Notes
Kinetic theory and the ideal gas law
- Temperature measures the average kinetic energy of particles: $\displaystyle K_{avg} = \frac{3}{2}k_BT$ (use kelvin).
- Ideal gas law: $PV = nRT = Nk_BT$.
- Pressure comes from molecules colliding with the container walls.
Heat and thermal equilibrium
- Heat ($Q$) is energy transferred because of a temperature difference. Objects in contact reach thermal equilibrium (the same temperature).
- Heat transfer by conduction (contact), convection (moving fluid), and radiation (electromagnetic waves).
- Conduction rate: $\displaystyle \frac{Q}{\Delta t} = \frac{kA\Delta T}{L}$. Thicker material or lower conductivity means slower heat flow.
- Specific heat: $Q = mc\Delta T$.
The first law of thermodynamics
where $Q$ is heat added to the gas and $W$ is work done on the gas. On a $PV$ diagram, the work done by the gas is the area under the curve.
| Process | Constant | Key fact |
|---|---|---|
| Isobaric | Pressure | $W = -P\Delta V$ |
| Isochoric (isovolumetric) | Volume | $W = 0$, so $\Delta U = Q$ |
| Isothermal | Temperature | $\Delta U = 0$, so $Q = -W$ |
| Adiabatic | No heat flow | $Q = 0$, so $\Delta U = W$ |
- For a cycle that returns to its start, $\Delta U = 0$.
The second law and entropy
- Heat flows naturally from hot to cold.
- The entropy of an isolated system never decreases.
- Heat engines can't convert all heat into work: efficiency $\displaystyle e = \frac{W}{Q_H} < 1$.
Practice
$K_{avg} \propto T$.
Constant volume means $W = 0$ and $\Delta U = Q$.
$\Delta U = Q + W = 200 + 50$.
Constant temperature means constant internal energy for an ideal gas.
Net work is the enclosed area.
Energy passes through the solid by particle collisions.
$P \propto T$ at constant volume.
$\displaystyle e = \frac{300}{1000}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A gas expands at a constant pressure of $2.0 \times 10^5$ Pa from 0.010 m³ to 0.030 m³ while 9,000 J of heat is added.
(a) Calculate the work done on the gas.
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$W = -P\Delta V = -(2.0 \times 10^5)(0.020) = -4{,}000$ J.
Check off the points you earned:
(b) Calculate the change in internal energy.
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$\Delta U = Q + W = 9{,}000 - 4{,}000 = 5{,}000$ J.
Check off the points you earned:
(c) Did the gas's temperature increase, decrease, or stay the same? Explain.
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Increased. Internal energy of an ideal gas depends on temperature, and $\Delta U > 0$. (Also, $PV$ tripled at constant $P$, so $T$ tripled.)
Check off the points you earned: