Tutorials

The Ideal Gas Law

Ideal Gas Law Core Concepts

In this tutorial, you will learn how the ideal gas law equation was derived and how to use it. You will also learn what defines an ideal gas, what the ideal gas constant is, ideal gas law units, and what assumptions we make to call a gas ideal – the ideal gas properties.

Topics Covered in Other Articles

Introduction

The ideal gas law is an equation of state that describes ideal gases. This equation of state relates a gas’s pressure, volume, temperature, and mass, and is very useful for describing how gases will behave in ideal conditions. This is the most common equation of state for gases.

A few notable other ones are the Van der Waal’s and the Virial equation of state, which both describe the state of gases in non-ideal states. See our article on Van der Waal’s Equation to learn more about this.

The ideal gas equation was first stated by Benoît Paul Émile Clapeyron in 1834 as a combination of Boyle’s law, Charles’s law, Avogadro’s law, and Gay-Lussac’s law. Clapeyton was a French engineer, and one of the founders of thermodynamics.

What are the Ideal Gas Properties?

Gases consist of a large number of particles constantly colliding with each other randomly. It order to model and predict the behavior of gases, the concept of an ideal gas was created. For a gas to be ideal, a few assumptions need to be made. These can also be viewed as the ideal gas properties.

  1. For one, we assume that the volume of the gas particles are negligible. This means that the volume of the container is much larger than the volume of the gas particles.
  2. The second assumption we make is that the gas particles are equally sized, and do not have intermolecular forces with other gas particles.
  3. Third, we assume that the gas particles move randomly according to Newton’s Laws of Motion.
  4. Lastly, we assume that all collisions are perfectly elastic and have no energy loss. This means that the collisions between gas particles and the walls don’t lose energy, and exert a constant pressure.

Although no gas is perfectly ideal most gases are close enough at room temperature, and are nearly ideal.

Combining the Gas Laws into the Ideal Gas Law Equation

If we consider the three basic gas laws, Charles’ Law, Avogadro’s Law, and Boyle’s Law, we can make relations between a gas’s pressure, volume, temperature, and quantity of moles. By taking each equation and combining them, we can derive the ideal gas law equation.

     \begin{align*} P&\propto\frac{1}{V}\\ V&\propto T\\ n&\propto V\\ \implies PV&\propto nT\\ \implies \frac{PV}{nT}&=c \end{align*}

Because this proportionality takes into account all changes of state of gases, it will be constant for an ideal gas. This constant is known as the Ideal Gas Constant, or Universal Gas Constant, and has a value of 0.0082057\frac{\text{atm L}}{\text{mol K}}. We can plug this constant, labeled R, into the equation to derive the ideal gas law, \boxed{PV=nRT}.

Ideal Gas Law Units

The following units are used in the ideal gas law equation, when SI units (international system of units) are used.

  • P is pressure measured in Pascals, Pa.
  • V is the volume measured in cubic meters, m3
  • n is the number of moles.
  • R = 8.3145 is the universal gas constant measured in J/(K · mol), or alternatively m3·Pa / (K · mol)
  • T is the temperature measured in Kelvin.

If you are using liters and atmospheres of pressure, instead of Pascals and cubic meters, then you have the following:

  • P is pressure measured in atmospheres
  • V is the volume measured in liters
  • n is the number of moles.
  • R = 0.08206 is the universal gas constant measured in L·atm /(K · mol)
  • T is the temperature measured in Kelvin.

Ideal Gas Law Example Problem

  1. A mixture of ethanol and methanol is burned in oxygen to produce 35 cm3 of CO2 and 55 cm3 of H2O. Complete combustion occurs and the volumes of both products are measured at 101 kPa and 120 degrees centigrade. What is the molar ratio, ethanol to methanol, in the mixture?
    \text{a)}\hspace{0.1in}1:3\hspace{0.2in}\text{b)}\hspace{0.1in}2:3\hspace{0.2in}\text{c)}\hspace{0.1in}3:2\hspace{0.2in}\text{d)}\hspace{0.1in}3:1\hspace{0.2in}

    \text{To start solving this problem, first the chemical reactions occurring must be balanced.}
    \text{Methanol: }\ch{CH3OH + 1.5 O2->2 H2O + CO2}
    \text{Ethanol: }\ch{CH3CH2OH + 3 O2 -> 3 H2O + 2 CO2}

    \text{Next, the volumes of gas produced are converted into moles using the ideal gas law.}

        \begin{align*}PV=nRT&\implies n=\frac{PV}{RT}\\n_{\ch{H2O}}&=\frac{PV_{\ch{H2O}}}{RT}\\n_{\ch{H2O}}&=\frac{\left(101\cdot10^3\text{ Pa}\right)\cdot\left(55\cdot10^{-6}\text{ m$^3$}\right)}{\left(8.314\text{ m$^3\cdot$Pa/K$\cdot$mol}\right)\left(120+273\text{ K}\right)}\\n_{\ch{H2O}}&=1.7\text{ mmol H$_2$O}\\n_{\ch{CO2}}&=\frac{\left(101\cdot10^3\text{ Pa}\right)\cdot\left(35\cdot10^{-6}\text{ m$^3$}\right)}{\left(8.314\text{ m$^3\cdot$Pa/K$\cdot$mol}\right)\left(120+273\text{ K}\right)}\\n_{\ch{CO2}}&=1.08\text{ mmol CO$_2$}\end{align*}


    \text{From these values, the ratio of ethanol to methanol can be calculated:}
    \varphi=\frac{n_{\ch{H2O}}}{n_{\ch{CO2}}}=\frac{1.7}{1.08}=1.57
    \text{Next, for each of the answer choices, the molar ratio is calculated.}

    a) Let the ratio of ethanol:methanol be 1:3. Then, the methanol reaction will happen 3 times as much as the ethanol reaction. Thus, the methanol products will be 3 times as much as the ethanol products.

        \begin{align*}\varphi&=\frac{1\cdot\ch{3 H2O}+3\cdot\ch{2 H2O}}{1\cdot\ch{2 CO2}+3\cdot\ch{CO2}}\\\varphi&=\frac{1\cdot3+3\cdot2}{1\cdot2+3\cdot1}=1.8\end{align*}



    b) Let the ratio be 2:3.

        \begin{align*}\varphi&=\frac{2\cdot\ch{3 H2O}+3\cdot\ch{2 H2O}}{2\cdot\ch{2 CO2}+3\cdot\ch{CO2}}\\\varphi&=\frac{2\cdot3+3\cdot2}{2\cdot2+3\cdot1}=1.71\end{align*}



    c) Let the ratio be 3:2.

        \begin{align*}\varphi&=\frac{3\cdot\ch{3 H2O}+2\cdot\ch{2 H2O}}{3\cdot\ch{2 CO2}+2\cdot\ch{CO2}}\\\varphi&=\frac{3\cdot3+2\cdot2}{3\cdot2+2\cdot1}=1.63\end{align*}



    d) Let the ratio be 3:1

        \begin{align*}\varphi&=\frac{3\cdot\ch{3 H2O}+1\cdot\ch{2 H2O}}{3\cdot\ch{2 CO2}+1\cdot\ch{CO2}}\\\varphi&=\frac{3\cdot3+1\cdot2}{3\cdot2+1\cdot1}=1.57\end{align*}



    From these values, we can see that option (d) has the same molar ratio as we calculated before, and therefore option (d) is the correct answer.

Leave a Reply

Your email address will not be published. Required fields are marked *