Thermodynamics (HL): The Mechanics of Heat and Work
Welcome to the deep dive into Thermodynamics! If you’ve successfully navigated Gas Laws and Thermal Energy Transfer, this is where we put it all together to understand how energy truly flows and changes form within a system, specifically gases.
This topic is the ultimate link between the microscopic world (the motion of particles) and the macroscopic world (measurable quantities like pressure, volume, and temperature). Don't worry if the concepts seem abstract—we will break down the crucial relationship between heat, internal energy, and mechanical work. This is core HL material, so let’s get focused!
B.4 Thermodynamics (HL Extension)
1. The First Law of Thermodynamics
The First Law of Thermodynamics is simply the principle of conservation of energy applied to thermal systems. It states that energy cannot be created or destroyed, only transferred or transformed.
When heat energy \(Q\) is supplied to a system (like a fixed amount of gas), that energy must be accounted for by two things:
- An increase in the system's internal energy (\(\Delta U\)).
- Work done (\(W\)) by the system on its surroundings.
The mathematical statement of the First Law is:
\[Q = \Delta U + W\]
Let’s define the key terms clearly:
- \(Q\) (Heat Transfer): The energy transferred due to a temperature difference.
- \(\Delta U\) (Change in Internal Energy): The change in the total random kinetic energy (and potential energy) of all the particles in the system. For an ideal gas, internal energy depends ONLY on temperature.
- \(W\) (Work Done): The mechanical energy transferred when the system (the gas) changes its volume against an external pressure.
Crucial Concept: The Sign Convention
Getting the signs right is the single biggest challenge in thermodynamics. We must be consistent about energy entering or leaving the system.
Quick Review: IB Sign Convention
The standard convention (which IB uses) defines all terms relative to the system:
- Heat \(Q\):
- \(Q > 0\) (Positive): Heat is supplied TO the system. (Energy enters)
- \(Q < 0\) (Negative): Heat is removed FROM the system. (Energy leaves)
- Internal Energy \(\Delta U\):
- \(\Delta U > 0\): Internal energy increases. Temperature rises.
- \(\Delta U < 0\): Internal energy decreases. Temperature falls.
- Work \(W\):
- \(W > 0\) (Positive): Work is done BY the system (e.g., gas expands). (Energy leaves as mechanical work)
- \(W < 0\) (Negative): Work is done ON the system (e.g., gas is compressed). (Energy enters as mechanical work)
Analogy Aid: Think of \(\Delta U\) as your bank account balance. \(Q\) and \(W\) are deposits or withdrawals. If the system does work (\(W > 0\)), it's like a withdrawal, reducing its internal energy unless heat (\(Q\)) is added to cover the cost.
Key Takeaway: The First Law is Energy Conservation: \(Q = \Delta U + W\). Master the sign conventions immediately!
2. Work Done by a Gas and P-V Diagrams
In thermodynamics, work done by a gas is a result of a change in its volume against an external pressure.
Work Done during a Volume Change
If the pressure \(P\) remains constant (isobaric process), the work done \(W\) when a volume changes by \(\Delta V\) is given by:
\[W = P \Delta V\]
If the volume increases (\(\Delta V > 0\)), the gas expands and does positive work (\(W > 0\)). If the volume decreases (\(\Delta V < 0\)), the gas is compressed and negative work is done by the gas (or positive work is done *on* the gas).
Interpreting P-V Diagrams
For any thermodynamic process, whether pressure is constant or changing, the work done by the gas is represented graphically on a Pressure-Volume (\(P-V\)) diagram.
The work done by the gas is equal to the area under the curve on the \(P-V\) graph.
- If the process moves from a small volume to a large volume (expansion, moving right on the graph), \(W\) is positive.
- If the process moves from a large volume to a small volume (compression, moving left on the graph), \(W\) is negative.
Work in Cyclic Processes
Many thermodynamic engines operate in a cycle, returning the gas to its initial state (A \(\rightarrow\) B \(\rightarrow\) A).
- In a cyclic process, since the initial and final states are the same, the temperature is the same, meaning the change in internal energy is zero: \(\Delta U = 0\).
- Therefore, the First Law simplifies to \(Q = W\). The net heat supplied equals the net work done.
- On a \(P-V\) diagram, the net work done during a cycle is the area enclosed by the loop.
- If the cycle is traversed clockwise (like in an engine), the net work \(W\) is positive (net work done BY the gas).
- If the cycle is traversed counter-clockwise (like in a refrigerator or heat pump), the net work \(W\) is negative (net work done ON the gas).
Key Takeaway: Work is the area under the \(P-V\) curve. For a cycle, the net work is the area inside the loop, and \(\Delta U\) is zero.
3. Specific Thermodynamic Processes
To analyze a gas system, we often study four specific processes where one variable is held constant. You must know how the First Law applies to each one.
| Process Name | Constant Variable | Defining Feature | First Law (\(Q = \Delta U + W\)) |
|---|---|---|---|
| Isochoric | Volume (\(V\)) | \(\Delta V = 0\); Work \(W = 0\) | \(Q = \Delta U\) |
| Isobaric | Pressure (\(P\)) | \(P = \text{constant}\); \(W = P\Delta V\) | \(Q = \Delta U + P\Delta V\) |
| Isothermal | Temperature (\(T\)) | \(\Delta T = 0\); Internal Energy \(\Delta U = 0\) | \(Q = W\) |
| Adiabatic | No Heat Transfer | \(Q = 0\) (Insulated system) | \(\Delta U = -W\) |
Detailing the Processes
1. Isochoric (Constant Volume)
- Meaning: The gas is in a rigid container.
- Since the volume doesn't change, no work is done (\(W=0\)).
- Any heat supplied goes entirely into changing the internal energy (and thus the temperature).
- P-V Diagram: A vertical line.
2. Isobaric (Constant Pressure)
- Meaning: The gas is in a container with a freely moving piston, exposed to constant atmospheric pressure.
- Heat supplied is used both to increase internal energy and to do work against the constant pressure.
- P-V Diagram: A horizontal line.
3. Isothermal (Constant Temperature)
- Meaning: The gas is kept in perfect thermal contact with a large heat reservoir (e.g., an ice bath) so its temperature never changes.
- Since temperature is constant, \(\Delta U = 0\).
- Any heat added must equal the work done. If the gas expands and does work (\(W>0\)), it must absorb heat (\(Q>0\)).
- P-V Diagram: A curve following Boyle's Law (\(PV = \text{constant}\)).
4. Adiabatic (No Heat Transfer)
- Meaning: The gas is perfectly insulated, or the process happens so quickly that heat doesn't have time to flow (e.g., quickly pumping a bicycle tire or engine compression).
- \(Q=0\). This means \(\Delta U = -W\).
- If the gas expands (\(W>0\)), its internal energy drops (\(\Delta U < 0\)), and the gas cools down.
- If the gas is compressed (\(W<0\)), its internal energy rises (\(\Delta U > 0\)), and the gas heats up.
-
P-V Diagram: A steeper curve than an isothermal process. The relationship is defined by:
\[PV^\gamma = \text{constant}\]
where \(\gamma\) (gamma) is the adiabatic index (ratio of specific heat capacities).
Common Mistake to Avoid: Confusing Isothermal (\(\Delta T=0\)) with Adiabatic (\(Q=0\)). They look similar on a P-V graph but represent completely different physical constraints!
4. Specific Heat Capacities for Gases (HL Only)
When we apply heat to a solid or liquid, the volume change is usually negligible. For gases, however, the volume change is significant, which means the amount of heat required to raise the temperature depends entirely on *how* you heat the gas.
This leads to two distinct molar specific heat capacities for gases:
a) Molar Specific Heat at Constant Volume, \(C_v\)
This is the heat needed to raise the temperature of one mole of gas by 1 K while keeping the volume constant (isochoric process).
- Since \(W=0\) (no volume change), all the heat supplied goes directly into increasing the internal energy (\(Q = \Delta U\)).
b) Molar Specific Heat at Constant Pressure, \(C_p\)
This is the heat needed to raise the temperature of one mole of gas by 1 K while keeping the pressure constant (isobaric process).
- Since the gas expands, it does positive work (\(W > 0\)).
- Therefore, the heat supplied (\(Q\)) must cover both the increase in internal energy (\(\Delta U\)) and the work done (\(W\)).
Why \(C_p\) is greater than \(C_v\):
To achieve the same temperature rise (\(\Delta T\)), the change in internal energy \(\Delta U\) is the same in both cases. However, at constant pressure, you need *extra* energy to push back the piston (do the work \(W\)). Thus:
\[C_p > C_v\]
Mayer's Relation
For an ideal gas, the difference between these two specific heat capacities is equal to the Ideal Gas Constant (\(R\)). This is known as Mayer's Relation:
\[C_p - C_v = R\]
where \(R \approx 8.31 \, \text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}\). This powerful equation connects the macroscopic measurement of specific heat capacity to fundamental constants.
Did you know?
For a monatomic ideal gas (like Helium), the internal energy only involves translational kinetic energy. Physics theory (equipartition theorem) dictates that for monatomic gases, \(C_v = \frac{3}{2} R\). Using Mayer's relation, we can predict that \(C_p\) must be \(\frac{5}{2} R\)! This theoretical prediction matches experimental observations beautifully.
Key Takeaway: Because a gas must do work when heated at constant pressure, \(C_p\) is always greater than \(C_v\). The difference is \(R\).
Final Study Checklist for Thermodynamics (HL)
If you can confidently answer the following, you are ready for your exam:
- State the First Law of Thermodynamics and use the correct sign conventions for \(Q\), \(\Delta U\), and \(W\).
- Calculate work done using the formula \(W = P \Delta V\) and by finding the area under curves in \(P-V\) diagrams.
- Analyze how \(Q\), \(\Delta U\), and \(W\) relate to each other in isochoric, isobaric, isothermal, and adiabatic processes.
- Explain why \(C_p\) must be greater than \(C_v\) for an ideal gas and recall Mayer's relation.
You've got this! Thermodynamics links heat, energy, and mechanics—it's challenging but incredibly rewarding once the pieces click together. Keep practicing those \(P-V\) diagrams!