Notes
Kinetic molecular theory and pressure
- Gas particles are far apart, move randomly, collide elastically and do not attract. Average kinetic energy is proportional to Kelvin temperature.
- Pressure comes from collisions with walls. $1 \text{ atm} = 760 \text{ mmHg} = 760 \text{ torr} = 101.3 \text{ kPa}$.
- Always convert temperature: $K = °C + 273$.
Gas laws
| Law | Constant | Equation | Relationship |
|---|---|---|---|
| Boyle | $T$, $n$ | $P_1V_1 = P_2V_2$ | inverse |
| Charles | $P$, $n$ | $\displaystyle \frac{V_1}{T_1} = \frac{V_2}{T_2}$ | direct |
| Gay-Lussac | $V$, $n$ | $\displaystyle \frac{P_1}{T_1} = \frac{P_2}{T_2}$ | direct |
| Combined | $n$ | $\displaystyle \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ | all three |
| Ideal | none | $PV = nRT$ | one set of conditions |
- $R = 0.0821$ L·atm/(mol·K). At STP (273 K, 1 atm), 1 mol of gas fills 22.4 L.
Mixtures
- Dalton's law: $P_{\text{total}} = P_1 + P_2 + \ldots$, and $\displaystyle P_A = \frac{n_A}{n_{\text{total}}} P_{\text{total}}$.
- Gas collected over water: subtract water vapor pressure.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \frac{1520}{760} = 2.00$ atm.
$\displaystyle V_2 = \frac{(2.0)(8.0)}{4.0} = 4.0$ L.
Use kelvins: 293 K and 313 K. $\displaystyle P_2 = 2.00 \times \frac{313}{293} = 2.14$ atm. 4.00 atm comes from using °C.
$\displaystyle n = \frac{PV}{RT} = \frac{(2.00)(10.0)}{(0.0821)(300)} = 0.812$ mol.
$\displaystyle V_2 = 10.0 \times \frac{1.20}{0.800} \times \frac{400}{300} = 20.0$ L.
$P_{\text{N}_2} = 0.78 \times 1.00 = 0.78$ atm.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A weather balloon is filled with 0.200 mol of helium at 25 °C and 1.00 atm.
(a) Find the volume of the balloon at launch.
Show the answer and scoring guide
$T = 25 + 273 = 298$ K. $\displaystyle V = \frac{nRT}{P} = \frac{(0.200)(0.0821)(298)}{1.00} = 4.89$ L.
Check off the points you earned:
(b) The balloon rises to where the pressure is 0.500 atm and the temperature is -23 °C. Assuming no helium escapes, find the new volume.
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$T_2 = -23 + 273 = 250$ K. $\displaystyle V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 4.89 \times \frac{1.00}{0.500} \times \frac{250}{298} = 8.21$ L. (Check with $PV = nRT$: $\displaystyle \frac{(0.200)(0.0821)(250)}{0.500} = 8.21$ L.)
Check off the points you earned:
(c) Use kinetic molecular theory to explain why the balloon expands as it rises, even though the temperature drops.
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Outside pressure drops by half, so fewer collisions push in on the balloon; the helium particles push the walls outward until the inside and outside pressures match. The lower temperature slows the particles and would shrink the balloon somewhat, but the pressure drop has a larger effect, so the volume increases overall.
Check off the points you earned: