Midterm 1 : 1st Law of Thermodynamics

Distinguish between intensive and extensive properties. Distinguish between open, isolated and closed systems.

Thermodynamic vocabulary

• System: the portion of the universe under study. Surroundings: everything else. Together they make the universe.

• The boundary (real or imaginary) determines what can cross between system and surroundings.

Intensive vs. extensive properties

• Extensive property: depends on the amount of matter (extent) present. If you double the sample, the value doubles.

  • Examples: mass ($m$), volume ($V$), moles ($n$), energy ($E$), enthalpy ($H$), entropy ($S$), Gibbs energy ($G$), heat capacity ($C$), length, weight.

• Intensive property: independent of the amount present. Mnemonic: INTensive = INDependent.

  • Examples: temperature ($T$), pressure ($P$), density ($\rho = m/V$), voltage ($V$ = J/C), color, hardness, luster, odor, melting/boiling/freezing point, conductivity, malleability, ductility, specific heat ($c_s$), molar heat capacity ($c_{molar}$).

• Shortcut rule: the ratio of two extensive properties is intensive. $\rho = m/V$, $c_s = C/m$, and $c_{molar} = C/n$ all follow this rule. Doubling the sample doubles both numerator and denominator, so the ratio is unchanged.

• Molecular meaning:

  • Temperature measures average kinetic energy per particle, so it does not scale with $N$.

  • Total energy sums over all particles, so it scales with $N$.

• Common pitfalls:

  • Weight vs. mass: both are extensive (weight depends on amount and gravity).

  • Concentration is intensive, but amount of solute is extensive. This matters for heterogeneous equilibria later: the “concentration” of a pure solid is constant regardless of how much solid is present.

  • Thermodynamic functions of the form $\Delta H$, $\Delta E$ and $\Delta G$ for a reaction are extensive: doubling the balanced equation doubles $\Delta H$. This is why $2\,\text{C}_2\text{H}_2 + 5\,\text{O}_2$ gives twice the $\Delta H$ of $\text{C}_2\text{H}_2 + \tfrac52\text{O}_2$.

Classifying systems by what crosses the boundary

System	Matter exchange	Energy exchange (heat/work)	Example
Open	Yes	Yes	Living plant, gasoline burning in a car engine, open beaker of boiling water
Closed	No	Yes	Coolant in a refrigerator coil, mercury in a thermometer, sealed flask in a water bath
Isolated	No	No	Ideal thermos of coffee, the universe

• Pitfall: “closed” does not mean fully sealed from energy. It is closed to matter only. The fully sealed case is isolated.

• The bomb calorimeter is treated as isolated: the sample burns inside a rigid sealed steel bomb (constant $V$, no matter exchange), and the insulated water jacket is assumed to prevent heat leaving. Strictly, heat flows from the bomb to the water inside the calorimeter, which is why we measure it.

• Related process labels (a constraint on a process, not on the system type):

  • Isothermal: $\Delta T = 0$

  • Isobaric: $\Delta P = 0$

  • Isochoric: $\Delta V = 0$

  • Adiabatic: $q = 0$

Worked classification (from your slide exercise)

1. Coffee in a very-high-quality thermos → isolated (no matter or energy exchange, ideal case).

2. Coolant in refrigerator coil → closed (sealed fluid, but it absorbs and releases heat and is worked on by the compressor).

3. Benzene burned in bomb calorimeter → isolated (sealed, insulated).

4. Gasoline burning in an automobile engine → open (fuel and air in, exhaust out, heat and work out).

5. Mercury in thermometer → closed (mercury cannot escape, but it exchanges heat).

6. A living plant → open (takes in CO₂, H₂O, light; releases O₂, water vapor).

Applying $E = \tfrac32 RT$ to isothermal changes.

Where the equation comes from

• For 1 mole of an ideal monatomic gas (He, Ne, Ar, Kr), the only energy is translational kinetic energy. There are 3 translational degrees of freedom ($x, y, z$), each carrying $\tfrac12 kT$ per atom ($\tfrac12 RT$ per mole). This is the equipartition theorem.

  • $E = 3 \times \tfrac12 RT = \tfrac32 RT$ (J/mol), where $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ and $T$ is in kelvin.

• Key assumptions (the ideal gas model):

  • Particles have no intermolecular attractions or repulsions, so there is no potential energy contribution.

  • Particles are point masses, so there is no rotation or vibration (true for monatomic gases at ordinary temperatures).

  • Electronic excitation is negligible.

Consequence: $E$ depends only on temperature

• $\Delta E = \tfrac32 R\,\Delta T$ (per mole), or $\Delta E = \tfrac32 nR\,\Delta T$ for $n$ moles. The $\tfrac32$ and $R$ are constants, so they factor out of the $\Delta$.

• Isothermal process: $\Delta T = 0$, therefore $\Delta E = 0$.

  • Molecular picture: if the average kinetic energy per particle does not change and there are no interparticle forces, the total energy cannot change. Compression or expansion changes how much space the particles have, not how fast they move.

• Since $E$ is a state function, $\Delta E = 0$ for any path between two states at the same $T$, whether sudden or gradual, one step or many.

Pitfalls and edge cases

• $\Delta E = 0$ for an isothermal process is exact only for ideal gases. For real gases, liquids and solids, $E$ also depends on volume (through intermolecular forces).

• Phase changes are isothermal but $\Delta E \neq 0$. Melting or boiling absorbs heat at constant $T$ because that energy goes into potential energy (breaking intermolecular attractions), not kinetic energy.

