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Thermodynamics

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Chapter Notes

Overview
This chapter introduces the fundamental principles of thermodynamics, focusing on energy transformations in chemical systems. It explains how to quantify energy changes (heat and work) during chemical reactions and physical processes, and introduces key state functions like internal energy (U) and enthalpy (H). The central idea is the First Law of Thermodynamics (conservation of energy) and its application to calculate enthalpy changes, including the use of Hess's Law. A student must understand that energy is neither created nor destroyed, and that enthalpy change (ΔH) is the heat exchanged at constant pressure, which can be measured or calculated from standard enthalpies of formation.
Key Concepts (33)
System
That part of the universe in which observations are made.
e.g. The reaction mixture in a beaker.
Surroundings
The rest of the universe other than the system, which can interact with the system.
e.g. The room where the beaker is kept.
Boundary
The wall that separates the system from the surroundings; it may be real or imaginary.
e.g. The walls of a beaker or an imaginary surface enclosing a gas.
Open System
A system in which there is exchange of energy and matter between the system and surroundings.
e.g. Reactants in an open beaker.
Closed System
A system in which there is no exchange of matter, but exchange of energy is possible between the system and surroundings.
e.g. Reactants in a closed copper vessel.
Isolated System
A system in which there is no exchange of energy or matter between the system and surroundings.
e.g. Reactants in a thermos flask or a closed insulated vessel.
State Functions (State Variables)
Properties whose values depend only on the state of the system and not on how it is reached.
e.g. Pressure (p), Volume (V), Temperature (T), Internal Energy (U), Enthalpy (H).
Internal Energy (U)
The sum of all forms of energy (chemical, electrical, mechanical, etc.) contained within a system.
e.g. The total energy of water in a beaker.
Adiabatic Process
A process in which there is no transfer of heat between the system and surroundings.
e.g. A reaction in a thermos flask or an insulated beaker.
Heat (q)
The exchange of energy resulting from a temperature difference between the system and surroundings.
e.g. Heat absorbed by water from a hot plate.
Work (w)
Energy transferred when a force moves an object; in thermodynamics, often pressure-volume work.
e.g. Work done by a gas expanding against a piston.
First Law of Thermodynamics
The energy of an isolated system is constant; energy can neither be created nor destroyed. Mathematically: ΔU = q + w.
e.g. If 100 J of heat is added to a system and it does 40 J of work, its internal energy increases by 60 J.
Enthalpy (H)
A thermodynamic state function defined as H = U + pV. It represents the heat content of the system at constant pressure.
e.g. The heat absorbed or released in a reaction at constant pressure is ΔH.
Heat Capacity (C)
The quantity of heat required to raise the temperature of a system by one degree Celsius (or one Kelvin).
e.g. Water has a large heat capacity, so it takes a lot of energy to raise its temperature.
Molar Heat Capacity (Cm)
The heat capacity for one mole of a substance; the quantity of heat needed to raise the temperature of one mole by one degree Celsius.
e.g. Cm for water is 75.3 J K⁻¹ mol⁻¹.
Specific Heat Capacity (c)
The quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius.
e.g. Specific heat of water is 4.18 J g⁻¹ °C⁻¹.
Extensive Property
A property whose value depends on the quantity or size of matter present in the system.
e.g. Mass, volume, internal energy, enthalpy, heat capacity.
Intensive Property
A property that does not depend on the quantity or size of matter present.
e.g. Temperature, density, pressure, molar volume, molar heat capacity.
Standard Enthalpy of Reaction (ΔrH⁰)
The enthalpy change for a reaction when all the participating substances are in their standard states (pure form at 1 bar, usually at 298 K).
e.g. C(graphite) + O₂(g) → CO₂(g); ΔrH⁰ = -393.5 kJ mol⁻¹.
Standard Enthalpy of Formation (ΔfH⁰)
The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (reference states).
e.g. H₂(g) + ½O₂(g) → H₂O(l); ΔfH⁰ = -285.8 kJ mol⁻¹.
Standard Enthalpy of Fusion (ΔfusH⁰)
The enthalpy change that accompanies melting of one mole of a solid substance in its standard state.
e.g. H₂O(s) → H₂O(l); ΔfusH⁰ = 6.00 kJ mol⁻¹.
Standard Enthalpy of Vaporisation (ΔvapH⁰)
The amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar).
e.g. H₂O(l) → H₂O(g); ΔvapH⁰ = +40.79 kJ mol⁻¹.
Standard Enthalpy of Sublimation (ΔsubH⁰)
The change in enthalpy when one mole of a solid substance sublimes at constant temperature and under standard pressure (1 bar).
e.g. CO₂(s) → CO₂(g); ΔsubH⁰ = 25.2 kJ mol⁻¹.
Thermochemical Equation
A balanced chemical equation together with the value of its ΔrH.
e.g. C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l); ΔrH⁰ = -1367 kJ mol⁻¹.
Hess's Law of Constant Heat Summation
If a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
e.g. C(graphite) + O₂(g) → CO(g) can be found by combining C + O₂ → CO₂ and CO₂ → CO + ½O₂.
Reversible Process
A process or change that is brought out in such a way that the process could, at any moment, be reversed by an infinitesimal change. It proceeds infinitely slowly by a series of equilibrium states.
e.g. Isothermal reversible expansion of an ideal gas against an external pressure that is always infinitesimally less than the gas pressure.
Irreversible Process
Any process that is not reversible; usually occurs in a single step or finite steps with a finite pressure difference.
e.g. Expansion of a gas against a constant external pressure of 1 atm.
Free Expansion
Expansion of a gas into vacuum (pex = 0). No work is done during free expansion of an ideal gas.
e.g. A gas expanding into an evacuated flask.
Isothermal Process
A process carried out at constant temperature.
e.g. Expansion of an ideal gas at 25°C.
Adiabatic Wall
A wall that does not permit the transfer of heat between the system and surroundings.
e.g. The walls of a thermos flask.
Calorimetry
An experimental technique to measure energy changes associated with chemical or physical processes.
e.g. Using a bomb calorimeter to measure heat of combustion.
Bomb Calorimeter
A sealed steel vessel used to measure heat changes at constant volume (ΔU).
e.g. Burning 1 g of graphite in a bomb calorimeter to find its enthalpy of combustion.
Reference State of an Element
The most stable state of aggregation of an element at 25°C and 1 bar pressure.
e.g. C(graphite), O₂(g), H₂(g), S(rhombic).
Formulas & Equations (20)
ΔU = q + w
Reads as: Delta U equals q plus w
Variables: ΔU = change in internal energy (J or kJ); q = heat exchanged (J or kJ); w = work done on/by the system (J or kJ).
Use when: Applying the First Law of Thermodynamics to any process.
⚠ Students forget the sign convention: q is positive when heat is absorbed by the system; w is positive when work is done on the system.
w = –pex ΔV
Reads as: w equals negative p-ex times delta V
Variables: w = work done on the system (J or L atm); pex = external pressure (atm or bar); ΔV = change in volume (Vf – Vi) (L or m³).
Use when: Calculating pressure-volume work in a single step (irreversible) process.
⚠ Students forget the negative sign. For compression (ΔV negative), w becomes positive (work done on system).
w_rev = –∫ p dV (from Vi to Vf)
Reads as: w reversible equals negative integral of p dV from V i to V f
Variables: w_rev = reversible work; p = internal pressure of gas; dV = infinitesimal change in volume.
Use when: Calculating work for a reversible process (infinitesimally slow).
⚠ Students forget that for reversible processes, pex is always nearly equal to p (internal pressure).
w_rev = –nRT ln(Vf/Vi) = –2.303 nRT log(Vf/Vi)
Reads as: w reversible equals negative n R T natural log of Vf over Vi, or negative 2.303 n R T log of Vf over Vi
Variables: n = number of moles; R = universal gas constant (0.08206 L atm mol⁻¹ K⁻¹ or 8.314 J mol⁻¹ K⁻¹); T = absolute temperature (K); Vf = final volume; Vi = initial volume.
Use when: Calculating work done in an isothermal reversible expansion/compression of an ideal gas.
⚠ Students forget to use absolute temperature (Kelvin) and the correct value of R (consistent units).
For isothermal irreversible change: q = –w = pex (Vf – Vi)
Reads as: For isothermal irreversible change, q equals negative w equals p-ex times (Vf minus Vi)
Variables: q = heat absorbed; w = work done; pex = constant external pressure; Vf, Vi = final and initial volumes.
Use when: Calculating heat and work for isothermal irreversible expansion against constant external pressure.
⚠ Students forget that for isothermal process, ΔU = 0, so q = –w.
For isothermal reversible change: q = –w = nRT ln(Vf/Vi) = 2.303 nRT log(Vf/Vi)
Reads as: For isothermal reversible change, q equals negative w equals n R T natural log of Vf over Vi
Variables: Same as above.
Use when: Calculating heat and work for isothermal reversible expansion.
⚠ Students forget the sign: q is positive (heat absorbed) when gas expands (Vf > Vi).
For adiabatic change: q = 0, ΔU = w_ad
Reads as: For adiabatic change, q equals zero, delta U equals w ad
Variables: w_ad = work done in adiabatic process.
Use when: Any process where no heat is exchanged with surroundings.
⚠ Students think ΔU is zero in adiabatic process; it is not — work changes internal energy.
H = U + pV
Reads as: H equals U plus p V
Variables: H = enthalpy (J or kJ); U = internal energy; p = pressure; V = volume.
Use when: Defining enthalpy as a state function.
⚠ Students forget that H is a state function, but q (heat) is path-dependent.
ΔH = ΔU + pΔV (at constant pressure)
Reads as: Delta H equals delta U plus p delta V
Variables: ΔH = change in enthalpy; ΔU = change in internal energy; p = constant pressure; ΔV = change in volume.
Use when: Relating enthalpy change to internal energy change at constant pressure.
⚠ Students forget that this equation is valid only at constant pressure.
ΔH = q_p
Reads as: Delta H equals q sub p
Variables: q_p = heat absorbed by the system at constant pressure.
Use when: Measuring enthalpy change directly from heat exchanged at constant pressure.
⚠ Students confuse q_p with q_V (heat at constant volume, which equals ΔU).
ΔU = q_V
Reads as: Delta U equals q sub V
Variables: q_V = heat exchanged at constant volume.
Use when: Measuring internal energy change directly from heat exchanged at constant volume (bomb calorimeter).
⚠ Students forget that at constant volume, no work is done (ΔV = 0).
pΔV = Δn_g RT
Reads as: p delta V equals delta n g R T
Variables: Δn_g = (number of moles of gaseous products) – (number of moles of gaseous reactants); R = gas constant; T = temperature.
Use when: Calculating the work term (pΔV) for reactions involving gases.
⚠ Students forget to count only gaseous moles, not solids or liquids.
ΔH = ΔU + Δn_g RT
Reads as: Delta H equals delta U plus delta n g R T
Variables: Same as above.
Use when: Converting between ΔH and ΔU for reactions involving gases.
⚠ Students forget to use the correct sign of Δn_g (products minus reactants).
q = C ΔT
Reads as: q equals C delta T
Variables: q = heat supplied (J); C = heat capacity (J/K); ΔT = change in temperature (K or °C).
Use when: Calculating heat from temperature change and heat capacity.
⚠ Students forget that C is for the entire system, not per mole.
q = c × m × ΔT
Reads as: q equals c times m times delta T
Variables: c = specific heat capacity (J g⁻¹ °C⁻¹); m = mass (g); ΔT = temperature change (°C or K).
Use when: Calculating heat using specific heat and mass.
⚠ Students forget to use consistent units (e.g., J vs kJ).
C_p – C_v = R
Reads as: C p minus C v equals R
Variables: C_p = molar heat capacity at constant pressure; C_v = molar heat capacity at constant volume; R = universal gas constant (8.314 J mol⁻¹ K⁻¹).
Use when: Relating molar heat capacities for an ideal gas.
⚠ Students forget this relation is for one mole of an ideal gas.
ΔrH⁰ = Σ a_i ΔfH⁰(products) – Σ b_i ΔfH⁰(reactants)
Reads as: Delta r H standard equals sum of a i times delta f H standard of products minus sum of b i times delta f H standard of reactants
Variables: a_i, b_i = stoichiometric coefficients; ΔfH⁰ = standard enthalpy of formation.
Use when: Calculating standard enthalpy change of a reaction from standard enthalpies of formation.
⚠ Students forget to multiply ΔfH⁰ by the stoichiometric coefficient.
ΔfH⁰ (element in reference state) = 0
Reads as: Delta f H standard of an element in its reference state equals zero
Variables: Reference state = most stable state at 1 bar and specified temperature (usually 298 K).
Use when: Using standard enthalpy of formation data; elements in their standard states have ΔfH⁰ = 0.
⚠ Students think all elements have ΔfH⁰ = 0, but it is only for the most stable allotrope (e.g., graphite, not diamond).
C(graphite) + O₂(g) → CO₂(g); ΔrH⁰ = –393.5 kJ mol⁻¹
Reads as: C graphite plus O2 gas gives CO2 gas; delta r H standard equals negative 393.5 kilojoules per mole
Variables: Standard enthalpy of combustion of graphite.
Use when: Example of a thermochemical equation and Hess's Law application.
⚠ Students forget that this is also the ΔfH⁰ of CO₂(g).
CO(g) + ½O₂(g) → CO₂(g); ΔrH⁰ = –283.0 kJ mol⁻¹
Reads as: CO gas plus half O2 gas gives CO2 gas; delta r H standard equals negative 283.0 kilojoules per mole
Variables: Standard enthalpy of combustion of CO.
Use when: Using Hess's Law to find ΔfH⁰ of CO(g).
⚠ Students forget to reverse the equation when needed.
How It Works
Measuring ΔU using a Bomb Calorimeter (Constant Volume)
  1. Step 1: Place a known mass of the combustible substance in the steel bomb.
  2. Step 2: Fill the bomb with pure oxygen and seal it.
  3. Step 3: Immerse the bomb in a known volume of water in the calorimeter.
  4. Step 4: Ignite the substance electrically; the heat released raises the temperature of the water.
