ChemistryThermodynamics

At a certain temperature T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then, change in internal energy of the system is :

Solution written and verified by Roshan, science educator with 5 years of experience teaching NEET and JEE aspirants. Last reviewed September 2026.

Options

700 J

300 J

400 J

500 J

Correct Answer

Option 2 : ΔU = 300 J

Step-by-Step Solution

First Law of Thermodynamics:

ΔU = q + W (IUPAC convention)

where q = heat absorbed by system, W = work done ON the system

Assign signs correctly:

q = +500 J (absorbed by system → positive)

W = −200 J (work done BY system → negative, because system loses energy)

ΔU = q + W = 500 + (−200) = 300 J

Internal energy of system increases by 300 J.

ΔU = q + W (IUPAC)

= +500 + (−200) = +300 J

System absorbs 500 J heat but loses 200 J as work → net gain = 300 J

Theory: Laws of Thermodynamics

1. First Law of Thermodynamics

Energy can neither be created nor destroyed — only converted from one form to another. Mathematically: ΔU = q + W (IUPAC convention). ΔU = change in internal energy of the system. q = heat transferred: positive if absorbed by system (endothermic), negative if released by system (exothermic). W = work done: positive if done ON system (compression), negative if done BY system (expansion). Important: there are two sign conventions. IUPAC/Modern: ΔU = q + W. Older convention: ΔU = q − W (where W = work done BY system). NEET uses the IUPAC convention.

2. Sign Conventions — Never Confuse!

📌 Heat (q): q > 0 = absorbed by system (endothermic); q < 0 = released by system (exothermic)

📌 Work (W) — IUPAC: W > 0 = done ON system (compression); W < 0 = done BY system (expansion)

📌 Work done by gas: W_by = PΔV (expansion). In IUPAC: W = −PΔV

📌 In this problem: "work done BY system = 200 J" → W = −200 J in IUPAC convention

📌 Common mistake: Option 1 (700J) = 500 + 200 (wrong sign for W)

📌 Internal energy U: state function (path-independent), extensive property

3. Enthalpy and Its Relation to Internal Energy

Enthalpy H = U + PV. Change: ΔH = ΔU + Δ(PV). At constant pressure: ΔH = ΔU + PΔV = ΔU − W_by = qₚ (heat at constant pressure). For reactions involving gases: ΔH = ΔU + ΔnᵍRT, where Δnᵍ = moles of gaseous products − moles of gaseous reactants. For solids and liquids: ΔH ≈ ΔU (PΔV ≈ 0 for condensed phases). Most chemistry is done at constant pressure → enthalpy change is directly measured as heat evolved/absorbed.

4. Thermodynamic Processes

📌 Isothermal (constant T): ΔU = 0 for ideal gas → q = −W

📌 Adiabatic (no heat exchange, q=0): ΔU = W only

📌 Isochoric (constant V): W = 0 → ΔU = q = qᵥ

📌 Isobaric (constant P): ΔH = qₚ (heat at constant P)

📌 For ideal gas isothermal expansion: ΔU = 0, q = nRT ln(V₂/V₁), W = −nRT ln(V₂/V₁)

📌 Reversible work > Irreversible work for same expansion (maximum work = reversible)

5. Hess's Law

The enthalpy change for a reaction is independent of the pathway — it depends only on the initial and final states (since H is a state function). So ΔH for a reaction = sum of ΔH for all steps in any pathway. Applications: calculate ΔH for reactions that cannot be directly measured (e.g., formation of CO: C + ½O₂ → CO, can't control the reaction to stop at CO). Standard enthalpy of formation (ΔHf°): enthalpy change when 1 mole of compound is formed from elements in their standard states. ΔH°reaction = Σ ΔHf°(products) − Σ ΔHf°(reactants).

6. Bond Enthalpy and Bond Energy

Bond enthalpy = energy required to break 1 mole of bonds homolytically in gaseous molecules. ΔH°reaction = Σ(Bond energies broken) − Σ(Bond energies formed). Breaking bonds requires energy (+ve), forming bonds releases energy (−ve). More energy released forming bonds than needed to break → exothermic reaction (ΔH < 0). Used to estimate ΔH when thermodynamic data is unavailable. Average bond enthalpies: C-H (414), C-C (347), C=C (611), C≡C (837), O-H (460), N≡N (946 — highest), H-H (436) kJ/mol.

7. Second Law and Entropy

Second Law: entropy (S) of universe always increases for spontaneous processes: ΔS_universe = ΔS_system + ΔS_surroundings > 0. Entropy = measure of disorder/randomness. ΔS = q_rev/T (at constant T). For phase changes: ΔS = ΔH_transition/T. Entropy increases when: gas is produced, temperature increases, mixing occurs, ordered → disordered. Third Law: entropy of a perfect crystal at 0 K = 0 (absolute zero entropy). This defines the absolute entropy scale.