• $\Delta E = 0$ does not mean $q = 0$ or $w = 0$. Both can be nonzero, but they must cancel (see Card 4).

• For polyatomic ideal gases, $E$ is larger ($\tfrac52RT$ or $3RT$, see Card 12), but it still depends only on $T$, so isothermal $\Delta E = 0$ still holds.

Worked example
Q: How much does the internal energy change when 2.00 mol of He is heated from 300 K to 400 K? What if the same gas is compressed isothermally at 300 K?

1. Heating: $\Delta E = \tfrac32 nR\,\Delta T = \tfrac32 (2.00\ \text{mol})(8.314\ \text{J K}^{-1}\text{mol}^{-1})(100\ \text{K})$

2. Units: mol × J K⁻¹ mol⁻¹ × K = J

3. $\Delta E = 2494\ \text{J} \approx$ +2.49 kJ

4. Isothermal compression: $\Delta T = 0 \Rightarrow$ $\Delta E = 0$

First law of thermodynamics (in words and in equation form).

Statement in words

• Energy cannot be created or destroyed; it can only be converted from one form to another or transferred. The energy of the universe is constant.

• For a closed system, the change in internal energy equals the heat added to it plus the work done on it.

Equation

• $\Delta E = q + w$

  • $\Delta E$ = change in internal energy of the system (J), the total of all kinetic and potential energy of the particles. It is a state function.

  • $q$ = heat transferred across the boundary (J), driven by a temperature difference. It is a path function.

  • $w$ = work done on or by the system (J). It is a path function.

• Sign convention (system’s point of view): $+$ means energy enters the system, $-$ means energy leaves (see Card 4).

Conceptual background

• Heat is not a substance (the old “caloric” theory is wrong). It is a mechanism of energy transfer driven by a temperature gradient, flowing hot → cold until thermal equilibrium.

• Joule’s experiment (1840s): a falling weight turned a paddle wheel in water, and the water warmed without contact with any hotter body. This showed mechanical work converts to heat. 1 cal = 4.184 J (mechanical equivalent of heat).

• Analogy to other “potential” differences:

  • Electrical current flows with a difference in electrical potential.

  • Heat flows with a difference in temperature.

  • Chemical reactions proceed with a difference in chemical potential ($\Delta G$).

Why $E$ is a state function but $q$ and $w$ are not

• State function: depends only on the current state (P, V, T, n), not on how it was reached. Examples: $E, H, S, G, P, V, T$.

• Path function: depends on the route. Only the sum $q + w$ is path-independent.

• Analogy: the elevation difference between two towns is fixed, but the distance walked depends on the road.

Pitfalls

• Writing $\Delta q$ or $\Delta w$: these are not changes in state functions, so write $q$ and $w$.

• Using the opposite sign convention. Some textbooks define $w$ as work done by the system, giving $\Delta E = q - w$. Your course uses $\Delta E = q + w$.

• The first law says nothing about direction or spontaneity. That requires the second law.

Worked example
*Q: A system’s internal energy increases by 80 J while 50 J of work is done on it. What is $q$?*

1. $$\Delta E = q + w \Rightarrow 80\ \text{J} = q + 50\ \text{J}$$

2. $q = 80 - 50 =$ +30 J

3. The positive sign means 30 J of heat flowed into the system.

Application of the first law to isothermal conditions and sign convention for heat and work.

Sign convention (system’s perspective)

• $q > 0$: heat flows into the system (endothermic).

• $q < 0$: heat flows out of the system (exothermic).

• $w > 0$: work done on the system (e.g., compression, $\Delta V < 0$).

• $w < 0$: work done by the system on the surroundings (e.g., expansion, $\Delta V > 0$).

• Mnemonic: think of a bank account. Whatever enters the system is $+$.

Isothermal ideal-gas process

1. For an ideal gas, $E = \tfrac32RT$ (monatomic), so $\Delta E = \tfrac32R\,\Delta T$.

2. Isothermal: $\Delta T = 0 \Rightarrow \Delta E = 0$.

3. First law: $\Delta E = q + w \Rightarrow 0 = q + w$.

4. Therefore $q = -w$.

Interpretation in words

• Heat absorbed by the system equals work done by the system on the surroundings. (And vice versa for compression.)

• Expansion ($w<0$): the gas does work and would cool, but heat flows in from the surroundings ($q>0$) to keep $T$ constant.

• Compression ($w>0$): work is put in, and the same amount of energy leaves as heat ($q<0$) so $T$ stays constant.

• Molecular picture: during expansion, molecules colliding with a receding piston lose kinetic energy. The thermal reservoir resupplies it.

Pitfalls

• $q = -w$ holds only when $\Delta E = 0$ (isothermal ideal gas). In general, $\Delta E \neq 0$.

• Isothermal does not mean no heat flow. Isothermal means constant $T$. No heat flow is adiabatic.

• $q$ and $w$ depend on the path (reversible vs. irreversible) even though $q = -w$ always holds on each path.

Worked example (from your slides): bicycle pump
*Q: Air is compressed in a pump of inner diameter 3.0 cm, depressed 20. cm against 2.00 atm. Find $w$ and its sign.*

1. $w = -P_{ext}\Delta V$, and $\Delta V = -(\pi r^2 \cdot \ell)$ (volume decreases).

2. $r = 1.5\ \text{cm}$, so $V = \pi(1.5)^2(20.)\ \text{cm}^3 = 141\ \text{cm}^3 = 0.141\ \text{L}$ (using $1\ \text{L} = 1000\ \text{cm}^3$).