  5. Step 5: Measure the temperature change (ΔT) of the water.
  6. Step 6: Calculate q_V = C_calorimeter × ΔT, where C_calorimeter is the heat capacity of the calorimeter.
  7. Step 7: Since ΔV = 0, ΔU = q_V.
💡 The bomb calorimeter measures heat at constant volume, so the heat measured equals ΔU, not ΔH. For reactions with gases, convert ΔU to ΔH using ΔH = ΔU + Δn_g RT.
Measuring ΔH using a Calorimeter at Constant Pressure
  1. Step 1: Place the reactants in a container open to the atmosphere (constant pressure).
  2. Step 2: Measure the initial temperature of the system.
  3. Step 3: Allow the reaction to occur; heat is absorbed or released.
  4. Step 4: Measure the final temperature of the system.
  5. Step 5: Calculate q_p = C_calorimeter × ΔT (or using specific heat and mass of solution).
  6. Step 6: Since pressure is constant, ΔH = q_p.
💡 This method is simpler but less accurate for reactions involving gases because some work may be done (pΔV). For solids and liquids, ΔH ≈ ΔU.
Calculating Enthalpy Change using Hess's Law
  1. Step 1: Write the target reaction whose ΔrH is unknown.
  2. Step 2: Identify known reactions (with known ΔrH) that involve the same reactants and products.
  3. Step 3: Manipulate the known reactions (reverse, multiply by coefficients) so that when added, they yield the target reaction.
  4. Step 4: When reversing a reaction, change the sign of ΔrH.
  5. Step 5: When multiplying a reaction by a coefficient, multiply ΔrH by the same coefficient.
  6. Step 6: Add the manipulated reactions and their ΔrH values to get the target ΔrH.
💡 Hess's Law works because enthalpy is a state function — the path does not matter. Always ensure the final equation is balanced and all intermediate species cancel.
Calculating ΔrH⁰ from Standard Enthalpies of Formation
  1. Step 1: Write the balanced chemical equation.
  2. Step 2: Look up the standard enthalpy of formation (ΔfH⁰) for each reactant and product from a data table.
  3. Step 3: Multiply each ΔfH⁰ by its stoichiometric coefficient in the balanced equation.
  4. Step 4: Sum the ΔfH⁰ values for products.
  5. Step 5: Sum the ΔfH⁰ values for reactants.
  6. Step 6: Subtract the reactant sum from the product sum: ΔrH⁰ = Σ ΔfH⁰(products) – Σ ΔfH⁰(reactants).
💡 ΔfH⁰ of any element in its standard state is zero. Do not forget to multiply by coefficients.
Calculating Work in Isothermal Reversible Expansion of an Ideal Gas
  1. Step 1: Identify n (moles), T (temperature in Kelvin), Vi (initial volume), Vf (final volume).
  2. Step 2: Use the formula w_rev = –2.303 nRT log(Vf/Vi).
  3. Step 3: Use R = 8.314 J mol⁻¹ K⁻¹ for work in Joules, or R = 0.08206 L atm mol⁻¹ K⁻¹ for work in L atm.
  4. Step 4: Calculate the numerical value; remember the negative sign indicates work done by the system (expansion).
  5. Step 5: For isothermal process, ΔU = 0, so q = –w.
💡 The log is base 10. Use consistent units. For compression (Vf < Vi), w is positive (work done on system).
Exam Traps — Don't Make These Mistakes
⚠ Sign Convention for Work (w)
✗ w = –pΔV, and they write w = –pΔV = –(positive) = negative for compression.
→ For compression, ΔV is negative (Vf < Vi). So w = –pex(negative) = positive. Students forget that ΔV itself carries a sign.
✓ For compression: ΔV = Vf – Vi is negative, so w = –pex(negative) = positive (work done on system). For expansion: ΔV is positive, so w = –pex(positive) = negative (work done by system).
⚠ Confusing ΔH and ΔU
✗ ΔH = ΔU for all reactions.
→ ΔH = ΔU only when no gases are involved or when Δn_g = 0. For reactions with gases, ΔH = ΔU + Δn_g RT.
✓ ΔH = ΔU + Δn_g RT. For solids and liquids, pΔV is negligible, so ΔH ≈ ΔU. For gases, calculate Δn_g carefully.
⚠ Standard Enthalpy of Formation of Elements
✗ ΔfH⁰ of O₂(g) is zero, and ΔfH⁰ of O₃(g) is also zero.
→ ΔfH⁰ = 0 only for the most stable allotropic form of an element at 1 bar and given temperature. O₂(g) is the reference state for oxygen; O₃(g) is not the most stable form, so its ΔfH⁰ is not zero.
✓ ΔfH⁰ of an element in its reference state (most stable form) is zero. For oxygen, it is O₂(g); for carbon, it is C(graphite); for sulfur, it is S(rhombic).
⚠ Hess's Law: Reversing Equations
✗ When reversing a reaction, they keep the same sign of ΔH.
→ Reversing a reaction changes the direction of energy flow. If forward is exothermic (negative ΔH), reverse is endothermic (positive ΔH).
✓ When reversing a chemical equation, change the sign of ΔH. For example, if A → B has ΔH = –x, then B → A has ΔH = +x.
⚠ Heat Capacity vs Specific Heat
✗ Using C (heat capacity) when the problem gives specific heat (c) and mass (m).
→ Heat capacity (C) is for the entire system (units J/K). Specific heat (c) is per unit mass (J/g°C). Using C = c × m is correct, but students often plug c directly into q = CΔT without multiplying by mass.
✓ If given specific heat (c) and mass (m), use q = c × m × ΔT. If given heat capacity (C) of the calorimeter, use q = C × ΔT.
⚠ Isothermal Process for Ideal Gas: ΔU = 0
✗ For an isothermal process, ΔU = 0, so q = –w. But they forget that this is true only for an ideal gas.
→ For an ideal gas, internal energy depends only on temperature. At constant temperature, ΔU = 0. For real gases or other systems, ΔU may not be zero even if temperature is constant.
✓ For an ideal gas undergoing an isothermal process, ΔU = 0, so q = –w. For non-ideal gases or phase changes, this may not hold.
Compare & Contrast
Exchange of matter with surroundings
Open System
Closed System
An open system allows exchange of both energy and matter (e.g., open beaker), while a closed system allows exchange of energy but not matter (e.g., sealed copper vessel).
Heat exchange with surroundings
Isothermal Process
Adiabatic Process
In an isothermal process, temperature remains constant and heat is exchanged (q ≠ 0, ΔU = 0 for ideal gas). In an adiabatic process, no heat is exchanged (q = 0), so any change in internal energy is due to work alone (ΔU = w).
How the process is carried out
Reversible Process
Irreversible Process
A reversible process occurs infinitely slowly through a series of equilibrium states, with the system and surroundings always nearly in equilibrium. An irreversible process occurs in a single step or finite steps, often with a finite pressure difference, and is not in equilibrium during the process.
Definition and measurement conditions
Enthalpy (H)
Internal Energy (U)
H = U + pV. ΔH is the heat exchanged at constant pressure (qp), while ΔU is the heat exchanged at constant volume (qV). For reactions involving gases, ΔH and ΔU differ by Δn_g RT.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 11-12 marks
Most asked: First Law of Thermodynamics and its mathematical expression (ΔU = q + w) · Calculation of work done in isothermal reversible and irreversible processes · Enthalpy change (ΔH) and its relation to ΔU (ΔH = ΔU + Δn_g RT) · Hess's Law of Constant Heat Summation · Standard enthalpy of formation and calculation of ΔrH⁰ · Distinction between extensive and intensive properties · Sign conventions for heat and work
Don't skip: Numerical problems on calculating w, q, ΔU, and ΔH for isothermal reversible/irreversible processes; Application of Hess's Law to find unknown enthalpy changes; Calculating ΔrH⁰ from standard enthalpies of formation; Deriving the relation Cp – Cv = R
One-Line Revision (read aloud before the exam)
  1. Thermodynamics deals with energy changes in macroscopic systems and is based on initial and final equilibrium states.
  2. A system is the part of the universe under study; surroundings are everything else; together they form the universe.
  3. Open systems exchange both energy and matter, closed systems exchange only energy, and isolated systems exchange neither.
  4. State functions (U, H, p, V, T) depend only on the current state, not on the path taken.
  5. Internal energy (U) is the total energy of a system; its absolute value cannot be measured, only changes (ΔU) can.
  6. The First Law of Thermodynamics states ΔU = q + w, meaning energy is conserved.
  7. Work done in a reversible isothermal expansion of an ideal gas is w = –2.303 nRT log(Vf/Vi).
  8. Enthalpy H = U + pV; at constant pressure, ΔH = qp.
  9. For reactions involving gases, ΔH = ΔU + ΔngRT, where Δng = moles of gaseous products – moles of gaseous reactants.
  10. Heat capacity C = q/ΔT; molar heat capacity Cm = C/n; specific heat c = q/(mΔT).
  11. For an ideal gas, Cp – Cv = R.
  12. Extensive properties (mass, volume, U, H) depend on amount of matter; intensive properties (T, p, density) do not.
  13. Standard enthalpy of reaction (ΔrH⁰) is the enthalpy change when all substances are in their standard states (1 bar, usually 298 K).
  14. Standard enthalpy of formation (ΔfH⁰) is the enthalpy change when 1 mole of a compound is formed from its elements in their reference states.
  15. ΔfH⁰ of an element in its most stable state (e.g., C graphite, O2 gas) is zero by convention.
  16. Hess's Law: The enthalpy change for a reaction is the same whether it occurs in one step or several steps.
  17. A bomb calorimeter measures ΔU (constant volume); a simple calorimeter measures ΔH (constant pressure).
  18. Reversible processes occur infinitely slowly through equilibrium states; irreversible processes occur in finite steps.
  19. Free expansion (into vacuum) does no work (w = 0) for an ideal gas.
  20. In an isothermal process for an ideal gas, ΔU = 0, so q = –w.

