**Thermodynamics** is the branch of chemistry that deals with energy changes accompanying chemical reactions and physical processes. It studies how energy is transformed from one form to another (heat, work, electrical, mechanical) during chemical reactions.
**Key Points:**
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**System** = The part of the universe where observations are made
**Surroundings** = Everything else in the universe that can interact with the system
**Universe = System + Surroundings**
**Boundary** = The real or imaginary wall separating the system from surroundings; designed to control movement of matter and energy
**Example:** When studying a chemical reaction in a beaker kept on a table:
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#### (1) **Open System**
#### (2) **Closed System**
#### (3) **Isolated System**
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The **state** of a thermodynamic system is completely described by specifying **macroscopic (bulk) properties:**
**State Variables/State Functions:** Properties whose values depend ONLY on the current state, not on how that state was reached
**Common State Functions:**
**Key Concept:** For a complete description of a system, only a **minimum number** of properties need to be specified. Once these are fixed, all other properties automatically have definite values.
**Example:** For an ideal gas, if you know p, V, and T, the state is completely defined. Other properties will have fixed values.
**Important Distinction:** The state of surroundings can NEVER be completely specified, and fortunately, it is NOT necessary to do so.
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**Internal Energy (U)** = Total energy of a system; sum of all chemical, electrical, mechanical, and other forms of energy
Internal energy changes when:
#### **(a) Work**
**Joule's Experiment (1840-1850):** Showed that a fixed amount of work done on an isolated system produces the same change in state regardless of the TYPE of work (mechanical paddle work or electrical work).
**Proof of Internal Energy as State Function:**
**Conclusion:** Since identical work produces identical state changes, internal energy U must be a **state function** depending only on initial and final states.
**Mathematical Expression:**
$$\Delta U = U_2 - U_1 = w_{ad}$$
**Sign Convention (IUPAC):**
#### **(b) Heat**
**Heat (q)** = Energy transfer resulting from temperature difference between system and surroundings
**Key Difference from Work:**
**Heat Transfer Setup:** System with thermally conducting walls in a heat reservoir
**Sign Convention (IUPAC):**
#### **(c) The General Case - First Law of Thermodynamics**
When a process involves BOTH work and heat:
$$\boxed{\Delta U = q + w}$$ **Equation 5.1**
where:
**Critical Insight:** Although q and w individually depend on the pathway taken, their SUM (q + w = ΔU) is **path-independent** and depends only on initial and final states.
**The First Law of Thermodynamics (Mathematical Statement):**
*The energy of an isolated system is constant*
**Stated as:** Law of Conservation of Energy - Energy can neither be created nor destroyed; it can only be converted from one form to another.
**For Isolated Systems:** w = 0 and q = 0 → **ΔU = 0**
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Most work in chemistry is **pressure-volume (PV) work** during gas expansion/compression.
#### **Derivation of Work Equation:**
Consider an ideal gas in a cylinder with movable frictionless piston:
When piston moves inward (compression):
**But** displacement l = ΔV/A = (Vf - Vi)/A
Therefore: w = pex × A × (Vf - Vi)/A = pex(Vf - Vi)
**Standard Form:**
$$\boxed{w = -p_{ex}\Delta V = -p_{ex}(V_f - V_i)}$$ **Equation 5.2**
**Sign Logic:**
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#### **Finite Steps (Irreversible):**
If pressure changes in finite steps, work is summed:
$$w = -\sum p \Delta V$$
#### **Reversible Process (Infinite Steps):**
A **reversible process** is one where the system moves infinitely slowly through a series of equilibrium states, always being infinitesimally close to equilibrium with surroundings.
$$\boxed{w_{rev} = -\int_{V_i}^{V_f} p_{ex} dV}$$ **Equation 5.3**
At each step: pex = pin + dp (compression) or pex = pin - dp (expansion)
**General form:** pex = pin + dp
For reversible conditions:
$$\boxed{w_{rev} = -\int_{V_i}^{V_f} p_{in} dV}$$ **Equation 5.4**
#### **For Ideal Gas at Constant Temperature (Isothermal Reversible Process):**
Using ideal gas law: pV = nRT → p = nRT/V
$$w_{rev} = -\int_{V_i}^{V_f} \frac{nRT}{V} dV = -nRT \ln\frac{V_f}{V_i} = -2.303nRT \log\frac{V_f}{V_i}$$
$$\boxed{w_{rev} = -nRT \ln\frac{V_f}{V_i}}$$ **Equation 5.5**
**Example Calculation:**
If 2 L of gas at 25°C expands to 10 L:
$$w = -nRT \ln(V_f/V_i) = -(1)(8.314)(298) \ln(10/2) = -(1)(8.314)(298)(1.609) = -3,994 \text{ J}$$
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#### **(1) Free Expansion**
**Free Expansion:** Expansion of a gas into vacuum (pex = 0)
Since pex = 0:
$$w = -p_{ex}\Delta V = 0$$ (no work done)
**For Ideal Gas:** Joule showed experimentally that q = 0
Therefore: **ΔU = 0** for isothermal free expansion of ideal gas
**Why?** Ideal gas molecules have no intermolecular forces. Expansion doesn't require breaking bonds, so no energy change.
