Skip to main content
Chemistry LibreTexts

10.4: Gas Mixtures

  • Page ID
    52150
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Introduction

    Gasses disperse throughout a container and form homogenous mixtures or solutions. Unlike liquid solutions where it is common to have a solvent that the solute actually interacts with and dissolves in, the component of a gaseous mixture are so dispersed that they essentially behave as free particles, and so there is no concept of a solute of solvent. The air that you are breathing is a mixture of gasses with 99.96% of dry air consisting of three substances; nitrogen, oxygen and argon.

    Table \(\PageIndex{1}\): Composition of air near the surface of the earth, percent values are based on moles . Note many of these gasses will vary based on anthropomorphic and geologic activities.
    N2 78.08% H2O 0-4% He 5 ppm H2 0.5 ppm
    O2 20.95% CO2 325 ppm CH4 2 ppm N2O 0.3 ppm
    Ar 0.93% Ne 18 ppm Kr 1 ppm CO 0.05-0.2 ppm

    As each gas interacts independently of the other gasses, each gas exerts its own pressure when its particles collide with a surface, which would be known as the partial pressure of that gas.

    Partial Pressure

    Consider a mixture of two gasses A & B (right side of Figure \(\PageIndex{1}\), which being in the same container have the same volume and pressure. If nA is the mole of gas A and nB is the moles of gas B, we could define a partial pressure of each gas (noting that V and T are constant).

    \[P_A=n_A\left ( \frac{RT}{V} \right ) \; \; \; and \; \; \; P_B=n_B\left ( \frac{RT}{V} \right )\]

    where PA= partial pressure of gas A and pB = partial pressure of gas B.

    clipboard_e77c7449596d26280ef65c3e4d63c9542.png

    Figure \(\PageIndex{1}\): In a mixture each gas has its own partial pressure, which is the pressure that gas would exert if it was alone. Note, these cylinders are at the same volumn and temperature

    If there were three gasses, the total pressure would be the sum of all three gases, and this leads to Dalton's Law of Partial Pressures.

    Dalton's Law of Partial Pressure

    Dalton's Law of Partial Pressures states the total pressure of a gas mixture is the sum of the partial pressures of the individual gasses

    \[P_{total}=P_1+P_2+P_3 ... = \sum_i P_i\]

    The Pressure of a Mixture of Gases

    A 10.0-L vessel contains 2.50 × 10−3 mol of H2, 1.00 × 10−3 mol of He, and 3.00 × 10−4 mol of Ne at 35 °C.

    1. What are the partial pressures of each of the gases?
    2. What is the total pressure in atmospheres?

    Solution

    The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using \(P=\dfrac{nRT}{V}\):

    \[P_\mathrm{H_2}=\mathrm{\dfrac{(2.50×10^{−3}\:mol)(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=6.32×10^{−3}\:atm}\]

    \[P_\ce{He}=\mathrm{\dfrac{(1.00×10^{−3}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=2.53×10^{−3}\:atm}\]

    \[P_\ce{Ne}=\mathrm{\dfrac{(3.00×10^{−4}\cancel{mol})(0.08206\cancel{L}atm\cancel{mol^{−1}\:K^{−1}})(308\cancel{K})}{10.0\cancel{L}}=7.58×10^{−4}\:atm}\]

    The total pressure is given by the sum of the partial pressures:

    \[P_\ce{T}=P_\mathrm{H_2}+P_\ce{He}+P_\ce{Ne}=\mathrm{(0.00632+0.00253+0.00076)\:atm=9.61×10^{−3}\:atm}\]

    Exercise \(\PageIndex{1}\)

    A 5.73-L flask at 25 °C contains 0.0388 mol of N2, 0.147 mol of CO, and 0.0803 mol of H2. What is the total pressure in the flask in atmospheres?

    Answer

    1.137 atm

    Mole Fractions and Gas Mixtures

    The mole fraction of a gas in a mixture of gasses is defined as the moles of that gas divided by moles of all gasses, for gas "A" given the symbol XA.

    \[X_A=\frac{n_A}{n_T} \; \; where \; \; n_T=\sum_{i=A}^{all \; i}n_i\]

    Exercise \(\PageIndex{2}\)

    Using the data from table 10.4.2, determine the mole fraction of nitrogen, oxygen and argon for earth's lower atmosphere.

    Answer

    Nitrogen = 0.7908, Oxygen = 0.2095 and Argon = 0.0093

    The mole fraction can also be expressed in terms of pressure

    \[X_A=\frac{P_A}{P_T}\]

    which is derived in the third step of the following derivation.

    Prove: \(P_A=X_AP_T\)

    Solution

    \[P_A = \left ( \frac{n_ART}{V} \right ) \; \; \; and \; \; \; P_T = \left ( \frac{n_TRT}{V} \right )\]

    so

    \[n_A=P_A\left ( \frac{V}{RT} \right ) \; \; \; and \; \; \; n_T=P_T\left ( \frac{V}{RT} \right )\]

    substituting nA and nT into

    \[x_A=\frac{n_A}{n_T}=\frac{P_A \cancel{\left ( \frac{V}{RT} \right )}}{P_T \cancel{\left ( \frac{V}{RT} \right )}}=\frac{P_A}{P_T}\]

    rearranging gives:

    \[P_A=X_AP_T\]

    The following video shows how \(P_A=X_AP_T\) for a system of three gasses.

    Video \(\PageIndex{1}\): 4'15" YouTube video showing how the partial pressure of a gas is the mole fraction of that gas times the total pressure (https://youtu.be/MD16T5yXvPE).  [Video Corregendum: about 2:04 in the video it states "3 moles" N2, when it is really 7 (and 7 is written)]

    Example \(\PageIndex{1}\)

    Calculate the Partial pressure if 78 g nitrogen and 42 g helium are in a pressure tank at 3.75 atm and 50.0 deg. C

    Solution

    Video \(\PageIndex{2}\): Solution to example \(\PageIndex{2}\) (https://youtu.be/V9JMfV4cV8Y).

    Contributors and Attributions


    This page titled 10.4: Gas Mixtures is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Robert Belford.

    • Was this article helpful?