodynamics and EnergeticsEASY
Two moles of an ideal gas is expanded isothermally from a volume of 300 cm3300\,\text{cm}^3 to 2.5 dm32.5\,\text{dm}^3 at 298 K against a constant pressure at 1.9 bar. Calculate the work done in joules.

[Q95 · 9th May Shift 1 · 2023]

Leaving the answer in dm³ bar when joules are asked

−12 dm³ bar is −1200 J. Options are printed a factor of 100 apart. In SI, cm³ must become m³ (× 10⁻⁶) before multiplying by pascals.
Drill 12 more on work against a constant external pressure: w = −p_ext δv

Concept 3 of 5

ΔU From Heat Absorbed and an Expansion

Intuition

Combine the two: the gas takes in q and spends P_ext ΔV on pushing back the surroundings. ΔU = q − P_ext ΔV. When the work exceeds the heat, ΔU is negative even though heat was absorbed.

Definition

  • ΔU=q−PextΔV\Delta U = q - P_{\text{ext}}\Delta V, the work converted to the same unit as q.
  • 800 J absorbed, 1 bar, 10 → 20 dm³: 800−1000=−200800 - 1000 = -200 J. 10 kJ absorbed, 2 bar, 5 → 8 L: 10000−600=940010000 - 600 = 9400 J.
  • 200 J absorbed, 2×1052 \times 10^5 Pa, 500 cm³: 200−100=+100200 - 100 = +100 J. 150 J, 300 cm³: 90 J. 210 J, 10⁵ Pa, 3 → 6 L: 210−300=−90210 - 300 = -90 J.
  • 302.6 J absorbed, 2 atm, 100 → 200 L: 302.6−20265=−19962302.6 - 20265 = -19962 J.

First law with expansion work

ΔU=q−Pext ΔV\Delta U = q - P_{\text{ext}}\,\Delta V

Worked example

A gas absorbs 450 J and expands from 4 dm³ to 7 dm³ against 1.2 bar. Find ΔU.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Chemical Thermodynamics and EnergeticsMODERATE
A gas expands isothermally against a constant external pressure of 1 bar from 10 dm310\,\text{dm}^3 to 20 dm320\,\text{dm}^3 by absorbing 800 J of heat from surrounding. Calculate value of ΔU\Delta U.

[Q85 · 10th May Shift 2 · 2023]

Adding the work instead of subtracting it

An EXPANSION is work done by the system: it is subtracted from q. 800 + 1000 = 1800 J is not on offer, but 800 − 100 (a slipped conversion) is.
Drill 6 more on δu from heat absorbed and an expansion

Concept 4 of 5

Work in a Gas-Phase Reaction: w = −Δn_g RT

Intuition

A reaction that makes more moles of gas pushes the atmosphere back and does work (negative); one that consumes gas has work done on it (positive); equal moles either side means zero work. At constant temperature PΔV = Δn_g RT. With volumes given at fixed P, use −PΔV directly.

Definition

  • w=−ΔngRTw = -\Delta n_g RT, Δng\Delta n_g = gaseous product moles − gaseous reactant moles, R = 8.314, T in K.
  • 4SO2+2O2→4SO34\text{SO}_2 + 2\text{O}_2 \to 4\text{SO}_3 at 300 K: Δng=−2\Delta n_g = -2, w=+2×8.314×300=+4988w = +2 \times 8.314 \times 300 = +4988 J.
  • Zero work: H2+Cl2→2HCl\text{H}_2 + \text{Cl}_2 \to 2\text{HCl}, CH4+Cl2→CH3Cl+HCl\text{CH}_4 + \text{Cl}_2 \to \text{CH}_3\text{Cl} + \text{HCl}. Negative work (gas made): 2H2O2(l)→2H2O(l)+O2(g)2\text{H}_2\text{O}_2(l) \to 2\text{H}_2\text{O}(l) + \text{O}_2(g). Positive: N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3.
  • Volumes at 1 bar: C2H4\text{C}_2\text{H}_4 (200 mL) + HCl (150 mL) → C2H5Cl\text{C}_2\text{H}_5\text{Cl}: 150 mL reacts with 150 mL to give 150 mL, ΔV=−150\Delta V = -150 mL, w=+0.15w = +0.15 dm³ bar =+15= +15 J.