3. $w = -(2.00\ \text{atm})(-0.141\ \text{L}) =$ +0.28 atm·L

4. In joules: $0.283\ \text{atm·L} \times 101.3\ \text{J/(atm·L)} \approx$ +28.7 J

5. Positive: work is done on the air.

Derivation of work for irreversible and reversible expansions.

Step 1: general expression for pressure-volume work

• Work = force × distance. For a gas pushing a piston of area $A$ against an external force $F_{ext}$ through a displacement $dz$:

  • $dw = -F_{ext}\,dz$ (minus sign: the system expands against a force, so it loses energy).

• Pressure = force/area, so $F_{ext} = P_{ext}A$:

  • $$dw = -P_{ext}A\,dz$$

• Area × distance = volume: $dV = A\,dz$

  • $dw = -P_{ext}\,dV$

• Integrate:

  • $w = -\displaystyle\int_{V_i}^{V_f} P_{ext}\,dV$ (i)

• Note: the work depends on $P_{ext}$, the pressure the surroundings exert. This is not necessarily the pressure of the gas.

Step 2: irreversible (sudden) expansion at constant $P_{ext}$

• $P_{ext}$ is constant, so it comes out of the integral:

  • $w_{irr} = -P_{ext}(V_f - V_i) = -P_{ext}\,\Delta V$

• Variables: $P_{ext}$ (Pa, atm or bar), $\Delta V$ (m³ or L).

• Units: Pa·m³ = J. With atm and L, 1 L·atm = 101.325 J (equivalently $8.314/0.08206 = 101.3$).

• This is “irreversible” because the system is not in equilibrium during the process (gas pressure > $P_{ext}$ throughout expansion).

Step 3: reversible (gradual) isothermal expansion of an ideal gas

• Reversible means infinitely slow change in which the system is always in equilibrium with its surroundings: $P_{ext} = P_{sys}$ at every instant.

• Substitute into (i) and use $PV = nRT$, so $P = nRT/V$:

  • $$w = -\displaystyle\int_{V_i}^{V_f}\frac{nRT}{V}\,dV$$

• $n$, $R$ and $T$ are constant (isothermal), so pull them out:

  • $$w = -nRT\displaystyle\int_{V_i}^{V_f}\frac{1}{V}\,dV$$

• Use $\int \frac1x dx = \ln x$:

  • $$w = -nRT\,[\ln V]_{V_i}^{V_f} = -nRT(\ln V_f - \ln V_i)$$

• Use $\ln A - \ln B = \ln(A/B)$:

  • $w_{rev} = -nRT\ln\dfrac{V_f}{V_i}$

• Since $P_iV_i = P_fV_f$ for an isothermal ideal gas, this can also be written $w_{rev} = -nRT\ln\dfrac{P_i}{P_f}$.

Corresponding heat

• For an isothermal ideal gas, $q = -w$, so $q_{rev} = nRT\ln\dfrac{V_f}{V_i}$ and $q_{irr} = +P_{ext}\Delta V$. Heat is also path-dependent.

• Dividing by $T$ gives $\Delta S = q_{rev}/T = nR\ln(V_f/V_i)$, the macroscopic entropy definition, which is a state function.

Assumptions and pitfalls

• The reversible formula requires an ideal gas and constant $T$.

• Reversibility is a theoretical limit. No real expansion is truly reversible, but it defines the maximum work.

• Sign check: expansion has $V_f > V_i$, so $\ln(V_f/V_i) > 0$ and $w < 0$. Compression gives $w > 0$.

• Do not use $w = -P\Delta V$ with the gas pressure unless the process is irreversible.

Worked example (from your slides)
*Q: 1.00 mol ideal gas at 298 K expands isothermally from 1.00 m³ to 11.0 m³. Find $w$ for (a) one-step irreversible expansion against the final pressure and (b) reversible expansion.*

1. $$nRT = (1.00)(8.314)(298) = 2478\ \text{J}$$

2. (a) Final pressure $P_f = nRT/V_f = 2478/11.0 = 225\ \text{Pa}$ (1 Pa = 1 J/m³)

   • $w_{irr} = -P_{ext}\Delta V = -(225\ \text{J/m}^3)(10.0\ \text{m}^3) =$ −2.25 × 10³ J

3. (b) $w_{rev} = -(2478\ \text{J})\ln(11.0/1.00) = -(2478)(2.398) =$ −5.94 × 10³ J (your slide: −5941 J)

4. The reversible path extracts more than twice the work.

⚠ Slide correction: the 5-step expansion total on your slide reads −4325 J with steps “−1625, −982…”. Recomputing each step ($2\ \text{m}^3 \times P$) gives −1652, −991, −708, −551, −450 J, totaling ≈ −4352 J. The slide has transposed digits. The trend (more steps → closer to −5941 J) is unaffected.

Work done under reversible conditions is greater than when done irreversibly.

Statement

• For a given initial and final state, the maximum work that a system can do on its surroundings is obtained when the process is carried out reversibly.

  • Expansion: $|w_{rev}| > |w_{irr}|$ (the reversible path gives the most negative $w$).

  • Compression: the reversible path requires the least work input ($w_{rev}$ is the smallest positive value).

• Your slide’s shorthand “$w_{rev} > w_{irr}$” refers to the magnitude of work done by the system.

Why: the physical argument

• Work of expansion = $-\int P_{ext}\,dV$. Magnitude = area under the $P_{ext}$ vs. $V$ curve.

• During expansion, $P_{ext} \leq P_{sys}$ (otherwise the gas wouldn’t expand).

