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The Ideal Gas Law

 

This is part of Year 11 HSC Chemistry course under the topic of Gas Laws.

HSC Chemistry Syllabus

  • Conduct investigations and solve problems to determine the relationship between the Ideal Gas Law and:
    • Gay-Lussac's Law (Temperature)
    • Boyle's Law
    • Charles' Law
    • Avogadro's Law

    The Combined Gas Law

    The Combined Gas Law unites Boyle's, Charles's, and Gay-Lussac's laws into a single expression:

    P1V1T1=P2V2T2

     

    Using the Combined Gas Law:

    Question 1: Given an initial volume of 8.33L, pressure of 1.82atm, and temperature of 286K, which changes to a temperature of 355K and a volume of 5.72L, what is the final pressure?

     

    Solution:

    Identify:

     

    V1=8.33L,P1=1.82atm,T1=286K

     

    V2=5.72L,P2=?atm,T2=355K

     

    Substitute and Solve: 

     

    P1×V1T1÷V2T2=P2

     

    P2=(1.82atm×8.33L)286K÷5.72L355K

     

    P2=3.29atm

     

    The Ideal Gas Law

    The Ideal Gas Law combines the relationships described by Boyle's law, Charles's law, and Avogadro's law. These laws explain how the various properties of gases affect volume:

    • Boyle's law on pressure: V1P
    • Charles's law on temperature: VT
    • Avogadro's Law on amounts: Vn

     

    Combining these individual effects, we obtain the equation known as the ideal gas law:

     

    VnTP  or  PVnT  or  PVnT=R

     

    Here R represents a proportionality constant known as the universal gas constant. The most common form of the ideal gas equation is 

     

    PV=nRT

     

    Where R=8.314Jmol-1K-1

     

    Solving Gas Law Problems

    To solve problems involving the gas laws, follow these steps:

    1. Summarize the changing gas variables – knowns and unknowns – and those held constant.
    2. Convert units, if necessary.
    3. Rearrange the ideal gas law to obtain the needed relationship of variables and solve for the unknown.


    Example:

    Question: A gas sample has an initial volume of 1.00 L, an initial pressure of 1.00 atm, and an initial temperature of 273 K. If the volume is increased to 2.00 L and the temperature is raised to 546 K, what is the final pressure, assuming the number of moles of gas remains constant? 

     

    Solution:

    • Identify the given information and what you need to find:

    V1=1.00L,P1=1.00atm,T1=273K

    V2=2.00L,T2=546K,P2=?

     

    • Convert units if necessary: No conversion needed

     

    • Use the combined gas law

    P1V1T1=P2V2T2

     

    1.00×1.00273=P2×200546

     

    P2=1.00×1.00×546273×2.00=1.00atm

     

    The final pressure of the gas is 1.00 atm

     

     

    BACK TO MODULE 2: INTRODUCTION TO QUANTITATIVE CHEMISTRY