MCQs — 10 Questions with Answers

Q1. A chemical reaction occurs in an open beaker on a laboratory bench. Which statement correctly describes this system?

  • A. The system is closed; energy can be exchanged but not matter.
  • B. The system is open; both matter and energy can be exchanged with the surroundings. ✓
  • C. The system is isolated; no exchange of matter or energy occurs.
  • D. The system is closed because the beaker provides a physical boundary.

Answer: B — An open beaker allows vapours to escape (matter loss) and heat transfer to the environment (energy loss), making it an open system.

Q2. Internal energy is a state function. This means that:

  • A. Its value depends on how quickly the reaction proceeds.
  • B. Its value depends only on the initial and final states, regardless of the path taken. ✓
  • C. It always increases when a chemical reaction occurs.
  • D. It is the same for all systems at the same temperature.

Answer: B — State functions are path-independent; only initial and final states determine ΔU, not the reaction route or speed.

Q3. Which of the following is NOT a state variable used to describe the state of a thermodynamic system?

  • A. Pressure and temperature
  • B. Volume and amount of substance (n)
  • C. The rate at which the system changes ✓
  • D. Composition of the system

Answer: C — Rate of change is a kinetic property, not a thermodynamic state variable; thermodynamics only uses p, V, T, n, and composition.

Q4. A thermos flask contains hot water sealed inside. Which type of system is this?

  • A. Open system
  • B. Closed system
  • C. Isolated system ✓
  • D. Equilibrium system

Answer: C — A sealed, insulated thermos flask prevents both matter and energy exchange, defining an isolated system.

Q5. In a closed steel vessel, a fuel burns and releases 500 kJ of heat. How does this affect the internal energy of the fuel?