• The reversible process sets $P_{ext}$ as large as possible at every step, equal to $P_{sys}$. Any irreversible path has $P_{ext}$ lower somewhere, so it traces less area.

• Graphical picture (your slides): with 1 step, the rectangle is small. As steps increase (5 → 40 → ∞), the rectangles approach the curve and the area approaches the maximum.

Numerical illustration (1 mol, 298 K, 1 → 11 m³)

Path	Work done by gas
1 sudden step	≈ 2250 J
5 steps	≈ 4350 J
40 steps	≈ 5672 J
Infinite steps (reversible)	5941 J (maximum)

Connection to entropy and the second law

• $\Delta S$ is a state function, equal to $q_{rev}/T$ for any path.

• Because $|q| = |w|$ at constant $T$ and $|w_{irr}| < |w_{rev}|$, we get $|q_{irr}| < |q_{rev}|$, so:

  • $\Delta S > \dfrac{q_{irr}}{T}$ for an irreversible process.

• Interpretation: the “missing” work is dissipated as increased disorder in the universe, which is why $\Delta S_{univ} > 0$ for irreversible processes.

Pitfalls

• A reversible process is not one that merely “can go backward.” It is one that proceeds through a continuous sequence of equilibrium states, infinitely slowly.

• Phase changes at the transition temperature (melting ice at 0 °C, 1 atm) are naturally reversible equilibrium processes. Gas expansions are not.

• $\Delta E$, $\Delta H$ and $\Delta S$ are the same for both paths. Only $q$ and $w$ differ.

Worked example
*Q: 1.00 mol ideal gas at 298 K expands from 1.00 L to 10.0 L. Compare $|w|$ for a single step against $P_{ext}$ = final pressure vs. reversible.*

1. $$nRT = 2478\ \text{J}$$

2. $$P_f = nRT/V_f = 2478\ \text{J}/(10.0\ \text{L}) = 247.8\ \text{J/L}$$

3. $$|w_{irr}| = P_f\Delta V = (247.8\ \text{J/L})(9.00\ \text{L}) = 2230\ \text{J}$$

4. $$|w_{rev}| = nRT\ln(10.0) = (2478)(2.303) = 5706\ \text{J}$$

5. Ratio ≈ 2.6: the reversible path does about 2.6× more work.

Calculate the work done by a gas due to expansion (not necessarily isothermal). Calculate the change in internal energy due to heat and work.

Choosing the right work formula

• Constant external pressure (irreversible): $w = -P_{ext}\Delta V$

• Reversible isothermal ideal gas: $w = -nRT\ln(V_f/V_i)$

• Free expansion into vacuum ($P_{ext} = 0$): $w = 0$

• Reaction producing or consuming gas at constant $P$ and $T$: using $P\Delta V = \Delta n_{gas}RT$:

  • $w = -\Delta n_{gas}\,RT$

  • $\Delta n_{gas}$ = (moles gas products) − (moles gas reactants). Solids and liquids contribute negligible volume.

Calculating $\Delta E$

• $\Delta E = q + w$, with each term in joules.

• Heat from an electric heater: $q = P_{power}\times t$, where 1 W = 1 J/s.

Unit management (the most common source of error)

• $1\ \text{L·atm} = 101.325\ \text{J}$; also $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} = 0.08206\ \text{L atm K}^{-1}\text{mol}^{-1}$.

• Convert kJ ↔ J before adding. Temperatures in K for $RT$ terms. $1\ \text{mL} = 10^{-3}\ \text{L}$.

• Sign check: expansion → $w$ negative; compression → $w$ positive.

Worked example 1: heater and expansion (your slide)
*Q: A 100. W heater runs for 10.0 min heating a gas that expands from 2.00 L to 10.0 L against a constant 0.975 atm. Find $\Delta E$.*

1. $$q = (100.\ \text{J/s})(10.0 \times 60\ \text{s}) = 6.00 \times 10^4\ \text{J} = +60.0\ \text{kJ}$$

2. $$\Delta V = 10.0 - 2.00 = 8.00\ \text{L}$$

3. $$w = -P_{ext}\Delta V = -(0.975\ \text{atm})(8.00\ \text{L})(101.33\ \text{J/L·atm}) = -790\ \text{J} = -0.790\ \text{kJ}$$

4. $\Delta E = q + w = 60.0 - 0.790 =$ +59.2 kJ

Worked example 2: simple expansion
*Q: A gas expands by 500. mL against 1.20 atm. Find $w$ in J.*

1. $$w = -(1.20\ \text{atm})(0.500\ \text{L}) = -0.600\ \text{L·atm}$$

2. $-0.600\ \text{L·atm} \times 101.3\ \text{J/(L·atm)} =$ −60.8 J

Worked example 3: work from a chemical reaction
*Q: Ni(s) + 4 CO(g) → Ni(CO)₄(g) for 1.00 mol Ni at 75 °C. Find $w$.*

1. $\Delta n_{gas} = 1 - 4 = -3\ \text{mol}$ (solid Ni is ignored)

2. $$T = 75 + 273 = 348\ \text{K}$$

3. $w = -\Delta n_{gas}RT = -(-3.00\ \text{mol})(8.31\ \text{J K}^{-1}\text{mol}^{-1})(348\ \text{K}) =$ +8.68 × 10³ J

4. Positive: the gas volume shrinks, so surroundings do work on the system.

Calorimetry. Use the heat capacity of a substance to calculate the heat required to raise its temperature by a given amount.

