PYQ Vault
EbooksTutor

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).

All Chemical Thermodynamics and Energetics notesMHT-CET Chemistry strategy

Why this matters

76 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

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.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q76Easy

Example 1 · Chemical Thermodynamics and Energetics · First Law of Thermodynamics, Internal Energy and Work

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?

'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

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?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q95Easy

Example 2 · Chemical Thermodynamics and Energetics · First Law of Thermodynamics, Internal Energy and Work

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.

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

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.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q88Moderate

Example 3 · Chemical Thermodynamics and Energetics · First Law of Thermodynamics, Internal Energy and Work

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.

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 5 more on δu from heat absorbed and an expansion

Concept 4 of 5: Work in a Gas-Phase Reaction: w = −Δn_g RT

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).
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q52Moderate

Example 4 · Chemical Thermodynamics and Energetics · First Law of Thermodynamics, Internal Energy and Work

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})

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)

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.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q80Moderate

Example 5 · Chemical Thermodynamics and Energetics · First Law of Thermodynamics, Internal Energy and Work

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}

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 7 more on reversible isothermal work: −2.303 nrt log(v₂/v₁)

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Watch out for (5)

Test yourself on Chemical Thermodynamics and Energetics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Thermodynamics and Energetics chapter testDrill all 76 questions