Dalton's Law of Partial Pressure, Molar Volume, and Ideal Gas Law

Dalton's Law of Partial Pressure
  • Dalton's law of partial pressure states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas.

  • This is analogous to calculating your grade, where each component contributes to the total.

Partial Pressure

  • Partial pressure is the individual pressure exerted by a gas in a mixture of gases.

  • Example: Air is composed of oxygen, nitrogen, argon, water vapor, and carbon dioxide, each exerting its own pressure.

Calculating Partial Pressure

  • The partial pressure of a gas can be calculated as:

    Partial Pressure of O<em>2=Pressure exerted by O</em>2Total PressurePartial\ Pressure\ of\ O<em>2 = \frac{Pressure\ exerted\ by\ O</em>2}{Total\ Pressure}

  • The total pressure in a system is the sum of the partial pressures of all gases:

    P<em>total=P</em>O<em>2+P</em>N<em>2+P</em>Ar+P<em>H</em>2O+P<em>CO</em>2P<em>{total} = P</em>{O<em>2} + P</em>{N<em>2} + P</em>{Ar} + P<em>{H</em>2O} + P<em>{CO</em>2}

  • Example values: Oxygen (20.9 kPa), Nitrogen, Argon, Water vapor, leading to a total pressure of 101.3 kPa.

Example Calculation

  • To find the percentage of total pressure exerted by oxygen:

    20.9 kPa101.3 kPa0.2063\frac{20.9\ kPa}{101.3\ kPa} \approx 0.2063

  • Converting to percentage:

    0.2063×100%=20.63%0.2063 \times 100\% = 20.63\%

  • This means 20.63% of the total pressure is due to oxygen gas.

Mole Fraction

  • Instead of using actual pressures, mole fraction can be used if the mole quantities are known.

  • Mole fraction is the ratio of the moles of a particular gas to the total moles of all gases in the mixture.

Application Example

  • Given: A container with hydrogen, nitrogen, and oxygen gas has a total pressure of 646 Torr.

  • Total pressure is the sum of individual partial pressures:

    P<em>total=P</em>H<em>2+P</em>N<em>2+P</em>O2=646 TorrP<em>{total} = P</em>{H<em>2} + P</em>{N<em>2} + P</em>{O_2} = 646\ Torr

  • Hydrogen makes up 23%, oxygen 67%, and nitrogen 10% of the total pressure.

Calculating Partial Pressures from Percentages

  • If hydrogen is 23% of the total pressure, its partial pressure is:

    P<em>H</em>2=0.23×646 Torr=148.58 TorrP<em>{H</em>2} = 0.23 \times 646 \ Torr = 148.58 \ Torr

  • Similarly, for oxygen:

    P<em>O</em>2=0.67×646 Torr=432.82 TorrP<em>{O</em>2} = 0.67 \times 646 \ Torr = 432.82 \ Torr

  • To find the partial pressure of nitrogen, subtract the partial pressures of hydrogen and oxygen from the total pressure:

    P<em>N</em>2=646 Torr(148.58 Torr+432.82 Torr)=64.6 TorrP<em>{N</em>2} = 646 \ Torr - (148.58 \ Torr + 432.82 \ Torr) = 64.6 \ Torr

Key Idea

  • Dalton's law states that the total pressure of a gas mixture is the sum of the partial pressures of each gas.

  • This is similar to calculating a grade, where knowing the percentages allows you to determine the value contributed by each component.

Collecting Gas Over Water

  • A common method to collect gas produced in a chemical reaction involves water displacement.

  • However, the collected gas is mixed with water vapor, affecting pressure readings.

Accounting for Water Vapor

  • The total pressure recorded when collecting gas over water is the sum of the gas's pressure and water vapor pressure:

    P<em>total=P</em>gas+P<em>H</em>2OP<em>{total} = P</em>{gas} + P<em>{H</em>2O}

  • To find the true pressure of the gas, subtract the partial pressure of water vapor.

  • Use a chart to find the partial pressure of water at a specific temperature.

Example: Nitrogen Gas Collection

  • Nitrogen gas collected over water at 25°C shows a total pressure of 98.9 kPa.

  • The chart indicates water's partial pressure is 23.756 Torr at 25°C.

  • Convert Torr to kPa:

    23.756 Torr×101.325 kPa760 Torr3.167 kPa23.756\ Torr \times \frac{101.325\ kPa}{760\ Torr} \approx 3.167\ kPa

  • Subtract water's partial pressure to find nitrogen's partial pressure:

    P<em>N</em>2=98.9 kPa3.167 kPa95.733 kPaP<em>{N</em>2} = 98.9\ kPa - 3.167\ kPa \approx 95.733\ kPa

Avogadro's Law

  • Avogadro's Law: As the number of moles of gas increases, the volume also increases, given constant pressure and temperature.

  • This relationship leads to the concept of molar volume.

Molar Volume

  • STP (Standard Temperature and Pressure): 273 K and 1 atm.

  • SATP (Standard Ambient Temperature and Pressure): 298 K (25°C) and 1 atm.

  • At STP, one mole of any gas occupies 22.4 liters.

  • At SATP, one mole of any gas occupies 24.8 liters.

Molar Volume Application

  • Problem: Find the volume of four moles of nitrogen monoxide gas at STP.

  • At STP, one mole occupies 22.4 liters, so:

    4 moles×22.4 L/mol=89.6 liters4\ moles \times 22.4\ L/mol = 89.6\ liters

Example Calculation 2

  • Problem: How many moles are in 25.4 liters of carbon dioxide at STP?

  • Using molar volume:

    25.4 L22.4 L/mol1.1339 moles\frac{25.4\ L}{22.4\ L/mol} \approx 1.1339\ moles

  • This result can be extended to find mass (using molar mass) or the number of molecules (using Avogadro's number).

Ideal Gas Law

  • Ideal gases are theoretical gases with negligible particle volume and no intermolecular interactions.

  • Collisions between ideal gas particles are perfectly elastic.

Ideal Gas Law Equation

  • The ideal gas law relates pressure, volume, number of moles, and temperature:

    PV=nRTPV = nRT

    Where:

    • P = Pressure

    • V = Volume (in liters)

    • n = Number of moles

    • R = Ideal gas constant

    • T = Temperature (in Kelvin)

Gas Constant Values

  • Two common values for R:

    • R=8.314 LkPamolKR = 8.314\ \frac{L \cdot kPa}{mol \cdot K}

    • R=0.08206 LatmmolKR = 0.08206\ \frac{L \cdot atm}{mol \cdot K}

Ideal Gas Law Application

  • Problem: What volume does 5 moles of O2O_2 occupy at 28°C (301.15 K) and 0.998 atm?

  • Using R = 0.08206:

    V=nRTP=5 mol×0.08206 LatmmolK×301.15 K0.998 atm123.8 litersV = \frac{nRT}{P} = \frac{5\ mol \times 0.08206\ \frac{L \cdot atm}{mol \cdot K} \times 301.15\ K}{0.998\ atm} \approx 123.8\ liters

Finding Mass with Ideal Gas Law

  • Problem: What mass of neon gas is needed to produce a pressure of 90 kPa in a 0.88 liter tube at 30°C (303.15 K)?

  • First, find the number of moles using the ideal gas law:

    n=PVRT=90 kPa×0.88 L8.314 LkPamolK×303.15 K0.03142 molesn = \frac{PV}{RT} = \frac{90\ kPa \times 0.88\ L}{8.314\ \frac{L \cdot kPa}{mol \cdot K} \times 303.15\ K} \approx 0.03142\ moles

  • Then, convert moles to mass using neon's molar mass (20.18 g/mol):

    0.03142 moles×20.18 gmol0.63 g0.03142\ moles \times 20.18\ \frac{g}{mol} \approx 0.63\ g

Choosing The Right Gas Law

  • Use Charles, Boyle, Gay-Lussac, or the combined gas law when a property changes in the system, and you want to see that change impact another property in the system.

  • Use the ideal gas law when trying to find a particular property in the system.