8. Gibbs Free Energy — Spontaneity

G = H − TS. ΔG = ΔH − TΔS (at constant T and P). Spontaneity: ΔG < 0 → spontaneous; ΔG > 0 → non-spontaneous; ΔG = 0 → equilibrium. Cases: ΔH < 0, ΔS > 0 → always spontaneous (ΔG always negative). ΔH > 0, ΔS < 0 → never spontaneous. ΔH < 0, ΔS < 0 → spontaneous at low T (when TΔS < ΔH). ΔH > 0, ΔS > 0 → spontaneous at high T (when TΔS > ΔH). Relationship to equilibrium: ΔG° = −RT ln K = −nFE°. Standard Gibbs energy connects thermodynamics, equilibrium, and electrochemistry.

Frequently Asked Questions

1. Why is work done by system negative in IUPAC convention? ⌄

IUPAC defines work from the system's perspective. When the system does work ON the surroundings (expansion), the system loses energy → work is negative for the system (W < 0). When surroundings do work ON the system (compression), the system gains energy → work is positive (W > 0). Think of it like a bank account: money leaving your account (work done by you) is negative; money coming in (work done for you) is positive. In old convention (used in many textbooks): ΔU = q − W_by, where W_by = work done by system. Both give the same ΔU — just different sign convention for W.

2. Why is option 1 (700 J) wrong? ⌄

Option 1 (700 J) = 500 + 200. This assumes BOTH q and W are positive: treating work done by system as positive energy for the system. But work done BY the system means the system LOSES energy (like a battery doing work on a circuit — the battery's energy decreases). Correctly: W = −200 J (system loses 200 J as work). ΔU = 500 + (−200) = 300 J. If you get 700 J, you made the sign error of treating work done BY system as +200 instead of −200.

3. What is internal energy? ⌄

Internal energy (U) is the total energy stored within the system — kinetic energy (translational, rotational, vibrational) + potential energy (intermolecular interactions, chemical bonds, nuclear energy) of all particles. It's a state function (depends only on current state, not how system got there). Absolute value of U cannot be measured — only ΔU can be measured. At constant volume: ΔU = qᵥ (heat absorbed at constant volume, measured in bomb calorimeter). For ideal gas: U depends only on temperature (not volume or pressure).

4. What is the difference between q and ΔH? ⌄

q = heat transferred — depends on the path (not a state function). At constant pressure: qₚ = ΔH (enthalpy change) — ΔH IS a state function. At constant volume: qᵥ = ΔU (internal energy change). So q equals a state function (ΔH or ΔU) only at specific conditions (constant P or constant V). In general: q is path-dependent. That's why we define H and U — to have state functions that equal q under specific conditions, making calculations path-independent.

5. For an ideal gas, why is ΔU = 0 in an isothermal process? ⌄

For an ideal gas: kinetic energy = (3/2)nRT (for monoatomic). There are no intermolecular forces → no potential energy. So U depends ONLY on temperature. In an isothermal process: ΔT = 0 → ΔU = 0. From first law: ΔU = q + W = 0 → q = −W = P_ext × ΔV. All heat absorbed is converted to work (or vice versa). For real gases: intermolecular forces exist → expanding the gas changes intermolecular PE → ΔU ≠ 0 even at constant T (Joule-Thomson effect).

6. What is a bomb calorimeter and what does it measure? ⌄

Bomb calorimeter = constant volume calorimeter. The sample burns inside a sealed steel "bomb" submerged in water. Since volume is constant: W_pV = 0, so ΔU = qᵥ. The heat released raises the temperature of the water bath. ΔU = −C_cal × ΔT, where C_cal = heat capacity of calorimeter + water. Used to measure: heat of combustion of foods (calories in food), heat of combustion of fuels, ΔH_combustion. To get ΔH: ΔH = ΔU + ΔnᵍRT. Bomb calorimeter is very accurate but measures ΔU, not ΔH directly.

7. When is ΔH = ΔU? ⌄

ΔH = ΔU + ΔnᵍRT. ΔH = ΔU when: (1) Δnᵍ = 0 (no change in moles of gas). Example: H₂(g) + I₂(g) → 2HI(g); Δnᵍ = 2−(1+1) = 0 → ΔH = ΔU. (2) Reactions involving only solids and liquids (PΔV ≈ 0). Example: CaCO₃(s) → CaO(s) + CO₂(g) has Δnᵍ = 1 → ΔH ≠ ΔU. For most organic reactions with gas phase: ΔH and ΔU differ by Δnᵍ × RT ≈ Δnᵍ × 2·5 kJ/mol at 300K.

8. What determines spontaneity when ΔH and ΔS have same sign? ⌄

ΔG = ΔH − TΔS. When ΔH < 0 and ΔS < 0: ΔG = (−) − T(−) = (−) + T(+). Spontaneous when |ΔH| > T|ΔS|, i.e., low temperature: at high T, the −TΔS term dominates → ΔG > 0 → non-spontaneous. When ΔH > 0 and ΔS > 0: ΔG = (+) − T(+). Spontaneous when T|ΔS| > |ΔH|, i.e., high temperature. Temperature where transition from spontaneous to non-spontaneous: T = ΔH/ΔS (where ΔG = 0). Examples: ice melting (ΔH > 0, ΔS > 0) spontaneous above 0°C.

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