Core concept

• Calorimetry measures heat flow by monitoring a temperature change in a known mass of a known substance (often water).

• Heat capacity ($C$): the heat needed to raise the whole sample’s temperature by 1 K (or 1 °C). Units: J/K (extensive).

  • $q = C\,\Delta T$

• Specific heat capacity ($c_s$): heat per gram per kelvin. Units: J g⁻¹ K⁻¹ (intensive).

  • $q = m\,c_s\,\Delta T$

• Molar heat capacity ($c_{molar}$): heat per mole per kelvin. Units: J mol⁻¹ K⁻¹ (intensive).

  • $q = n\,c_{molar}\,\Delta T$

• $\Delta T = T_{final} - T_{initial}$. A temperature difference in °C equals that in K, so no conversion is needed for $\Delta T$.

Molecular meaning

• Substances with many ways to store energy (rotation, vibration, hydrogen bonds) have large heat capacities. Water has a high $c_s$ (4.184 J g⁻¹ °C⁻¹) due to extensive hydrogen bonding, which absorbs energy.

Calorimetry principle (isolated system)

• Energy conservation: heat lost by the hot object = heat gained by the cold object.

  • $-q_{hot} = +q_{cold}$, or equivalently $\sum q = 0$.

• Assumptions: no heat exchange with the surroundings or the container (or the container’s heat capacity is included explicitly), and no phase changes.

• Coffee-cup calorimeter: include the calorimeter heat capacity: $q_{total} = C_{cal}\Delta T + m c_s \Delta T$.

Pitfalls

• Using $T$ in the wrong units in $RT$ but not here. For $\Delta T$ alone, °C and K are interchangeable.

• Sign errors: the metal’s $\Delta T$ is negative (cooling), so $q_{metal} < 0$.

• ⚠ Your slide has a typo in its first example, writing 4.148 instead of 4.184 for water’s $c_s$. The final answer (0.434) uses the correct value.

Worked example 1: heat required
*Q: How much energy is needed to raise an 8.50 × 10² g aluminum block from 22.8 °C to 94.6 °C? ($c_s$ = 0.900 J g⁻¹ °C⁻¹)*

1. $$\Delta T = 94.6 - 22.8 = 71.8\ °\text{C}$$

2. $$q = (8.50 \times 10^2\ \text{g})(0.900\ \text{J g}^{-1}\text{°C}^{-1})(71.8\ \text{°C})$$

3. Units: g × J g⁻¹ °C⁻¹ × °C = J

4. $q = 5.49 \times 10^4\ \text{J} =$ 54.9 kJ (3 sig. figs.)

Worked example 2: final temperature
*Q: 840 J is absorbed by 10.0 g of water at 25.0 °C. Find $T_f$.*

1. $$\Delta T = q/(m c_s) = 840/(10.0 \times 4.18) = 20.1\ °\text{C}$$

2. $T_f = 25.0 + 20.1 =$ 45.1 °C

Calculate energy changes from calorimetry data. Difference between $c_{molar}$ and $c_{specific}$. How $C$ is related to $c$. Relation between $C_P$ and $C_V$.

Three related quantities

Symbol	Name	Units	Extensive?
$C$	Heat capacity	J K⁻¹	Extensive
$c_s$	Specific heat capacity	J g⁻¹ K⁻¹	Intensive
$c_{molar}$	Molar heat capacity	J mol⁻¹ K⁻¹	Intensive

• Relationships: $C = m\,c_s = n\,c_{molar}$

• Conversion between the intensive forms: $c_{molar} = c_s \times M$ (molar mass, g/mol).

• Rule: lowercase $c$ = intensive; uppercase $C$ = extensive.

$C_V$ vs. $C_P$

• $C_V$ (constant volume): $C_V = \dfrac{\Delta E}{\Delta T}$, since $\Delta E = q_V = C_V\Delta T$.

• $C_P$ (constant pressure): $C_P = \dfrac{\Delta H}{\Delta T}$, since $\Delta H = q_P = C_P\Delta T$.

• Derivation of the ideal-gas relation:

  1. $H = E + PV$ and $PV = nRT$ gives $H = E + nRT$.

  2. $$\Delta H = \Delta E + nR\,\Delta T$$

  3. Divide by $\Delta T$: $\dfrac{\Delta H}{\Delta T} = \dfrac{\Delta E}{\Delta T} + nR$

  4. $C_P = C_V + nR$ (extensive form) and $c_P = c_V + R$ (per mole).

• Why $C_P > C_V$: at constant $P$, part of the heat supplied goes into expansion work against the surroundings ($P\Delta V = nR\Delta T$), so more heat is needed for the same $\Delta T$. At constant $V$, all heat raises internal energy.

• This relation is exact for ideal gases. For solids and liquids $C_P \approx C_V$ because they barely expand.

Energy changes from calorimetry data

1. Identify what absorbs heat (water, solution, calorimeter hardware).

2. Compute $q = C\Delta T$ or $mc_s\Delta T$ for each, and add them.

3. Apply energy conservation: $q_{rxn} = -(q_{water} + q_{cal})$.

4. Divide by moles for kJ/mol.

Worked example (from your slides): unknown metal
*Q: 18.0 g metal at 100 °C is placed in 50.2 g water at 22.0 °C. Final T = 24.8 °C. Find $c_s$ of the metal.*

1. $$-q_{metal} = +q_{water}$$

2. $$-[18.0\ \text{g}\ c_{metal}(24.8 - 100)] = +[50.2\ \text{g}\,(4.184)(24.8 - 22.0)]$$

3. $$c_{metal}(18.0)(75.2) = (50.2)(4.184)(2.8)$$

4. $c_{metal} = \dfrac{588}{1354} =$ 0.434 J g⁻¹ °C⁻¹

From internal energy to enthalpy. Why do we need enthalpy? Relation to calorimetry.

