Thermodynamics

Big Picture

Thermodynamics studies the relationship between heat, work, and internal energy. The First Law says energy is conserved (you cannot win). The Second Law says entropy always increases (you cannot break even). The Zeroth Law defines temperature. Together they explain why heat flows from hot to cold, why perpetual-motion machines are impossible, and why time has a direction.

ELI-10: Explain It Like I'm 10

Heat is energy that flows from hot things to cold things. Temperature measures how hot something is. The First Law is the energy accountant: energy can change form (motion to heat, heat to motion) but can never appear or disappear. The Second Law is why hot coffee cools down but cold coffee never heats itself back up — disorder (entropy) always increases. A heat engine takes heat from something hot, turns some of it into useful work, and dumps the rest into something cold.

9.1 Temperature, Heat, and Thermal Expansion

Core Ideas

Temperature measures the average random kinetic energy of particles in a substance. Heat is energy transferred due to a temperature difference.

Key Definitions

• Thermal equilibrium: two objects in contact eventually reach the same temperature (Zeroth Law).
• Temperature scales: Celsius (°C), Kelvin (K). \(T_K = T_C + 273.15\). The Kelvin scale starts at absolute zero.
• Specific heat \(c\): energy to raise 1 kg by 1 K. \(Q = mc\Delta T\). Water: \(c \approx 4186\ \text{J/(kg·K)}\).
• Thermal expansion: \(\Delta L = \alpha L_0 \Delta T\). Most materials expand when heated.

Common Misconception: "Heat Is a Substance Contained in Objects"

Until the late 18th century, scientists believed heat was a weightless fluid called caloric that flowed from hot objects to cold ones. This "caloric theory" treated heat as a kind of invisible substance that objects could store. It explained calorimetry reasonably well and even produced useful results — but it is wrong.

What is wrong with the caloric view? Heat is not a substance at all. An object does not "contain" a certain amount of heat. Instead:

• Heat is energy in transit. It only exists during a transfer, not as a static property of an object. You can add energy to an object (raising its internal energy), but the object does not "have heat" — it has internal energy.
• Friction disproves caloric. Count Rumford's cannon-boring experiments (1798) showed that friction could produce seemingly unlimited heat — far more than any finite caloric reservoir could supply. The mechanical work of boring was being converted directly into thermal energy.
• Joule's paddle-wheel experiment (1843) quantified the mechanical equivalent of heat, definitively showing heat is a form of energy, not a substance.

Correct understanding: When you touch a hot stove, energy transfers from the stove to your hand because of the temperature difference. The stove loses internal energy; your hand gains it. No "heat substance" flows — energy is transferred via molecular collisions at the boundary. After the transfer, we say heat was exchanged, but neither object "has" that heat anymore — the energy has become part of each object's internal energy.

Why this matters: The caloric misconception leads students to confuse heat with internal energy and to think objects can be "filled" with heat. Mastering thermodynamics starts with internalizing that heat, like work, is a process quantity — it describes energy crossing a boundary, not energy stored inside.

ELI-10: Explain It Like I'm 10

Temperature is a number that tells you how vigorously the tiny invisible particles in a substance are jiggling. Hotter = faster jiggling. Heat is energy moving from a hot thing to a cold thing until they both jiggle at the same speed. Water has a huge specific heat — it takes a lot of energy to warm it up, which is why oceans moderate climate. Railroad tracks have gaps so they do not buckle on hot days — thermal expansion.

9.2 Phase Changes and Calorimetry

Core Ideas

Phase changes (melting, boiling) occur at constant temperature while heat is added. The energy goes into breaking bonds, not raising temperature.

Latent heat of fusion \(L_f\): energy to melt/freeze (water: \(3.34 \times 10^5\ \text{J/kg}\)). Latent heat of vaporization \(L_v\): energy to boil/condense (water: \(2.26 \times 10^6\ \text{J/kg}\)). \[ Q = mL_f \quad\text{or}\quad Q = mL_v \]

Calorimetry: conservation of energy in heat exchanges: \(\sum Q = 0\) in an isolated system.

Worked Example: Mixing Hot and Cold Water

Problem: You pour 200 g of water at 80°C into an insulated cup containing 300 g of water at 20°C. What is the final equilibrium temperature? Assume no heat is lost to the surroundings and the cup absorbs negligible heat.