  • A. ΔU increases by 500 kJ because heat is released.
  • B. ΔU decreases by 500 kJ because energy leaves the system.
  • C. ΔU remains unchanged; heat release does not affect internal energy.
  • D. ΔU depends on both heat and work done; without work information, it cannot be determined. ✓

Answer: D — ΔU = q + w; heat alone (q = -500 kJ) is insufficient; work done must be known to calculate total ΔU.

Q6. Using the first law of thermodynamics (ΔU = q + w), if a gas expands against external pressure and absorbs 200 J of heat while doing 150 J of work, what is the change in internal energy?

  • A. ΔU = +350 J
  • B. ΔU = +50 J ✓
  • C. ΔU = +200 J
  • D. ΔU = -150 J

Answer: B — ΔU = q + w = (+200 J) + (-150 J) = +50 J; work done BY the gas is negative in sign convention.

Q7. Which assertion correctly describes the boundary of a thermodynamic system?

  • A. The boundary must always be a physical, tangible wall.
  • B. The boundary can be either real or imaginary and is used to track energy and matter movement. ✓
  • C. The boundary is always defined by the container's physical limits.
  • D. The boundary must be made of insulating material for all systems.

Answer: B — Boundaries are conceptual tools—real (like beaker walls) or imaginary (Cartesian coordinates)—to monitor what enters/leaves the system.

Q8. A student studies a reaction by placing reactants in a closed copper vessel. Which processes can occur? (I) Heat transfer between system and surroundings (II) Exchange of matter between system and surroundings (III) Change in internal energy of the system

  • A. Only I
  • B. Only I and III ✓
  • C. Only II and III
  • D. I, II, and III

Answer: B — Closed systems allow energy/heat transfer (I) and internal energy change (III), but NOT matter exchange; conductors like copper enable heat transfer.

Q9. The universe in thermodynamic terms is defined as. (State whether the following assertion is correct.) Assertion: The universe consists only of the system and the immediate surroundings. Reason: In practical thermodynamics, we ignore the rest of the universe because it does not directly interact with the system.

  • A. Both assertion and reason are correct; reason explains assertion. ✓
  • B. Both are correct but reason does not explain assertion.
  • C. Assertion is correct; reason is incorrect.
  • D. Assertion is incorrect; reason is irrelevant.

Answer: A — Universe = System + Surroundings is the thermodynamic definition; the reason correctly justifies why distant universe portions are ignored in practice.

Q10. HOTS: A student is given two scenarios: (Scenario 1) Methane burns in an open flame producing heat and light. (Scenario 2) Methane burns inside a sealed engine cylinder, doing mechanical work. In which scenario is the system's internal energy change more directly related to heat transfer alone, and why?

  • A. Scenario 1; because the open system allows all energy to escape as heat.
  • B. Scenario 1; because no work is done, only heat is released.
  • C. Scenario 2; because the closed system prevents heat loss.
  • D. Neither; ΔU always depends on both q and w according to the first law. ✓

Answer: D — First law ΔU = q + w applies universally; Scenario 1 involves minimal work (heat-dominated), Scenario 2 maximizes work (ΔU affected significantly by both), but both must consider both terms.

Flashcards

What is a thermodynamic system?

The portion of the universe we observe for energy changes, separated from surroundings by a real or imaginary boundary.

Distinguish between closed and isolated systems.

Closed system: energy exchange allowed but no matter exchange; isolated system: neither energy nor matter exchange with surroundings.

Define internal energy (U) of a system.

The sum of all forms of energy (chemical, mechanical, electrical, thermal) contained within the system.

What does 'state function' mean in thermodynamics?

A property whose value depends only on the current state of the system, not on the path taken to reach that state.

How does work change internal energy?

When work is done on a system, internal energy increases; when work is done by the system, internal energy decreases.

What is the relationship between heat and work?

Both heat and work are mechanisms by which energy enters or leaves a system and changes its internal energy.

Why is the state of surroundings never completely specified?

Because surroundings are so large and complex that tracking all their properties is impractical; only changes affecting the system matter.

What determines the state of a thermodynamic system?

Macroscopic properties like pressure (p), volume (V), temperature (T), and composition; once minimum properties are fixed, others are automatically determined.

Give one example of an open system.

A reaction mixture in an open beaker where both matter (vapours) and energy (heat) can exchange with the room.

How is the universe defined in thermodynamics?

Universe = System + Surroundings; represents the total closed entity in which all thermodynamic processes occur.