The problem with $\Delta E$

• First law at constant pressure for expansion work only: $\Delta E = q - P\Delta V$.

• If volume is constant ($\Delta V = 0$), $\Delta E = q_V$, so heat measures $\Delta E$ directly (bomb calorimeter).

• But most laboratory and biological chemistry occurs at constant pressure (open beakers, 1 atm). Constant volume is experimentally difficult to maintain.

• At constant $P$, the heat flow $q_P$ equals $\Delta E$ plus the expansion work term, so $q_P \neq \Delta E$. We need a state function whose change equals $q_P$.

Definition of enthalpy

• $H = E + PV$ (state function, extensive, J). Since $E$, $P$ and $V$ are state functions, so is $H$.

• At constant $P$: $\Delta H = \Delta E + P\Delta V$.

• Substitute $\Delta E = q - P\Delta V$: $\Delta H = q - P\Delta V + P\Delta V$

  • $\Delta H = q_P$

• Meaning: $\Delta H$ is the heat absorbed or released at constant pressure. $\Delta H < 0$ is exothermic; $\Delta H > 0$ is endothermic.

• $PV$ term physically: the energy needed to “make room” for the system by pushing back the atmosphere.

Relating $\Delta H$ and $\Delta E$ for reactions

• For ideal gases, $P\Delta V = \Delta n_{gas}RT$:

  • $\Delta H = \Delta E + \Delta n_{gas}RT$

• If $\Delta n_{gas} = 0$ (or only solids/liquids), $\Delta H \approx \Delta E$.

Relation to calorimetry

• Constant volume (bomb): $q_V = nc_V\Delta T = mc_{s,V}\Delta T = \Delta E$

• Constant pressure (coffee cup): $q_P = nc_P\Delta T = mc_{s,P}\Delta T = \Delta H$

Absolute vs. relative values

• Absolute $E$ and $H$ can’t be measured, only differences. We choose a reference: $\Delta H_f^\circ = 0$ for elements in their standard state (298 K, 1 atm). For hydrogen the standard state is H₂(g), not H(g).

Pitfall: $\Delta H = q_P$ is valid only at constant pressure with only expansion work.

Worked example
*Q: For ½H₂(g) + ½I₂(s) → HI(g), $\Delta H = +39.7$ kJ at 298 K. Find $\Delta E$.*

1. $\Delta n_{gas} = 1 - \tfrac12 = +\tfrac12$ mol (I₂(s) not counted)

2. $$\Delta H = \Delta E + \Delta n_{gas}RT \Rightarrow \Delta E = \Delta H - \Delta n_{gas}RT$$

3. $$\Delta n_{gas}RT = (0.5)(8.314)(298) = 1239\ \text{J} = 1.24\ \text{kJ}$$

4. $\Delta E = 39.7 - 1.24 =$ 38.5 kJ

Distinguish between the internal energy and the enthalpy of a process and describe how each is measured.

Comparison

	Internal energy $\Delta E$	Enthalpy $\Delta H$
Definition	Total kinetic + potential energy change of the system	$$\Delta H = \Delta E + \Delta(PV)$$
Equals heat at…	Constant volume: $\Delta E = q_V$	Constant pressure: $\Delta H = q_P$
Instrument	Bomb calorimeter (rigid sealed steel vessel)	Coffee-cup (constant-pressure) calorimeter
Heat capacity used	$$C_V$$	$$C_P$$
State function?	Yes	Yes
Includes expansion work?	No ($w_{PV}=0$ at constant $V$)	Accounts for it automatically

Measuring $\Delta E$: bomb calorimeter

• The sample burns in a sealed rigid “bomb” (constant $V$) surrounded by water in an insulated container. An ignition wire triggers the reaction.

• $q_{rxn} = -(q_{water} + q_{bomb}) = -(m_{w}c_{w} + C_{cal})\Delta T$, and $\Delta E = q_V = q_{rxn}$.

Measuring $\Delta H$: constant-pressure calorimeter

• A reaction occurs in solution in an insulated cup with a loosely fitting lid (open to atmosphere at ~1 atm).

• $q_{rxn} = -(m_{soln}c_s + C_{cal})\Delta T$, and $\Delta H = q_P = q_{rxn}$.

• $\Delta H$ per mole = $q_{rxn}/n_{\text{limiting reagent}}$.

Pitfalls

• Sign: if the temperature of the solution rises, the reaction is exothermic, so $\Delta H$ is negative.

• Limiting reagent: divide by moles of the limiting reactant, not by moles of “solution.”

• ⚠ Slide correction: your KOH(s) + HCl(aq) calorimetry example computes $q = 6762$ J correctly but (1) divides by 0.08 mol labeled “moles of solution” (it is the limiting reagent HCl: 0.800 L × 0.100 M), and (2) reports +84.5 kJ/mol, but since $T$ rose, the correct answer is ΔH = −84.5 kJ/mol.

