Day 4: Molar Mass & Dalton's Law

Application of the Ideal Gas Law

Determination of Molar Mass Using Ideal Gas Law

• Ideal Gas Law Equation:

  • The ideal gas law can be expressed as:
    $PV = nRT$
    where:

  • $P$ = pressure

  • $V$ = volume

  • $n$ = number of moles

  • $R$ = ideal gas constant (0.0821 L·atm/(K·mol))

  • $T$ = temperature (in Kelvin)

• Concept of Molar Mass:

  • Molar mass is defined as the mass of the substance divided by the amount of substance (moles) with the formula:
    $ext{Molar Mass} = rac{ ext{mass (grams)}}{n ext{ (moles)}}$

  • Units of molar mass: grams per mole (g/mol).

Example 1

• Given Data:

  • Mass (m) = 0.311 grams

  • Volume (V) = 0.255 liters

  • Temperature (T) = 55 °C = 328 K

  • Pressure (P) = 886 mmHg

• Pressure Conversion to Atmospheres:

  • $P = rac{886 ext{ mmHg}}{760 ext{ mmHg/atm}} = 1.17 ext{ atm}$

• Calculate Moles (n) Using the Ideal Gas Law:

  • Rearranging the ideal gas law gives:
    $n = rac{PV}{RT}$

  • Plugging in the values:
    $n = rac{(1.17 ext{ atm})(0.255 ext{ L})}{(0.0821 ext{ L·atm/(K·mol)})(328 ext{ K})}$

  • Calculating:

  • Result: $n = 0.00978 ext{ moles}$

• Calculating Molar Mass:

  • Using the molar mass formula:
    $ext{Molar Mass} = rac{0.311 ext{ grams}}{0.00978 ext{ moles}} = 31.8 ext{ g/mol}$

Example 2

• Given Data:

  • Mass (m) = 0.827 grams

  • Volume (V) = 0.270 liters

  • Temperature (T) = 88 °C = 361 K

  • Pressure (P) = 1.28 atm (already converted)

• Calculate Moles (n):

  • Using the ideal gas law:
    $n = rac{(1.28 ext{ atm})(0.270 ext{ L})}{(0.0821 ext{ L·atm/(K·mol)})(361 ext{ K})}$

  • Result: $n = 0.01166 ext{ moles}$

• Calculating Molar Mass:

  • Using the molar mass formula:
    $ext{Molar Mass} = rac{0.827}{0.01166} = 70.9 ext{ g/mol}$

Ideal vs. Non-Ideal Gas Behavior

• Ideal Gas Behavior:

  • Ideal gases conform to all postulates of the kinetic molecular theory:

  • No intermolecular forces

  • Gas particles have negligible volume compared to the volume of the container.

  • Ideal conditions emphasize:

  • High temperatures (particles move rapidly)

  • Low pressures (particles are far apart)

• Non-Ideal (Real) Gas Behavior:

  • Occurs under:

  • Low temperatures (particles move slower)

  • High pressures (particles are closely packed)

  • In non-ideal conditions:

  • Size of particles can be significant

  • Intermolecular forces can influence the behavior

Gas Mixtures and Partial Pressures

• Each gas in a mixture behaves independently.

• Volume Consideration:

  • The total volume of the gas mixture is equal to the volume of the container regardless of the number of gases present.

  • Example: In a 1L container with two gases, each can still occupy the entire volume.

• Partial Pressure Concept:

  • The pressure exerted by an individual gas in a mixture is known as its partial pressure.

  • The total pressure in a mixture is the sum of the partial pressures of all components:
    $P{ ext{total}} = PA + PB + PC$

• Calculating Partial Pressure:

  • For a gas in a mixture:
    $Pi = ext{Fractional Composition} imes P{ ext{total}}$

  • Fractional composition is the number of moles of gas divided by total number of moles.

Example of Partial Pressure Calculation

• Mixture Example:

  • Given: 80% Helium and 20% Neon

  • Total Pressure: 1 atm

  • Partial Pressure of Helium (PHe):
    $P_{He} = 0.80 imes 1 ext{ atm} = 0.8 ext{ atm}$

  • Partial Pressure of Neon (PNe):
    $P_{Ne} = 0.20 imes 1 ext{ atm} = 0.2 ext{ atm}$

• Partial Pressure Calculation from Total Pressure:

  • If the mixture total pressure (Ptotal) = 10 atm and considering 2% Oxygen:
    $P{O2} = 0.02 imes 10 ext{ atm} = 0.2 ext{ atm}$

Dalton's Law of Partial Pressures

• Statement of Dalton’s Law:

  • The total pressure in a mixture of gases is equal to the sum of the partial pressures of each gas.
    $P{ ext{total}} = PA + PB + PC$

Pressure Calculation Example

• Example with Given Values:

  • Total pressure = 558 mmHg

  • Partial pressures:

  • $P_{He} = 341 ext{ mmHg}$

  • $P_{Ne} = 112 ext{ mmHg}$

  • To find $P{Ar}$: $P{ ext{total}} = P{He} + P{Ne} + P{Ar}$ $558 = 341 + 112 + P{Ar}$ Resulting in: $P_{Ar} = 105 ext{ mmHg}$

Finding Mass of Argon in a Gas Mixture

• Given Data:

  • Mixture: He, Ne, Ar

  • Total pressure = 662 mmHg

  • $P<em>{He} = 341 ext{ mmHg}$, $P</em>{Ne} = 112 ext{ mmHg}$

• Calculate Partial Pressure of Argon (PAr): $P<em>{ ext{total}} = P</em>{He} + P<em>{Ne} + P</em>{Ar}$

  • Rearranging gives:
    $P_{Ar} = 662 - (341 + 112) = 209 ext{ mmHg}$

• Convert Pressure to Atmospheres:

  • $P_{Ar} = rac{209}{760} = 0.275 ext{ atm}$

• Using Ideal Gas Law to Find Moles (n):

  • For a 1-liter volume:
    $n = rac{PV}{RT}$
    $n = rac{(0.275)(1)}{(0.0821)(298)} ightarrow n = 0.0112$

• Calculate Mass of Argon:

  • Molar mass of Argon = 39.95 g/mol
    $ext{Mass} = n imes ext{Molar Mass} = 0.0112 imes 39.95 = 0.449 ext{ grams}$