#### **(2) Isothermal Expansion of Ideal Gas**
**Reversible Isothermal Expansion:**
$$w = -nRT \ln\frac{V_f}{V_i} = q$$
**Irreversible Isothermal Expansion (against constant external pressure):**
$$w = -p_{ex}(V_f - V_i)$$
$$q = -w = p_{ex}(V_f - V_i)$$
#### **(3) Adiabatic Process**
**Adiabatic Process:** No heat transfer (q = 0)
$$\Delta U = w_{ad}$$
All work done on system increases internal energy (or vice versa).
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$$\Delta U = q + w = q - p_{ex}\Delta V$$
#### **Case 1: Constant Volume Process**
At constant volume: ΔV = 0 → pex ΔV = 0
$$\boxed{\Delta U = q_V}$$
where qV denotes heat at constant volume
**Significance:** Calorimeter experiments measure qV directly
#### **Case 2: Constant Pressure Process**
At constant pressure:
$$\Delta U = q_p - p_{ext}\Delta V$$
$$\Delta U = q_p - p\Delta V$$
This will be crucial for defining **enthalpy** (next section)
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**Enthalpy (H)** = Total heat content of a system at constant pressure
For a system at constant pressure, we need a state function that directly relates to heat. Consider:
$$\Delta U = q_p - p\Delta V$$
**Rearranging:**
$$q_p = \Delta U + p\Delta V$$
For constant pressure:
$$q_p = \Delta U + p(V_f - V_i)$$
$$q_p = (U_f - U_i) + p(V_f - V_i)$$
$$q_p = (U_f + pV_f) - (U_i + pV_i)$$
**Define:** H = U + pV (Enthalpy)
$$q_p = \Delta H = H_f - H_i$$
$$\boxed{\Delta H = \Delta U + \Delta(pV)}$$
**At constant pressure:**
$$\boxed{\Delta H = \Delta U + p\Delta V}$$ **Equation 5.6**
**Key Point:** At constant pressure, **heat absorbed equals change in enthalpy** (ΔH = qp)
From ΔH = ΔU + Δ(pV):
For constant pressure:
$$\Delta H = \Delta U + p\Delta V$$
For ideal gas: pΔV = ΔnRT (where Δn = moles of gas products - moles of gas reactants)
$$\boxed{\Delta H = \Delta U + \Delta n_{gas}RT}$$
where Δngas = change in moles of gaseous substances
**Sign Relationship:**
**Numerical Example:**
N2(g) + 3H2(g) → 2NH3(g)
Δngas = 2 - (1 + 3) = -2 moles
ΔH = ΔU + (-2)RT = ΔU - 2RT
At 25°C: ΔH = ΔU - 2(8.314)(298) = ΔU - 4,951 J
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1. **State Function:** H depends only on current state (p, V, T, composition)
2. **Extensive Property:** ΔH scales with amount of substance
3. **Measured at Constant Pressure:** Chemical processes typically occur at constant pressure (1 atm)
4. **Standard Enthalpy of Formation:** ΔHf° = enthalpy change when 1 mole of compound forms from elements in standard states
**Standard States:**
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#### **Bomb Calorimeter (for ΔU):**
**Used for:** Measuring enthalpy of combustion
**Principle:** Combustion occurs in rigid vessel (constant volume)
**Process:**
1. Burn known mass of substance
2. Measure temperature rise of surrounding water
3. q = mass × specific heat × ΔT
4. ΔU = q/moles of substance
#### **Coffee Cup Calorimeter (for ΔH):**
**Used for:** Measuring enthalpy of dilution, neutralization, reactions in solution
**Principle:** Reaction occurs in container open to atmosphere (constant pressure)
**Process:**
1. Mix reactants in insulated cup
2. Measure temperature change
3. q = mcΔT
4. ΔH = q (at constant pressure)
---
#### **(1) Exothermic Reactions**
**Example:** CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH = -890 kJ/mol
#### **(2) Endothermic Reactions**
**Example:** N2(g) + O2(g) → 2NO(g) ΔH = +180 kJ/mol
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**Definition:** Enthalpy change when reactants in standard states convert to products in standard states
$$\boxed{\Delta_r H° = \sum \Delta_f H°_{products} - \sum \Delta_f H°_{reactants}}$$
**Working Formula:**
$$\Delta_r H° = \sum n \Delta_f H°(\text{products}) - \sum n \Delta_f H°(\text{reactants})$$
**Calculation Example:**
For reaction: 2H2(g) + O2(g) → 2H2O(l)
Using standard enthalpy of formation values:
$$\Delta_r H° = [2 × (-285.8)] - [2(0) + 1(0)] = -571.6 \text{ kJ}$$
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**Statement:** If a chemical equation can be expressed as a sum of two or more equations, the enthalpy change of the net equation equals the sum of the enthalpy changes of individual equations.
**Mathematical Expression:**
$$\Delta H_{net} = \sum \Delta H_i$$
**Principle:** Enthalpy is a state function; it depends only on initial and final states, not on the pathway.
**Application Steps:**
1. Write target equation
2. Manipulate given equations to obtain target (reverse, multiply, add)
3. Apply same manipulations to ΔH values
4. Sum all ΔH values
**Worked Example:**
**Given:**
**Find:** ΔH for C(s) + CO2(g) → 2CO(g)
**Solution:**
Reverse equation (2): 2CO2(g) → 2CO(g) + O2(g) ΔH = +566 kJ
Add to equation (1):
Net: C(s) + CO2(g) → 2CO(g)
ΔH = -393.5 + 566 = **+172.5 kJ**
**Important Rules for Hess's Law:**
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The formula $$\Delta_r H° = \sum \Delta_f H° (products) - \sum \Delta_f H° (reactants)$$ is actually Hess's Law applied using standard enthalpies of formation as the basis.
**Advantage:** No need to manipulate equations; direct calculation from tabulated values.
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#### **(1) Spontaneous Process**
#### **(2) Non-Spontaneous Process**
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**Definition:** Entropy measures the degree of disorder or randomness in a system; it is the measure of unavailable energy.
**Symbol:** S
**Units:** J/(mol·K) or J·K⁻¹
**Fundamental Concept (Statistical):**
**Statement:** The entropy of an isolated system always increases for any spontaneous process.
$$\boxed{\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0}$$
**For spontaneous process:** ΔSuniverse > 0
**For non-spontaneous process:** ΔSuniverse < 0
**For equilibrium:** ΔSuniverse = 0
**Key Implications:**
1. **Irreversible/Spontaneous Process:** ΔSuniverse > 0 ✓ Occurs naturally
2. **Reversible Process (at equilibrium):** ΔSuniverse = 0 → System and surroundings exchange infinitesimal energy
3. **Impossible Process:** ΔSuniverse < 0 ✗ Cannot occur
---
#### **Standard Entropy of Substances (S°)**
Unlike internal energy and enthalpy, entropy has an **absolute value** because:
**Third Law of Thermodynamics:** The entropy of a perfect crystal at absolute zero (0 K) is zero.
S°298K values are tabulated and measured from this reference.
**Characteristic Entropy Values (Relative):**
#### **Standard Entropy of Reaction (ΔrS°)**
$$\boxed{\Delta_r S° = \sum S°_{products} - \sum S°_{reactants}}$$
**Example:**
2H2(g) + O2(g) → 2H2O(l)
S°[H2(g)] = 130.7 J/(mol·K)
S°[O2(g)] = 205.1 J/(mol·K)
S°[H2O(l)] = 69.9 J/(mol·K)
ΔrS° = [2(69.9)] - [2(130.7) + 205.1]
ΔrS° = 139.8 - 466.5 = **-326.7 J/K**
**Interpretation:** Entropy decreases (negative ΔS) because gas → liquid (disorder decreases)
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When system releases heat (exothermic, q < 0):
$$\Delta S_{surroundings} = -\frac{q_{system}}{T} = -\frac{\Delta H_{system}}{T}$$
**At constant pressure:**
$$\boxed{\Delta S_{surroundings} = -\frac{\Delta H}{T}}$$
**Logic:**
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**Definition:** G = H - TS
A thermodynamic state function that combines enthalpy and entropy to predict spontaneity of processes at constant temperature and pressure.