Solution:

Step 1 — Set up conservation of energy. In an isolated system, the total heat exchange sums to zero: \[ Q{\text{hot}} + Q{\text{cold}} = 0 \]

Step 2 — Express each heat term. Both substances are water, so \(c = 4186\ \text{J/(kg·K)}\) for both. Let \(Tf\) be the final temperature. The hot water cools down (\(\Delta T\) is negative, so \(Q{\text{hot}}\) is negative — heat leaves it). The cold water warms up (\(\Delta T\) is positive, so \(Q{\text{cold}}\) is positive — heat enters it): \[ m{\text{hot}} c (Tf - T{\text{hot}}) + m_{\text{cold}} c (Tf - T{\text{cold}}) = 0 \]

Step 3 — Plug in numbers. Convert grams to kilograms: \(m{\text{hot}} = 0.200\ \text{kg}\), \(m{\text{cold}} = 0.300\ \text{kg}\). Since \(c\) is the same for both, it cancels out: \[ 0.200 (T_f - 80) + 0.300 (T_f - 20) = 0 \]

Step 4 — Solve for \(T_f\): \[ 0.200 T_f - 16 + 0.300 T_f - 6 = 0 \\ 0.500 T_f - 22 = 0 \\ T_f = \frac{22}{0.500} = 44\degree\text{C} \]

Shortcut — weighted average: When the same substance is mixed, the final temperature is the mass-weighted average of the initial temperatures: \[ T_f = \frac{m_1 T_1 + m_2 T_2}{m_1 + m_2} = \frac{(0.200)(80) + (0.300)(20)}{0.200 + 0.300} = \frac{16 + 6}{0.500} = 44\degree\text{C} \]

Check: The final temperature (44°C) is closer to 20°C than 80°C because there is more cold water (300 g vs. 200 g). This makes intuitive sense — the greater mass dominates.

ELI-10: Explain It Like I'm 10

Ice at 0°C melts into water at 0°C. You add heat but the temperature does not rise — the energy goes into breaking the crystal structure, not speeding up molecules. That is latent ("hidden") heat. Boiling requires even more latent heat than melting. A steam burn is worse than a boiling-water burn because steam carries huge latent heat that it releases when condensing on your skin.

9.3 Kinetic Theory and the Ideal Gas Law

Core Concepts

Ideal gas assumptions: molecules are point particles with no intermolecular forces, undergoing elastic collisions. This models dilute gases well.

Ideal Gas Law: \[ PV = nRT \]

• \(P\) = pressure (Pa)
• \(V\) = volume (m³)
• \(n\) = number of moles
• \(R = 8.31\ \text{J/(mol·K)}\) = universal gas constant
• \(T\) = temperature (K)

Kinetic theory connects microscopic motion to macroscopic pressure: pressure arises from countless molecular collisions with container walls. Average kinetic energy per molecule: \(\langle K \rangle = \frac{3}{2}k_B T\), where \(k_B = 1.38 \times 10^{-23}\ \text{J/K}\).

Explicit Assumptions of the Ideal Gas Model

The ideal gas law is a model — it is not an exact description of any real gas. It assumes:

1. Point particles: Gas molecules have negligible volume compared to the container. Real gas molecules do occupy space, and this becomes significant at high pressures.
2. No intermolecular forces: Molecules do not attract or repel each other except during collisions. Real gases have van der Waals forces; these become significant at low temperatures when molecules move slowly enough to "feel" each other.
3. Elastic collisions: Collisions between molecules and with container walls conserve kinetic energy — no energy is lost to deformation, vibration, or radiation.
4. Random motion: Molecules move in random directions with a distribution of speeds (Maxwell-Boltzmann distribution).
5. Large numbers: There are enough molecules that statistical averages are meaningful.

When do real gases deviate? At low temperatures and high densities, intermolecular attractions cannot be ignored, and molecular volume becomes a significant fraction of the container volume. The van der Waals equation corrects for these effects: \((P + a n^2/V^2)(V - nb) = nRT\).

Proportional Reasoning with the Ideal Gas Law

The ideal gas law encodes three historical gas laws as special cases. You can solve many problems without plugging into \(PV = nRT\) by reasoning proportionally.

Case 1 — Boyle's Law (constant \(T\), fixed \(n\)): \(P \propto 1/V\)

If temperature and amount of gas are fixed, pressure and volume are inversely proportional. Example: A gas at 100 kPa occupies 2.0 L. If the volume is compressed to 0.5 L at constant temperature, what is the new pressure?

\[ P_1 V_1 = P_2 V_2 \quad\Rightarrow\quad (100)(2.0) = P_2 (0.5) \quad\Rightarrow\quad P_2 = 400\ \text{kPa} \]

The volume decreased by a factor of 4, so the pressure increased by a factor of 4. Molecules hit the walls four times as often in the smaller space.

