- Last updated
- Save as PDF
- Page ID
- 53833
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vectorC}[1]{\textbf{#1}}\)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}}\)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}\)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
Sulfur dioxide is a by-product of many processes, both natural and human-made. Massive amounts of this gas are released during volcanic eruptions. Humans produce sulfur dioxide by burning coal. When in the atmosphere, the gas has a cooling effect by reflecting sunlight away from the earth. However, sulfur dioxide is also a component of smog and acid rain, both of which are harmful to the environment. Many efforts have been made to reduce \(\ce{SO_2}\) levels to lower acid rain production. However, \(\ce{SO_2}\) reduction efforts have an unforeseen complication:as we lower the concentration of this gas in the atmosphere, we lower its ability to cool, and consequently have global warming concerns.
Mole Fraction
One way to express relative amounts of substances in a mixture is with the mole fraction. Mole fraction \(X\) is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances, \(\ce{A}\) and \(\ce{B}\), the mole fractions of each would be written as follows:
\[X_A = \frac{\text{mol} \: \ce{A}}{\text{mol} \: \ce{A} + \text{mol} \: \ce{B}} \: \: \: \text{and} \: \: \: X_B = \frac{\text{mol} \: \ce{B}}{\text{mol} \: \ce{A} + \text{mol} \: \ce{B}}\nonumber \]
If a mixture consists of \(0.50 \: \text{mol} \: \ce{A}\) and \(1.00 \: \text{mol} \: \ce{B}\), then the mole fraction of \(\ce{A}\) would be \(X_A = \frac{0.5}{1.5} + 0.33\). Similarly, the mole fraction of \(\ce{B}\) would be \(X_B = \frac{1.0}{1.5} = 0.67\). Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton's law of partial pressures. Consider the following situation... A 20.0 liter vessel contains \(1.0 \: \text{mol}\) of hydrogen gas at a pressure of \(600 \: \text{mm} \: \ce{Hg}\). Another 20.0 liter vessels contains \(3.0 \: \text{mol}\) of helium at a pressure of \(1800 \: \text{mm} \: \ce{Hg}\). These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton's law, we can express the partial pressures as follows:
\[P_{H_2} = X_{H_2} \times P_\text{Total} \: \: \: \text{and} \: \: \: P_{He} = X_{He} \times P_\text{Total}\nonumber \]
The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:
\[X_{H_2} = \frac{1.0 \: \text{mol}}{1.0 \: \text{mol} + 3.0 \: \text{mol}} = 0.25 \: \: \: \text{and} \: \: \: X_{He} = \frac{3.0 \: \text{mol}}{1.0 \: \text{mol} + 3.0 \: \text{mol}} = 0.75\nonumber \]
The total pressure according to Dalton's law is \(600 \: \text{mm} \: \ce{Hg} + 1800 \: \text{mm} \: \ce{Hg} = 2400 \: \text{mm} \: \ce{Hg}\). So, each partial pressure will be:
\[P_{H_2} = 0.25 \times 2400 \: \text{mm} \: \ce{Hg} = 600 \: \text{mm} \: \ce{Hg}\nonumber \]
\[P_{He} = 0.75 \times 2400 \: \text{mm} \: \ce{Hg} = 1800 \: \text{mm} \: \ce{Hg}\nonumber \]
The partial pressures of each gas in the mixture do not change, since they were mixed into the same size vessel and the temperature was not changed.
Example \(\PageIndex{1}\)
A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is \(104 \: \text{kPa}\), what is the partial pressure of each gas?
Solution
Step 1: List the known quantities and plan the problem.
Known
- \(1.24 \: \text{mol} \: \ce{H_2}\)
- \(2.91 \: \text{mol} \: \ce{O_2}\)
- \(P_\text{Total} = 104 \: \text{kPa}\)
Unknown
- \(P_{H_2} = ? \: \text{kPa}\)
- \(P_{O_2} = ? \: \text{kPa}\)
First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.
Step 2: Solve.
\[\begin{array}{ll} X_{H_2} = \frac{1.24 \: \text{mol}}{1.24 \: \text{mol} + 2.91 \: \text{mol}} = 0.299 & X_{O_2} = \frac{2.91 \: \text{mol}}{1.24 \: \text{mol} + 2.91 \: \text{mol}} = 0.701 \\ P_{H_2} = 0.299 \times 104 \: \text{kPa} = 31.1 \: \text{kPa} & P_{O_2} = 0.701 \times 104 \: \text{kPa} = 72.9 \: \text{kPa} \end{array}\nonumber \]
Step 3: Think about your result.
The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.
