Thermodynamics: Work, Heat Capacity, and Processes

Superheating and Supercooling and Review of Thermodynamics

Superheating and Supercooling

• Supercooling: A phenomenon where a liquid, such as highly purified water, remains in a liquid state below its normal freezing point (e.g., below $0^ ext{o}C$).

  • Mechanism: Requires the absence of nucleation sites (imperfections, dust particles, air bubbles) for ice crystals to form.

  • Initiation of Freezing: Tapping the container or pouring the supercooled water over a small piece of ice provides these nucleation sites, causing rapid crystallization.

  • Demonstration: Pouring supercooled water over a bit of ice (which has an imperfect structure) creates an ice sculpture, whereas pouring it over a smooth plastic surface (lacking imperfections) does not.

  • DIY Tips: Use unopened, smooth-sided water bottles (like Fiji water) and place them carefully in a freezer. Note that supercooling might not always occur or could be partial (some liquid, some ice).

• Superheating: A phenomenon where a liquid is heated above its boiling point without boiling.

  • Safety Warning: This is dangerous and should not be attempted at home, as sudden boiling can lead to explosive release of hot liquid (e.g., hot water exploding from a microwave).

  • Safety Precaution: When heating water in a microwave, insert a wooden skewer or stir the water before removing it to create nucleation sites and enable safe boiling.

Review of Heat Capacities

• Context: For calculations of heat changes, especially with gases, moles ($n$) are often used rather than mass in kilograms.

• Molar Heat Capacities: These are specific heat capacities per mole.

  • $C_v$: Molar heat capacity at constant volume.

    • Monatomic Gas: $C_v = \frac{3}{2}R$

    • Diatomic Gas: $C_v = \frac{5}{2}R$ (due to two additional degrees of freedom compared to monatomic)

    • Polyatomic Gas: $C_v = 3R$ or $\frac{6}{2}R$ (due to one additional degree of freedom compared to diatomic)

    • $R$ is the ideal gas constant $(8.31 \text{ J/(mol·K)})$.

  • $C_p$: Molar heat capacity at constant pressure.

    • Relationship: $C<em>p = C</em>v + R$

    • Monatomic Gas: $C_p = \frac{5}{2}R$

    • Diatomic Gas: $C_p = \frac{7}{2}R$

    • Polyatomic Gas: $C_p = 4R$

  • Note: These theoretical values for gases are easy to remember based on degrees of freedom and are very close to experimental values at typical temperatures.

Example Problem: Heating a Cabin

• Scenario: A cabin (room dimensions $6\text{ m} \times 4\text{ m} \times 3\text{ m}$) starts at $0^ ext{o}C$. An electric heater of $2\text{ kW}$ power is used to raise the room temperature to $21^ ext{o}C$.

• Goal: Calculate the time ($t$) required to heat the room.

• Assumptions for Calculation: (Note: These are ideal and not physically realistic)

  • The room is perfectly airtight, meaning the volume of air is constant.

  • All heat from the heater is absorbed solely by the air, with no heat loss through walls or absorption by furniture.

• Approach: Time is related to power and total heat absorbed ($Q$) by $\text{Power} = \frac{\text{Energy}}{\text{Time}}$, so $t = \frac{Q}{P}$.

  • Process Type: Constant volume (airtight room), so $C_v$ is used.

  • Gas Type: Air, which is primarily nitrogen and oxygen, so it's treated as a diatomic gas. Thus, $C_v = \frac{5}{2}R$.

• Calculations:

  1. Volume of the room: $6 \text{ m} \times 4 \text{ m} \times 3 \text{ m} = 72 \text{ m}^3$.

  2. Convert volume to liters: $72 \text{ m}^3 \times \frac{1000 \text{ L}}{1 \text{ m}^3} = 72,000 \text{ L}$.

  3. Number of moles ($n$) of air: Assuming 1 mole of ideal gas occupies $22.4 \text{ L}$ at STP:
     $n = \frac{72,000 \text{ L}}{22.4 \text{ L/mol}} \approx 3.2 \times 10^3 \text{ mol}$ .

  4. Heat required ($Q$): Using $Q = nC_v \Delta T$
     $Q = (3.2 \times 10^3 \text{ mol}) \times (\frac{5}{2} \times 8.31 \text{ J/(mol·K)}) \times (21^ ext{o}C - 0^ ext{o}C) \approx 1.4 \times 10^6 \text{ J}$.

  5. Time ($t$) to heat the room: Using $t = \frac{Q}{P}$
     $t = \frac{1.4 \times 10^6 \text{ J}}{2 \text{ kW}} = \frac{1.4 \times 10^6 \text{ J}}{2000 \text{ J/s}} = 700 \text{ s} \approx 12 \text{ minutes}$.

• Realism Check: The calculated time of $12$ minutes is highly unrealistic.

  • Personal Experience: In reality, it took several hours (3-4 hours) to heat a similar room.

  • Unrealistic Assumptions: The key simplifying assumptions are major sources of error:

    • Heat Capacity of Furniture/Walls: Furniture, walls, and other objects in the room have significant heat capacities and absorb substantial amounts of energy, which was ignored.

    • Heat Loss: Heat inevitably escapes through walls, windows, and cracks, which was ignored.

    • Airtight Room: A truly airtight room would experience a significant increase in internal pressure as temperature rises (according to the ideal gas law $P \propto T$ at constant $V$), potentially creating immense forces on walls. A more realistic model would assume the room is not perfectly airtight and some air escapes, allowing pressure to remain relatively constant (an isobaric process).

