AvoGadro's Law

Avogadro's Law

Definition and Overview

• Avogadro's Law describes the relationship between the amount of gas (in moles) and the volume of a gas sample.

• A gas sample refers to a quantity of gas contained in any container, such as a balloon holding air.

Key Concepts

• Amount of Gas: Represented as $n$ (in moles), indicating the quantity of gas present.

• Volume of Gas: Represented as $V$, indicating the space the gas occupies.

Relationship Between Amount of Gas and Volume

• The relationship observed is such that:

  • Direct Relationship: When the amount of gas increases, the volume increases as well. Similarly, when the amount of gas decreases, the volume decreases.

Example Scenario - Balloon

• Imagine a balloon filled with air:

  • When air is added (increasing $n$), the balloon expands (increasing $V$).

  • When air is released (decreasing $n$), the balloon contracts (decreasing $V$).

Graphical Representation

• On a graph:

  • X-axis: Amount of gas (moles)

  • Y-axis: Volume

• When the amount of gas doubles, the volume also doubles, showing the proportional relationship.

Constants in Avogadro's Law

• While observing the relationship between amount and volume:

  • Temperature must be held constant. No changes in temperature should occur during the experiment.

  • Pressure must also be constant (e.g., not changing the altitude which affects gas pressure).

Apparatus for Demonstrating Avogadro’s Law

• A common apparatus used to visualize Avogadro's Law is a canister with a slidable top, such as a modified soda bottle:

  • When gas is added, the top slides up to increase the volume without changing the pressure.

  • When gas is removed, the top slides down, thus reducing the volume while maintaining constant pressure.

Mathematical Representation

• The most common equation representing Avogadro's Law is:
  $\frac{N_1}{V_1} = \frac{N_2}{V_2}$

• Where:

  • $N_1$: Initial amount of gas (moles)

  • $V_1$: Initial volume

  • $N_2$: Final amount of gas (moles)

  • $V_2$: Final volume

Example Problem

• Given: 50 grams of Oxygen ($O_2$) occupies 48 liters.

• Pressure must remain constant, and we need to determine how many grams of gas are present when the volume increases to 79 liters.

• Assumptions:

  • Initial calculation for moles is necessary, as grams must be converted to moles using molar mass.

Conversion of Grams to Moles

• The molar mass of oxygen ($O_2$):
  $32 ext{ g/mol} = 2 imes 16 ext{ g/mol}$ for the two oxygen atoms.

• Converting 50 grams to moles: $n_1 = 50 ext{ g} \times \frac{1 ext{ mol}}{32 ext{ g}} = 1.5625 ext{ moles}$

  • Rounded value: 1.6 moles (use 2 significant figures).

Variables for Calculation

• Before Change:

  • $N_1 = 1.6 ext{ moles}$

  • $V_1 = 48 ext{ liters}$

• After Change:

  • $V_2 = 79 ext{ liters}$

  • Solve for $N_2$.

Rearranging the Equation

• Rearranged equation to solve for $N_2$:
  $N_2 = \frac{V_2 \times N_1}{V_1}$

• Substituting known values:
  $N_2 = \frac{79 ext{ liters} \times 1.6 ext{ moles}}{48 ext{ liters}} = 2.6 ext{ moles}$

Final Conversion: Moles to Grams

• To convert moles back to grams:

  • Use the molar mass of $O_2$ again:
    $2.6 ext{ moles} \times 32 ext{ g/mol} = 83.2 ext{ g}$

  • Rounded to two significant figures: 83 grams.

Conclusion

• Final answer is 83 grams of oxygen is present in the container after the volume change.

• Additional resources: For more on manipulating gas equations, refer to videos on Rearranging Gas Equations and Gas Laws Extra Help.