Important Board Questions

Define an isolated system and give one practical example. Why is a thermos flask considered an isolated system? [2 marks]

State definition: no exchange of matter or energy. Example: sealed insulated container. Reason: walls prevent both heat transfer and mass loss.

Explain why internal energy (U) is called a state function. Show how ΔU differs from path-dependent properties and calculate ΔU for a system that absorbs 300 J of heat and has 200 J of work done on it. [5 marks]

State function depends only on initial & final states (p, V, T), not path. Path-dependent examples: q, w are separate. Apply ΔU = q + w = 300 J + 200 J = +500 J; show all sign conventions.

Classify the following into open, closed, or isolated systems with brief justification: (i) A sealed glass bottle containing hot water, (ii) A reaction in an open beaker in the lab, (iii) The Earth's atmosphere. For each, explain what exchanges are allowed or prevented and how this affects energy tracking in thermodynamics. [6 marks]

Sealed glass bottle: closed (energy yes, matter no). Open beaker: open (both exchanges). Atmosphere: isolated on large scale. Justify using definitions of matter/energy exchange; link to ΔU = q + w applicability and why system type determines which terms matter in calculations.

Assertion-Reason Questions for Thermodynamics

Q1. Assertion (A):

Internal energy (U) is a state function of a thermodynamic system.

Reason (R):

The value of internal energy depends only on the initial and final states of the system, not on the path followed to reach those states.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why U is a state function. The path-independence of state functions is the defining characteristic.

Q2. Assertion (A):

In a closed system, the first law of thermodynamics can be written as ΔU = q, provided the system undergoes changes at constant volume.

Reason (R):

At constant volume, no PV work is done by or on the system, so the work term (w) becomes zero.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true and logically connected. When ΔV = 0, the expansion work w = −PΔV = 0, making ΔU = q + 0 = q. This is a correct application of the first law.

Q3. Assertion (A):

A thermos flask containing a chemical reaction represents an isolated system.

Reason (R):

The reaction mixture in the flask cannot exchange matter with surroundings because the flask is sealed, though heat exchange may occur through the flask walls.

Answer: C — A is true but R is false — The assertion is true: a well-insulated thermos flask approximates an isolated system. However, the reason is incomplete/false: an isolated system must prevent both matter AND energy exchange. A regular thermos allows some heat transfer, making it more accurately a closed system with reduced energy exchange.

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Case-Based Questions (CBQ) for Thermodynamics

Case Study:

A chemistry student in Delhi observes that during the winter season, a hot cup of tea cools down gradually when left on a table. The cup is made of ceramic material with a wall thickness of 0.5 cm. Initially, the tea is at 75°C, and the room temperature is 15°C. After 30 minutes, the tea temperature drops to 35°C. The student wants to understand the thermodynamic nature of this cooling process and the energy changes involved. She measures that the cup contains 250 mL of tea with an approximate density and specific heat capacity similar to water (4.18 J/g·°C).

Q1. What type of thermodynamic system does the open cup represent?

  • A. An isolated system because the cup is stationary
  • B. A closed system because the cup is made of ceramic
  • C. An open system because mass (water vapor) and heat can exchange with the room air ✓
  • D. A closed system because the tea is a liquid enclosed within the cup

Answer: C

Q2. Calculate the heat released by the tea as it cools. (Assume density of tea ≈ 1 g/mL)

  • A. −2.61 × 10⁴ J
  • B. −2.09 × 10⁴ J ✓
  • C. −5.22 × 10⁴ J
  • D. −1.05 × 10⁴ J

Answer: B

Q3. If the ceramic cup itself has a mass of 150 g and undergoes the same temperature change as the tea (specific heat of ceramic ≈ 0.84 J/g·°C), what is the total heat lost by the tea and cup combined?

  • A. −2.31 × 10⁴ J
  • B. −2.71 × 10⁴ J
  • C. −3.11 × 10⁴ J ✓
  • D. −3.51 × 10⁴ J

Answer: C

Q4. In this cooling process, which statement about internal energy change (ΔU) is most accurate?

  • A. ΔU equals the heat released because volume change is negligible and work done is minimal ✓
  • B. ΔU is greater than the heat released because cooling does work on the system
  • C. ΔU cannot be determined without knowing the exact atmospheric pressure
  • D. ΔU is zero because the system reaches a lower temperature

Answer: A

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Class 11 Chemistry Formula Sheet

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