Case 2 — Charles's Law (constant \(P\), fixed \(n\)): \(V \propto T\)

If pressure and amount of gas are fixed, volume is directly proportional to absolute temperature. Example: A balloon at 300 K occupies 1.0 L. If the temperature increases to 600 K at constant pressure, the volume doubles to 2.0 L. Crucial: Always use Kelvin. If the temperature goes from 27°C (300 K) to 127°C (400 K), the volume increases by a factor of 400/300 = 4/3, not 127/27.

Case 3 — Gay-Lussac's Law (constant \(V\), fixed \(n\)): \(P \propto T\)

If volume and amount of gas are fixed, pressure is directly proportional to absolute temperature. Example: A sealed rigid container of gas at 300 K and 200 kPa is heated to 450 K. The new pressure is \(200 \times (450/300) = 300\ \text{kPa}\). The temperature increased by a factor of 1.5, so the molecules hit the walls 1.5× harder on average.

General proportional reasoning: When \(n\) is fixed, \(PV/T = \text{constant}\). If any two of \(P, V, T\) change, you can write: \[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \] Always use Kelvin temperatures and absolute pressures. This combined form handles problems where more than one variable changes simultaneously.

ELI-10: Explain It Like I'm 10

Temperature is the average jiggling energy of gas molecules. If you heat a sealed container of gas, the molecules jiggle harder, hit the walls harder and more often, and the pressure rises. If you squeeze the container (reduce volume), the molecules hit the walls more often and pressure rises. The ideal gas law puts these relationships into one equation. Real gases deviate when they are cold and dense enough for molecules to stick to each other.

9.4 First Law of Thermodynamics

Core Idea

Energy is conserved. The change in internal energy equals heat added minus work done by the system.

\[ \Delta U = Q - W \]

• \(\Delta U\) = change in internal energy (J)
• \(Q\) = heat added to the system (J)
• \(W\) = work done by the system (J)

Sign convention: This is the physics convention. Some chemistry texts use \(\Delta U = Q + W\) (with \(W\) being work done on the system). Know which convention is in use.

Work in a PV process: \(W = \int P\,dV\) — the area under the pressure-volume curve.

Understanding PV Diagrams

A PV diagram plots pressure (\(y\)-axis) against volume (\(x\)-axis) for a thermodynamic system. It is the most important visual tool in thermodynamics because:

• Every point on the diagram represents a unique equilibrium state of the system (specified by \(P\) and \(V\); \(T\) is then determined by the ideal gas law).
• Every path between two points represents a thermodynamic process — the sequence of states the system passes through.
• The work done during a process from \(V_1\) to \(V2\) is exactly the area under the \(P(V)\) curve: \(W = \int{V_1}^{V_2} P\,dV\). If the volume increases (expansion, rightward motion), the work is positive — the system does work on the surroundings. If the volume decreases (compression, leftward motion), the work is negative — work is done on the system.

Reading a PV diagram — key features:

Cycles: A closed loop on a PV diagram represents a cyclic process (the system returns to its initial state). The net work done by the system over one complete cycle is the area enclosed by the loop. If the cycle runs clockwise, net work is positive (a heat engine). If counterclockwise, net work is negative (a refrigerator or heat pump). Since \(\Delta U = 0\) for a full cycle (the system returns to the same state), the First Law gives \(Q{\text{net}} = W{\text{net}}\) — the net heat added equals the net work done.

Assumptions for Reversible Processes

The process descriptions and equations below assume quasi-static, reversible processes. A quasi-static process happens slowly enough that the system remains in internal equilibrium at every instant — it passes through a continuous sequence of equilibrium states, tracing a well-defined path on the PV diagram. A reversible process is one that can be reversed by an infinitesimal change in conditions without dissipating energy (no friction, no turbulence, no finite temperature gradients). Real processes are irreversible to some degree, but reversible processes provide an idealized upper bound and are mathematically tractable.

Thermodynamic Processes (ideal gas)

For a monatomic ideal gas: \(C_V = \frac{3}{2}R\), \(C_P = \frac{5}{2}R\), \(\gamma = C_P/C_V = 5/3\). For a diatomic ideal gas (e.g., N₂, O₂ at moderate temperatures): \(C_V = \frac{5}{2}R\), \(C_P = \frac{7}{2}R\), \(\gamma = 1.4\).