The First Law of Thermodynamics

• General Formulation: The change in internal energy $\Delta E$ of a system is equal to the heat ($Q$) added to the system plus the work ($W$) done on the system.

  • Expressed as: $\Delta E = Q + W$

  • For infinitesimal changes: $dE = dQ + dW$

• Meaning: This law represents the conservation of energy. Any energy input (heat or work) changes the internal energy of the system.

• Assumption for this course: The number of particles ($n$) in the system remains constant.

• If $n$ changes: The first law includes a chemical potential term: $dE = dQ + dW + \mu dn$, where $\mu$ (chemical potential) accounts for energy change from adding or removing particles. This is typically not covered in this course.

• Work ($W$) Definition: Work done on a gas is generally given by the integral: $W = -\int<em>{V</em>{initial}}^{V_{final}} P dV$

  • The negative sign indicates that work done by the gas (expansion) is positive, so work done on the gas is negative.

  • Pressure ($P$) in the integral can be a constant or a function of volume ($V$), depending on the process.

• Goal: To calculate $\Delta E$, $Q$, and $W$ for various thermodynamic processes.

Thermodynamic Processes

1. Isobaric Process (Constant Pressure)

• Definition: A process where the pressure ($P$) of the system remains constant.

• PV Diagram: A graph showing pressure (y-axis) vs. volume (x-axis) is called a PV diagram.

  • For an isobaric process, it's a horizontal line at a constant pressure ($P<em>0$) from an initial volume ($V</em>{initial}$) to a final volume ($V_{final}$), with an arrow indicating the direction of the process.

• Work ($W$) Calculation: Work done on the gas during an isobaric process is straightforward:

  • $W = -\int<em>{V</em>{initial}}^{V<em>{final}} P</em>0 dV = -P<em>0 \int</em>{V<em>{initial}}^{V</em>{final}} dV$

  • $W = -P<em>0 (V</em>{final} - V<em>{initial}) = -P</em>0 \Delta V$

  • Interpretation: If the gas expands (\Delta V > 0), $W$ is negative, meaning work is done by the gas. If the gas is compressed (\Delta V < 0), $W$ is positive, meaning work is done on the gas.

• Maintaining Constant Pressure During Expansion: When a gas expands, its pressure would naturally decrease. To keep the pressure constant during expansion, heat must be added to the system.

• Heat ($Q$) Calculation (e.g., for a monatomic ideal gas):

  1. Start with the First Law: $\Delta E = Q + W$

  2. For a monatomic ideal gas, the change in internal energy is: $\Delta E = \frac{3}{2} nR \Delta T$ (Internal energy depends only on temperature and degrees of freedom, not on constant pressure/volume conditions).

  3. Substitute $\Delta E$ and $W$ into the First Law: $\frac{3}{2} nR \Delta T = Q + (-P_0 \Delta V)$.

  4. Apply the Ideal Gas Law ($PV = nRT$) to relate $\Delta V$ and $\Delta T$ at constant pressure: $P_0 \Delta V = nR \Delta T$.

  5. Substitute into the First Law equation: $\frac{3}{2} nR \Delta T = Q - nR \Delta T$.

  6. Solve for $Q$: $Q = \frac{3}{2} nR \Delta T + nR \Delta T = (\frac{3}{2} + 1) nR \Delta T = \frac{5}{2} nR \Delta T$.

  7. Connection to $C<em>p$: This result ($\frac{5}{2} nR \Delta T$) matches the heat absorbed at constant pressure for a monatomic gas, where $Q = nC</em>p \Delta T$ and $C_p = \frac{5}{2}R$ for a monatomic gas.

2. Isothermal Process (Constant Temperature)

• Definition: A process where the temperature ($T$) of the system remains constant.

• PV Diagram: Since $T$ is constant, the Ideal Gas Law ($PV = nRT$) implies that $PV = \text{constant}$. This means $\text{Pressure} \propto \frac{1}{\text{Volume}}$.

  • The graph on a PV diagram is a hyperbola (a curve), often referred to as an isotherm.

  • As volume increases (expansion), pressure decreases along the curve.

• Work ($W$) Calculation: For an isothermal process, pressure is not constant, so a full integral is required.

  1. Start with the work formula: $W = -\int<em>{V</em>{initial}}^{V_{final}} P dV$

  2. From the Ideal Gas Law, express $P$ as a function of $V$: $P = \frac{nRT}{V}$.

  3. Substitute $P$ into the integral: $W = -\int<em>{V</em>{initial}}^{V_{final}} \frac{nRT}{V} dV$

  4. Since $n$, $R$, and $T$ are constant in an isothermal process, they can be taken out of the integral:
     $W = -nRT \int<em>{V</em>{initial}}^{V_{final}} \frac{1}{V} dV$

  5. The integral of $\frac{1}{V}$ is $\ln|V|$. Apply the limits:
     $W = -nRT [\ln V]<em>{V</em>{initial}}^{V_{final}}$

  6. Evaluate the definite integral:
     $W = -nRT (\ln V<em>{final} - \ln V</em>{initial})$

  7. Using logarithm properties, this simplifies to:
     $W = -nRT \ln \left(\frac{V<em>{final}}{V</em>{initial}}\right)$

• Note: This formula calculates the work done on the gas during an isothermal process.