Worked example 1: constant-pressure (coffee cup)
*Q: 0.0300 mol NaOH(s) is added to 1000 mL of 0.0300 M HNO₃ at 23.000 °C. $T_f$ = 23.639 °C. $C_{cal}$ = 451 J/°C; $c_s$ = 4.18 J g⁻¹ °C⁻¹; density = 1.00 g/mL. Find $\Delta H$ per mole.*

1. $$\Delta T = 0.639\ °\text{C}$$

2. $$q_{cal} = (451)(0.639) = 288\ \text{J}$$

3. $$q_{soln} = (1000\ \text{g})(4.18)(0.639) = 2671\ \text{J}$$

4. $q_{total\ gained} = 2959\ \text{J}$, so $q_{rxn} = -2959\ \text{J}$

5. $\Delta H = -2959\ \text{J}/0.0300\ \text{mol} = -98{,}600\ \text{J/mol} =$ −98.6 kJ/mol

Worked example 2: constant-volume (bomb)
*Q: 5.000 g methanol (32.04 g/mol) burns in a bomb with 4000 g water ($T$: 24.000 → 29.765 °C); $C_{cal}$ = 2657 J/°C. Find $\Delta E$ per mole.*

1. $$\Delta T = 5.765\ °\text{C}$$

2. $q_{cal} = 2657(5.765) = 15{,}318\ \text{J}$; $q_{water} = 4000(4.184)(5.765) = 96{,}486\ \text{J}$

3. Total heat released by reaction = 111.8 kJ per 5.000 g

4. Per mole: $111.8 \times (32.04/5.000) = 716\ \text{kJ/mol}$

5. Exothermic: $\Delta E = -716\ \text{kJ/mol}$

Contributions to internal energy in the case of monoatomic, polyatomic (linear), polyatomic (non-linear) species.

Foundation: equipartition

• Each quadratic degree of freedom contributes $\tfrac12 kT$ per molecule, i.e., $\tfrac12RT$ per mole.

  • $k = 1.381\times10^{-23}\ \text{J K}^{-1}$ (Boltzmann’s constant); $N_A k = R$.

• Ideal gas assumption: no intermolecular potential energy, so $E$ is purely the sum of molecular motions.

Types of motion

• Translation: 3 degrees of freedom for any gas particle (movement along $x$, $y$, $z$).

• Rotation: present for molecules, not atoms (point masses have negligible moment of inertia).

• Vibration: only significantly excited at high temperatures, ignored at room temperature (the energy spacing is large compared to $kT$; quantum mechanically the mode is “frozen out”).

Results

Species	Translational	Rotational	Total $E$ (per mol)	$$C_V$$	$$C_P = C_V + R$$
Monoatomic (He, Ar)	3 × ½RT	0	$\tfrac32RT$	$$\tfrac32R$$	$$\tfrac52R$$
Linear polyatomic (H₂, N₂, CO₂)	3 × ½RT	2 × ½RT	$\tfrac52RT$	$$\tfrac52R$$	$$\tfrac72R$$
Non-linear polyatomic (H₂O, NH₃, CH₄)	3 × ½RT	3 × ½RT	$3RT$ (= $\tfrac62RT$)	$$3R$$	$$4R$$

Why linear molecules have only 2 rotations

• A linear molecule can tumble about two axes perpendicular to the molecular axis. Rotation about the bond axis involves essentially no change in the positions of the atomic nuclei (point-like masses on the axis), so its moment of inertia is negligibly small.

• Quantized rotational energy $\propto 1/I$, so the first excited level of this mode is enormously high, so it is never populated at accessible temperatures. Your slide expresses this as requiring energies so large that the molecule would dissociate first.

• Non-linear molecules have atoms off any single axis, so all 3 rotational axes have appreciable moments of inertia.

Pitfalls

• These values hold at moderate temperatures. At high $T$, vibrations contribute (each vibrational mode adds $kT$: ½ kinetic + ½ potential), so measured $C_V$ rises above these values.

• The $C_P = C_V + R$ shortcut applies to 1 mole of ideal gas.

• In liquids and solids, intermolecular potential energy matters, so these formulas do not apply.

Worked example
*Q: Estimate the molar internal energy of N₂(g) and CO₂(g) at 298 K, and $\Delta E$ for heating 1.00 mol of H₂O vapor by 50.0 K (ideal, no vibration).*

1. N₂ and CO₂ are both linear: $E = \tfrac52RT = 2.5(8.314)(298) = 6194\ \text{J/mol} \approx$ 6.19 kJ/mol

2. H₂O is non-linear: $C_V = 3R = 24.94\ \text{J K}^{-1}\text{mol}^{-1}$

3. $\Delta E = nC_V\Delta T = (1.00)(24.94)(50.0) =$ 1247 J ≈ 1.25 kJ

Use standard enthalpies of formation to calculate the standard enthalpy of a reaction.

Definitions

• Standard enthalpy of formation ($\Delta H_f^\circ$): the enthalpy change when 1 mole of a compound forms from its elements in their standard states at 1 bar/1 atm (and usually 298 K). Units: kJ/mol.

• Elements in their standard states have $\Delta H_f^\circ = 0$: O₂(g), H₂(g), N₂(g), C(graphite), Fe(s), Hg(l), Br₂(l). Not H(g), O(g) or C(diamond).

• Formation of a compound is written with a coefficient of 1 on the product, e.g. C(s) + 2H₂(g) → CH₄(g).

Formula

• $\Delta H^\circ_{rxn} = \sum n_p\,\Delta H_f^\circ(\text{products}) - \sum n_r\,\Delta H_f^\circ(\text{reactants})$

  • $n_p, n_r$ = stoichiometric coefficients (mol) from the balanced equation.

• Explicitly, for $aA + bB \to cC + dD$:

  • $\Delta H^\circ_{rxn} = c\Delta H_f^\circ(C) + d\Delta H_f^\circ(D) - [a\Delta H_f^\circ(A) + b\Delta H_f^\circ(B)]$

Why it works

• $H$ is a state function. Imagine decomposing all reactants into elements (enthalpy change $= -\Delta H_f^\circ$ of reactants) and then forming all products from elements ($+\Delta H_f^\circ$ of products). The elements cancel, and the net change equals the formula above (a special case of Hess’s law).

• Bond picture: breaking bonds is endothermic ($\Delta H > 0$), forming bonds is exothermic ($\Delta H < 0$). Products − reactants = final − initial.

Pitfalls

• Multiply each $\Delta H_f^\circ$ by its coefficient.

• Phase matters: $\Delta H_f^\circ$[H₂O(l)] = −285.8 vs. $\Delta H_f^\circ$[H₂O(g)] = −241.8 kJ/mol. The difference is the enthalpy of vaporization (≈44 kJ/mol).

• Order: products minus reactants (not the reverse).

• Units: “kJ/mol” in this context means kJ per mole of reaction as written.

• Extensive: doubling the equation doubles $\Delta H^\circ_{rxn}$.

• ⚠ Slide corrections (acetylene examples): C₂H₂ is acetylene (ethylene is C₂H₄). The accepted $\Delta H_f^\circ$[C₂H₂(g)] is +226.7 kJ/mol (consistent with your own Hess’s-law question, 226.7), not +277. With H₂O(g) at −241.8, the correct combustion enthalpy is $2(-393.5) + (-241.8) - 226.7 = -1255.5$ kJ/mol.

Worked example (from your slides)
*Q: Calculate $\Delta H^\circ$ for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Given $\Delta H_f^\circ$: CH₄ = −74.8, CO₂ = −393.5, H₂O(l) = −285.8 kJ/mol.*

1. Products: $(1)(-393.5) + (2)(-285.8) = -393.5 - 571.6 = -965.1\ \text{kJ}$

2. Reactants: $(1)(-74.8) + (2)(0) = -74.8\ \text{kJ}$ (O₂ is an element)

3. $\Delta H^\circ_{rxn} = -965.1 - (-74.8) =$ −890.3 kJ/mol (exothermic)

Calculate an overall reaction enthalpy from the enthalpies of the reactions in a reaction sequence by using Hess’s law.

Hess’s law

• The enthalpy change for a reaction is the same whether it occurs in one step or through a series of steps; the overall $\Delta H$ is the sum of the $\Delta H$ values of the individual steps.

• Basis: $H$ is a state function, so $\Delta H$ depends only on initial and final states: $\Delta H_{RP} = \Delta H_{RI} + \Delta H_{IP}$ (reactants → intermediates → products).

• Utility: lets us find $\Delta H$ for reactions that are hard to measure directly (e.g., C → CH₄, diamond → graphite) from combustion or other easily measured reactions.

Three manipulation rules

1. Reverse a reaction → change the sign of $\Delta H$.

2. Multiply a reaction by a factor $k$ → multiply $\Delta H$ by $k$ ($\Delta H$ is extensive).

3. Add equations → add their $\Delta H$ values. Species that appear on both sides cancel.

Problem-solving strategy

1. Write the target equation clearly.

2. Locate each target reactant/product in exactly one given equation, and orient and scale that equation to match.

3. Add the modified equations. Check that all intermediates cancel and the result equals the target.

4. Sum the modified $\Delta H$ values. Keep track of signs and units (kJ).

Pitfalls

• Forgetting to flip the sign when reversing.

• Forgetting to multiply $\Delta H$ when scaling coefficients.

• Mismatched phases (H₂O(l) vs. H₂O(g)).

• The same $\Delta G$ logic also follows Hess’s law, which is why coupled biochemical reactions (e.g., ATP hydrolysis driving glucose phosphorylation) simply add their $\Delta G$ values.

Worked example 1 (from your slides): ΔH_f of propane
Target: 3C(gr) + 4H₂(g) → C₃H₈(g)
Given:
(a) C₃H₈ + 5O₂ → 3CO₂ + 4H₂O(l)   $\Delta H = -2220$ kJ
(b) C(gr) + O₂ → CO₂   $\Delta H = -394$ kJ
(c) H₂ + ½O₂ → H₂O(l)   $\Delta H = -286$ kJ

1. Reverse (a) (propane must be a product): 3CO₂ + 4H₂O → C₃H₈ + 5O₂   $\Delta H = +2220$ kJ

2. Multiply (b) by 3 (need 3 C): 3C + 3O₂ → 3CO₂   $\Delta H = 3(-394) = -1182$ kJ

3. Multiply (c) by 4 (need 4 H₂): 4H₂ + 2O₂ → 4H₂O   $\Delta H = 4(-286) = -1144$ kJ

4. Add: CO₂, H₂O and O₂ cancel (5 O₂ produced on the right; 3 + 2 = 5 O₂ consumed on the left).

5. $\Delta H = +2220 - 1182 - 1144 =$ −106 kJ

Worked example 2: diamond → graphite (from your slides)

1. C(diamond) + O₂ → CO₂   $-395.4$ kJ (as given)

2. 2CO₂ → 2CO + O₂   $+566.0$ kJ (as given)

3. 2CO → C(graphite) + CO₂   $-172.5$ kJ (as given)

4. Adding: C(diamond) → C(graphite), $\Delta H = -395.4 + 566.0 - 172.5 =$ −1.9 kJ

5. Slightly exothermic, so graphite is the more stable form.