Worked Example: Isobaric Process (Constant \(P\))

Setup: \(n = 2.00\ \text{mol}\) of a monatomic ideal gas at \(P = 1.00 \times 10^5\ \text{Pa}\) is heated from \(T_1 = 300\ \text{K}\) to \(T_2 = 600\ \text{K}\) while pressure is held constant.

Step 1 — Find \(\Delta V\). At constant pressure, \(V \propto T\): \[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\Rightarrow\quad \Delta V = \frac{nR}{P}(T_2 - T_1) = \frac{nR\Delta T}{P} \]

Numerically: \(\Delta V = \frac{(2.00)(8.31)(300)}{1.00 \times 10^5} = 0.0499\ \text{m}^3\).

Step 2 — Calculate work. \(W = P\Delta V = nR\Delta T\): \[ W = (2.00)(8.31)(300) = 4.99 \times 10^3\ \text{J} \] The gas expands, pushing the piston outward — it does positive work on the surroundings.

Step 3 — Calculate heat added. For an isobaric process: \(Q = nC_P\Delta T\). For monatomic gas, \(C_P = \frac{5}{2}R = 20.8\ \text{J/(mol·K)}\): \[ Q = (2.00)(20.8)(300) = 1.25 \times 10^4\ \text{J} \]

Step 4 — Find \(\Delta U\) using the First Law: \[ \Delta U = Q - W = 12,500 - 4,990 = 7.51 \times 10^3\ \text{J} \]

Verification using \(\Delta U = nC_V\Delta T\): \(C_V = \frac{3}{2}R = 12.5\ \text{J/(mol·K)}\): \[ \Delta U = (2.00)(12.5)(300) = 7.50 \times 10^3\ \text{J} \] Matches (within rounding). Of the 12,500 J of heat added, 7,500 J increases internal energy and 5,000 J goes into expansion work.

Worked Example: Isochoric Process (Constant \(V\))

Setup: The same 2.00 mol of monatomic ideal gas is heated from 300 K to 600 K in a rigid sealed container (volume cannot change).

Step 1 — Work. Since \(\Delta V = 0\), the gas does no work: \[ W = 0 \]

Step 2 — Heat added. For constant volume: \(Q = nC_V\Delta T\). With \(C_V = \frac{3}{2}R = 12.5\ \text{J/(mol·K)}\): \[ Q = (2.00)(12.5)(300) = 7.50 \times 10^3\ \text{J} \]

Step 3 — First Law: \[ \Delta U = Q - W = 7,500 - 0 = 7.50 \times 10^3\ \text{J} \]

Comparison with isobaric: Heating the same gas over the same \(\Delta T\) requires less heat at constant volume (7,500 J vs. 12,500 J) because no energy is spent on expansion work. All the heat goes into raising internal energy. The temperature rises faster per joule of heat input at constant volume. This is why \(C_P > C_V\): at constant pressure, some of the added energy leaks out as work.

Worked Example: Isothermal Process (Constant \(T\))

Setup: \(n = 2.00\ \text{mol}\) of an ideal gas at \(T = 300\ \text{K}\) expands slowly from \(V_1 = 0.0100\ \text{m}^3\) to \(V_2 = 0.0200\ \text{m}^3\) while in contact with a thermal reservoir that keeps the temperature constant.

Step 1 — Internal energy change. For an ideal gas, internal energy depends only on temperature. Since \(\Delta T = 0\): \[ \Delta U = 0 \]

Step 2 — Work. For an isothermal process of an ideal gas, \(P = nRT/V\), so: \[ W = \int_{V_1}^{V2} P\,dV = \int{V_1}^{V_2} \frac{nRT}{V}\,dV = nRT \ln\left(\frac{V_2}{V_1}\right) \] \[ W = (2.00)(8.31)(300) \ln(2.00) = (4,986)(0.6931) = 3.46 \times 10^3\ \text{J} \]

Step 3 — Heat. From the First Law with \(\Delta U = 0\): \[ Q = W = 3.46 \times 10^3\ \text{J} \]

Physical interpretation: The gas expands and does work on the surroundings. Since internal energy is constant (temperature is fixed), the energy for that work must come entirely from heat flowing in from the reservoir. The gas acts as an energy conduit: heat flows in from the hot reservoir and immediately leaves as work.

