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State Boyle's Law, Avogadro's Law, and Charles's Law. List three postulates of the kinetic molecular theory of gases and state the mathematical expressions for Boyle's Law, Charles's Law, and Avogadro

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Answer

g(ii)

1.39×10−7 mol/L\mathbf{1.39 \times 10^{-7} \text{ mol/L}}

Step-by-step solution

9 steps

Question One:

Part a(i)

Boyle's Law:** For a fixed amount of gas at constant temperature, the pressure of the gas is inversely proportional to its volume.

P∝1VorPV=constantP \propto \frac{1}{V} \quad \text{or} \quad PV = \text{constant}

Part a(ii)

Avogadro's Law:** For a fixed temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas.

V∝norVn=constantV \propto n \quad \text{or} \quad \frac{V}{n} = \text{constant}

Part a(iii)

Charles's Law:** For a fixed amount of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature.

V∝TorVT=constantV \propto T \quad \text{or} \quad \frac{V}{T} = \text{constant}

Part b

Three postulates of the kinetic molecular theory of gases:

  • Gases consist of a large number of tiny particles (atoms or molecules) that are in continuous, random motion.
  • The volume occupied by the gas particles themselves is negligible compared to the total volume of the container.
  • There are no attractive or repulsive forces between gas particles.
  • Collisions between gas particles and between particles and the container walls are perfectly elastic.
  • The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas.

Part c· 4 steps

  1. State the mathematical expressions for each law.

    From Boyle's Law: V∝1PV \propto \frac{1}{P} (at constant n,Tn, T)
    From Charles's Law: V∝TV \propto T (at constant n,Pn, P)
    From Avogadro's Law: V∝nV \propto n (at constant P,TP, T)

  2. Combine the proportionalities.

    Combining these three relationships, we get:

    V∝nTPV \propto \frac{nT}{P}

  3. Introduce the proportionality constant.

    Introducing the proportionality constant RR (the ideal gas constant), we obtain the ideal gas law:

    V=nRTPV = \frac{nRT}{P}

    Rearranging this equation gives:

    PV=nRTPV = nRT

  4. Show that for a fixed mass of gas, PVT\frac{PV}{T} is constant.

    For a fixed mass of a gas, the number of moles (nn) is constant. Since RR is also a constant, the product nRnR is constant.
    Therefore, for a fixed mass of a gas:

    PVT=nR=constant\frac{PV}{T} = nR = \text{constant}

Part d

The two assumptions of the kinetic molecular theory of gases that lead to deviations from ideal gas behavior are:

  • Negligible volume of gas particles: The KMT assumes that gas particles have negligible volume compared to the volume of the container. In real gases, especially at high pressures, the volume of the gas particles themselves becomes significant and cannot be ignored.
  • No intermolecular forces: The KMT assumes there are no attractive or repulsive forces between gas particles. In real gases, attractive forces exist, especially at low temperatures, causing particles to collide with the container walls with less force and reducing the observed pressure.

Part e

i) Chemical equilibrium is a dynamic state in a reversible chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction. At equilibrium, the concentrations of reactants and products remain constant over time, although the reactions continue to occur.

Part f(ii)· 3 steps

Consider the equilibrium system: 2SO2(g)+O2(g)⇌2SO3(g)+Heat energy2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} + \text{Heat energy}

i) Le Chatelier's Principle: If a change of condition (such as temperature, pressure, or concentration) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress and re-establishes a new equilibrium.

  1. Determine the number of moles of gas on each side of the reaction.

    Reactant side: 2 moles of SO2+1 mole of O2=3 moles of gas2 \text{ moles of } SO_2 + 1 \text{ mole of } O_2 = 3 \text{ moles of gas}
    Product side: 2 moles of SO3=2 moles of gas2 \text{ moles of } SO_3 = 2 \text{ moles of gas}

  2. Apply Le Chatelier's Principle for an increase in pressure.

    Increasing the pressure on a gaseous equilibrium system will cause the equilibrium to shift towards the side with fewer moles of gas to reduce the stress.

  3. Conclude the effect.

    Since the product side (2 moles of SO32 \text{ moles of } SO_3) has fewer moles of gas than the reactant side (3 moles of SO2+O23 \text{ moles of } SO_2 + O_2), increasing the pressure will shift the equilibrium to the right, favoring the formation of SO3SO_3.

Part g(i)

Molar solubility** is the number of moles of a solute that can dissolve in one liter of a solution to form a saturated solution. It is typically expressed in units of moles per liter (mol/L\text{mol/L}).

Part g(ii)

1.39×10−7 mol/L\mathbf{1.39 \times 10^{-7} \text{ mol/L}}

The solubility of lead chromate (PbCrO4\text{PbCrO}_4) is 4.5×10−5g/L4.5 \times 10^{-5} \text{g/L}. Calculate the molar solubility.

  1. Determine the molar mass of PbCrO4\text{PbCrO}_4.

    Molar mass of Pb\text{Pb} = 207.2 g/mol207.2 \text{ g/mol}
    Molar mass of Cr\text{Cr} = 52.00 g/mol52.00 \text{ g/mol}
    Molar mass of O\text{O} = 16.00 g/mol16.00 \text{ g/mol}
    Molar mass of PbCrO4=207.2+52.00+(4×16.00)=207.2+52.00+64.00=323.2 g/mol\text{PbCrO}_4 = 207.2 + 52.00 + (4 \times 16.00) = 207.2 + 52.00 + 64.00 = 323.2 \text{ g/mol}

  2. Convert the solubility from g/L\text{g/L} to mol/L\text{mol/L}.

    Molar solubility=Solubility in g/LMolar mass in g/mol\text{Molar solubility} = \frac{\text{Solubility in g/L}}{\text{Molar mass in g/mol}} Molar solubility=4.5×10−5 g/L323.2 g/mol\text{Molar solubility} = \frac{4.5 \times 10^{-5} \text{ g/L}}{323.2 \text{ g/mol}} Molar solubility=1.392×10−7 mol/L\text{Molar solubility} = 1.392 \times 10^{-7} \text{ mol/L}

    The molar solubility of lead chromate is 1.39×10−7 mol/L\boxed{\mathbf{1.39 \times 10^{-7} \text{ mol/L}}}.

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