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: Cp = Cv + 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:
Volume of the room: 6 \text{ m} \times 4 \text{ m} \times 3 \text{ m} = 72 \text{ m}^3.
Convert volume to liters: 72 \text{ m}^3 \times \frac{1000 \text{ L}}{1 \text{ m}^3} = 72,000 \text{ L}.
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} .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}.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{V{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 (P0) from an initial volume (V{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{V{initial}}^{V{final}} P0 dV = -P0 \int{V{initial}}^{V{final}} dV
W = -P0 (V{final} - V{initial}) = -P0 \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):
Start with the First Law: \Delta E = Q + W
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).
Substitute \Delta E and W into the First Law: \frac{3}{2} nR \Delta T = Q + (-P_0 \Delta V).
Apply the Ideal Gas Law (PV = nRT) to relate \Delta V and \Delta T at constant pressure: P_0 \Delta V = nR \Delta T.
Substitute into the First Law equation: \frac{3}{2} nR \Delta T = Q - nR \Delta T.
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.
Connection to Cp: This result (\frac{5}{2} nR \Delta T) matches the heat absorbed at constant pressure for a monatomic gas, where Q = nCp \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.
Start with the work formula: W = -\int{V{initial}}^{V_{final}} P dV
From the Ideal Gas Law, express P as a function of V: P = \frac{nRT}{V}.
Substitute P into the integral: W = -\int{V{initial}}^{V_{final}} \frac{nRT}{V} dV
Since n, R, and T are constant in an isothermal process, they can be taken out of the integral:
W = -nRT \int{V{initial}}^{V_{final}} \frac{1}{V} dVThe integral of \frac{1}{V} is \ln|V|. Apply the limits:
W = -nRT [\ln V]{V{initial}}^{V_{final}}Evaluate the definite integral:
W = -nRT (\ln V{final} - \ln V{initial})Using logarithm properties, this simplifies to:
W = -nRT \ln \left(\frac{V{final}}{V{initial}}\right)
Note: This formula calculates the work done on the gas during an isothermal process.