Worked Example: Adiabatic Process (\(Q = 0\))

Setup: The same 2.00 mol of monatomic ideal gas at \(T_1 = 300\ \text{K}\) and \(V_1 = 0.0100\ \text{m}^3\) expands adiabatically (perfectly insulated — no heat exchange) to \(V_2 = 0.0200\ \text{m}^3\).

Step 1 — Find final temperature. For an adiabatic process of an ideal gas: \(TV^{\gamma - 1} = \text{constant}\). For monatomic gas, \(\gamma = 5/3\), so \(\gamma - 1 = 2/3\): \[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = 300 \left(\frac{1}{2}\right)^{2/3} \] \[ T_2 = 300 \times (0.500)^{0.667} = 300 \times 0.630 = 189\ \text{K} \] The gas cools significantly — expansion without heat input requires the gas to draw on its own internal energy.

Step 2 — Heat. \[ Q = 0 \quad\text{(by definition of adiabatic)} \]

Step 3 — Internal energy change. \[ \Delta U = nC_V\Delta T = (2.00)(12.5)(189 - 300) = (2.00)(12.5)(-111) = -2.78 \times 10^3\ \text{J} \]

Step 4 — Work (from First Law). \[ \Delta U = Q - W \quad\Rightarrow\quad -2,780 = 0 - W \quad\Rightarrow\quad W = 2.78 \times 10^3\ \text{J} \]

Physical interpretation: The gas does 2,780 J of work expanding against the surroundings. Since no heat flows in, this energy comes entirely from the gas's internal energy, causing its temperature to drop from 300 K to 189 K. Adiabatic expansion is how refrigerators and air conditioners cool — compress a gas (it heats up), let it shed that heat, then let it expand adiabatically (it cools below ambient).

Comparison across processes (same 2 mol, \(\Delta V\) from 0.01 to 0.02 m³):

ELI-10: Explain It Like I'm 10

The First Law is an energy balance sheet. Internal energy is the total energy stored in a substance's molecules. You can change it two ways: add heat OR do work on it (or let it do work on something else). Squeeze a gas and it heats up — you did work on it so its internal energy increased. Let it expand and it cools — it did work on the surroundings. Heat and work are both ways to transfer energy.

9.5 Second Law of Thermodynamics and Entropy

Core Idea

Heat flows spontaneously from hot to cold, never the reverse. Entropy — a measure of disorder — always increases in an isolated system.

Key Concepts

• Entropy \(S\): \(\Delta S = \int \frac{dQ}{T}\) (reversible path). Increases in all real (irreversible) processes.
• Second Law: \(\Delta S_{\text{universe}} \geq 0\) for any process.
• Heat engines: take heat \(Q_H\) from a hot reservoir, do work \(W\), dump waste heat \(Q_C\) to a cold reservoir. Efficiency: \(e = W/Q_H = 1 - Q_C/Q_H\).
• Maximum (Carnot) efficiency: \(e_{\text{Carnot}} = 1 - T_C/T_H\) (temperatures in Kelvin). No real engine can exceed this.
• Refrigerators and heat pumps are heat engines run backward; they use work to move heat from cold to hot.

Detailed Worked Example: Heat Engine Efficiency

Problem: A heat engine operates between a hot reservoir at \(T_H = 600\ \text{K}\) and a cold reservoir at \(T_C = 300\ \text{K}\). In each cycle, it extracts \(Q_H = 500\ \text{J}\) of heat from the hot reservoir, performs \(W = 180\ \text{J}\) of useful work, and rejects the remainder to the cold reservoir.

(a) How much heat is rejected to the cold reservoir per cycle?

From energy conservation (First Law applied to a cycle, where \(\Delta U = 0\)): \[ Q_H = W + Q_C \quad\Rightarrow\quad Q_C = Q_H - W = 500 - 180 = 320\ \text{J} \]

(b) What is the actual thermal efficiency?

\[ e = \frac{W}{Q_H} = \frac{180}{500} = 0.36 = 36\% \]

This means 36% of the heat extracted from the hot reservoir becomes useful work; the remaining 64% is wasted to the cold reservoir.

(c) What is the maximum possible (Carnot) efficiency for these reservoir temperatures?

\[ e_{\text{Carnot}} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 1 - 0.50 = 0.50 = 50\% \]

(d) How does the actual engine compare to the Carnot ideal?