Reaction work

w=−Δng RTw = -\Delta n_g\,RT

Worked example

Find the work done at 27 °C when 2 mol of N2O4(g)\text{N}_2\text{O}_4(g) decompose to NO2(g)\text{NO}_2(g).
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Chemical Thermodynamics and EnergeticsMODERATE
What is the work done during oxidation of 4 moles of SO2(g)\text{SO}_2(g) to SO3(g)\text{SO}_3(g) at 27∘C27^{\circ}\text{C} (R=8.314 J K−1mol−1R = 8.314\text{ J K}^{-1}\text{mol}^{-1})

[Q52 · 11th May Shift 2 · 2024]

Counting liquids and solids in Δn

Only GASES do PV work. In C2H2+52O2→2CO2+H2O(l)\text{C}_2\text{H}_2 + \tfrac{5}{2}\text{O}_2 \to 2\text{CO}_2 + \text{H}_2\text{O}(l) the water is liquid: Δng=2−3.5\Delta n_g = 2 - 3.5, not 3−3.53 - 3.5.
Drill 6 more on work in a gas-phase reaction: w = −δn_g rt

Concept 5 of 5

Reversible Isothermal Work: −2.303 nRT log(V₂/V₁)

Intuition

Expanding in infinitesimal steps against a pressure always just below the gas's own extracts the maximum work, and the sum becomes a logarithm. For expansion the log is positive and w negative; for compression the same magnitude comes out positive. Pressures invert the ratio: V₂/V₁ = P₁/P₂.

Definition

  • wrev=−nRTln⁡V2V1=−2.303 nRTlog⁡10V2V1=−2.303 nRTlog⁡10P1P2w_{\text{rev}} = -nRT\ln\dfrac{V_2}{V_1} = -2.303\,nRT\log_{10}\dfrac{V_2}{V_1} = -2.303\,nRT\log_{10}\dfrac{P_1}{P_2}.
  • 2 mol, 300 K, 20 → 40 L: −2.303×2×8.314×300×0.301=−3458-2.303 \times 2 \times 8.314 \times 300 \times 0.301 = -3458 J. Compression 40 → 20 L: +3.46 kJ.
  • 1 mol, 300 K, 10 bar → 1 bar: −5744-5744 J. Compression x → 2x bar or 12 → 6 dm³: +1729 J.
  • Equal MASSES of gases expanding by the same ratio: work ∝ n = m/M, so the LIGHTEST gas does the most work — H₂ among H₂, N₂, Cl₂, O₂; NH₃ among NH₃, N₂, Cl₂, H₂S.
  • 2.303×8.314×300=57442.303 \times 8.314 \times 300 = 5744 per mole per decade — worth remembering.

Maximum (reversible) work

wrev=−2.303 nRTlog⁡10V2V1=−2.303 nRTlog⁡10P1P2w_{\text{rev}} = -2.303\,nRT\log_{10}\frac{V_2}{V_1} = -2.303\,nRT\log_{10}\frac{P_1}{P_2}

Worked example

3 mol of an ideal gas expand reversibly and isothermally at 400 K from 4 dm³ to 40 dm³. Find w.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Chemical Thermodynamics and EnergeticsMODERATE
Two moles of an ideal gas are compressed isothermally and reversibly from 40 L to 20 L at 300 K. What is the work done? R=8.314 J K−1 mol−1R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}

[Q80 · 11th May Shift 2 · 2023]

Using the pressure ratio the same way as the volume ratio

V2/V1=P1/P2V_2/V_1 = P_1/P_2, inverted. For 10 bar → 1 bar the log argument is 10 (expansion, w negative), not 0.1. Options with the opposite sign are always there.
Drill 8 more on reversible isothermal work: −2.303 nrt log(v₂/v₁)

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