$$\boxed{G = H - TS}$$
**Change in Gibbs Free Energy:**
$$\boxed{\Delta G = \Delta H - T\Delta S}$$
where:
**Units:** Ensure consistency; typically kJ
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**Criterion for Spontaneity at Constant T and P:**
| ΔG Value | ΔSuniverse | Process | Nature |
|----------|-----------|---------|--------|
| **ΔG < 0** | > 0 | **Spontaneous** | Occurs naturally; Irreversible |
| **ΔG = 0** | = 0 | **Equilibrium** | System at equilibrium; Reversible |
| **ΔG > 0** | < 0 | **Non-spontaneous** | Does not occur; requires work |
**Proof of Relationship:**
$$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$$
$$\Delta S_{universe} = \Delta S - \frac{\Delta H}{T}$$
Multiply by -T:
$$-T\Delta S_{universe} = \Delta H - T\Delta S = \Delta G$$
For spontaneous process: ΔSuniverse > 0
$$\therefore -T\Delta S_{universe} < 0$$
$$\therefore \Delta G < 0$$ ✓
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#### **Case 1: ΔH < 0, ΔS > 0 (Exothermic, Entropy Increases)**
ΔG = ΔH - TΔS = (negative) - T(positive) = **Always negative**
**Spontaneous at all temperatures**
**Examples:**
#### **Case 2: ΔH > 0, ΔS < 0 (Endothermic, Entropy Decreases)**
ΔG = ΔH - TΔS = (positive) - T(negative) = **Always positive**
**Non-spontaneous at all temperatures**
**Examples:**
#### **Case 3: ΔH < 0, ΔS < 0 (Exothermic, Entropy Decreases)**
ΔG = ΔH - TΔS = (negative) - T(negative)
**Spontaneous at LOW temperatures** (ΔH term dominates)
**Non-spontaneous at HIGH temperatures** (TΔS term dominates)
**Examples:**
#### **Case 4: ΔH > 0, ΔS > 0 (Endothermic, Entropy Increases)**
ΔG = ΔH - TΔS = (positive) - T(positive)
**Non-spontaneous at LOW temperatures** (ΔH term dominates)
**Spontaneous at HIGH temperatures** (TΔS term dominates)
**Examples:**
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**Transition Temperature:** For cases 3 and 4, spontaneity changes at equilibrium (ΔG = 0)
$$\Delta G = 0 = \Delta H - T_{eq}\Delta S$$
$$\boxed{T_{equilibrium} = \frac{\Delta H}{\Delta S}}$$
**Example - Melting of Ice:**
Ice ⇌ Water (at 0°C = 273.15 K)
ΔH (fusion) = +6.01 kJ/mol
ΔS (fusion) = +22.0 J/(mol·K) = +0.022 kJ/(mol·K)
Teq = 6.01 / 0.022 = **273 K ≈ 0°C** ✓
Below 0°C: ΔG < 0 (ice melting is non-spontaneous, reverse spontaneous - freezing)
Above 0°C: ΔG > 0 (ice melting is spontaneous)
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$$\boxed{\Delta_r G° = \sum \Delta_f G°_{products} - \sum \Delta_f G°_{reactants}}$$
**Alternative formula (using Hess's Law):**
$$\boxed{\Delta_r G° = \Delta_r H° - T\Delta_r S°}$$
where ΔrH° and ΔrS° are calculated from tabulated standard values.
**Worked Example:**
N2(g) + 3H2(g) ⇌ 2NH3(g) at 298 K
ΔrH° = -92.4 kJ
ΔrS° = -198.3 J/K = -0.1983 kJ/K
ΔrG° = ΔrH° - TΔrS°
ΔrG° = -92.4 - (298)(-0.1983)
ΔrG° = -92.4 + 59.1
ΔrG° = **-33.3 kJ**
**Interpretation:** Spontaneous at 298 K; reaction proceeds forward.
---
$$\boxed{\Delta G° = -2.303RT \log K_c}$$
or equivalently
$$\boxed{\Delta G° = -RT \ln K_c}$$
where:
**Derivation Logic:**
At equilibrium: ΔG = 0
At non-equilibrium: ΔG = ΔG° + RT ln Q
At equilibrium (Q = Kc): 0 = ΔG° + RT ln Kc
Therefore: ΔG° = -RT ln Kc
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| ΔG° Value | Kc Value | Equilibrium Position | Spontaneity |
|-----------|----------|-------------------|-------------|
| **ΔG° < 0** | **Kc > 1** | Products favored | Spontaneous (forward) |
| **ΔG° = 0** | **Kc = 1** | Equal amounts | At equilibrium |
| **ΔG° > 0** | **Kc < 1** | Reactants favored | Non-spontaneous (reverse possible) |
**Examples:**
**Example 1:**
H2(g) + I2(g) ⇌ 2HI(g) Kc = 50 at 25°C
Since Kc > 1 → ΔG° < 0 → Reaction is spontaneous; products favored
**Example 2:**
N2(g) + O2(g) ⇌ 2NO(g) Kc ≈ 10⁻³¹ at 298 K
Since
Q1. A chemical reaction occurs in an open beaker on a laboratory bench. Which statement correctly describes this system?
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:
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?
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?
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?
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?
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?
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
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.
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?
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.
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.
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.
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