The actual engine achieves 36% out of a theoretically possible 50%. It operates at \(36/50 = 72\%\) of the Carnot limit. The gap (from 36% to 50%) is due to irreversibilities — friction, turbulence, finite temperature differences during heat transfer, and other dissipative effects that generate entropy.

(e) What is the entropy change of the universe per cycle?

For a cyclic engine, \(\Delta S_{\text{engine}} = 0\) (it returns to the same state). The reservoirs, however, experience entropy changes:

• Hot reservoir loses \(Q_H = 500\ \text{J}\) at \(T_H = 600\ \text{K}\): \(\Delta S_H = -500/600 = -0.833\ \text{J/K}\).
• Cold reservoir gains \(Q_C = 320\ \text{J}\) at \(T_C = 300\ \text{K}\): \(\Delta S_C = +320/300 = +1.067\ \text{J/K}\).

\[ \Delta S_{\text{universe}} = \Delta S_H + \Delta SC + \Delta S{\text{engine}} = -0.833 + 1.067 + 0 = +0.233\ \text{J/K} \]

The positive sign confirms the Second Law: every real engine cycle increases the entropy of the universe. For a Carnot engine, we would have \(Q_C = Q_H \times (T_C/TH) = 500 \times (300/600) = 250\ \text{J}\), and \(\Delta S{\text{universe}} = -500/600 + 250/300 = -0.833 + 0.833 = 0\) — the Carnot cycle is reversible and produces zero net entropy, giving the maximum possible efficiency.

ELI-10: Explain It Like I'm 10

The Second Law is why things get messier. An egg can break and scramble, but a scrambled egg never unscrambles. A cup of hot coffee cools to room temperature; a cup of room-temperature coffee never spontaneously gets hot. Entropy is the universe's tendency toward disorder. You can create local order (building a house) but only by making even more disorder somewhere else (burning fuel, heating the air). A perfect engine — one that turns ALL heat into work — is impossible because you must always dump some waste heat into a cold reservoir. Even the best possible engine (the Carnot engine) cannot achieve 100% efficiency unless the cold reservoir is at absolute zero — which is unreachable.

Topic Summary

• Temperature reflects average molecular kinetic energy. Heat is energy flow due to \(\Delta T\).
• Zeroth Law: thermal equilibrium is transitive → temperature is well-defined.
• Heat is not a substance (caloric theory is wrong). Heat is energy in transit; internal energy is what objects store.
• Ideal Gas Law: \(PV = nRT\). Kinetic theory: \(P\) arises from molecular collisions. Proportional reasoning with the combined gas law: \(P_1V_1/T_1 = P_2V_2/T_2\).
• First Law: \(\Delta U = Q - W\). Energy is conserved; internal energy changes via heat and work.
• Processes: isobaric (\(P\) constant, \(W = P\Delta V\)), isochoric (\(V\) constant, \(W = 0\)), isothermal (\(T\) constant, \(\Delta U = 0\)), adiabatic (\(Q = 0\), \(PV^\gamma = \text{constant}\)).
• PV diagrams: work is the area under the \(P(V)\) curve. Cycle net work is the enclosed area.
• Second Law: entropy never decreases in an isolated system. Heat flows hot → cold.
• Heat engine efficiency: \(e = 1 - Q_C/QH\). Carnot max: \(e{\text{max}} = 1 - T_C/T_H\).

Essential Equations

Concept Check

1. Why does ice at 0°C cool a drink more effectively than water at 0°C?
2. A gas expands isothermally. What happens to its internal energy? Where does the energy for the work come from?
3. Can a heat engine ever be 100% efficient? Why, specifically?
4. If you leave a refrigerator door open in a closed room, does the room get warmer or cooler? Explain.
5. Two identical blocks — one at 400 K, one at 200 K — are brought into thermal contact in an isolated box. What is the final temperature? Did entropy increase?
6. A sealed syringe contains air at room temperature. You pull the plunger out rapidly (adiabatic expansion). Does the air inside warm up, cool down, or stay the same temperature? Explain using the First Law.
7. An ideal gas is compressed isobarically to half its original volume. By what factor does its temperature change?
8. Why is \(C_P\) always greater than \(C_V\) for an ideal gas? (Hint: think about where the extra energy goes.)

Open Educational References

• OpenStax, College Physics, Chapters 13–15: Thermodynamics
• OpenStax, University Physics, Volume 2, Chapters 1–4: Thermodynamics