11.3: Ideal Gas Law
- Page ID
- 369924
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objectives
- To use the ideal gas law to describe the behavior of a gas.
- Calculate T, V, P, or n of the ideal gas law: PV=nRT
A flat tire is not very useful. It does not cushion the rim of the wheel and creates a very uncomfortable ride. When air is added to the tire, the pressure increases as more molecules of gas are forced into the rigid tire. The amount of air that should be put into a tire depends on the pressure rating for that tire. Too little pressure and the tire will not hold its shape. Too much pressure and the tire could burst.
Avogadro's Law
You have learned about Avogadro's hypothesis: equal volumes of any gas at the same temperature and pressure contain the same number of molecules. It follows that the volume of a gas is directly proportional to the number of moles of gas present in the sample. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles of gas, when the temperature and pressure are held constant. The mathematical expression of Avogadro's Law is:
\[V = k \times n \: \: \: \text{and} \: \: \: \frac{V_1}{n_1} = \frac{V_2}{n_2}\]
(Where \(n\) is the number of moles of gas and \(k\) is a constant). Avogadro's Law is in evidence whenever you blow up a balloon. The volume of the balloon increases as you add moles of gas to the balloon by blowing it up.
If the container holding the gas is rigid rather than flexible, pressure can be substituted for volume in Avogadro's Law. Adding gas to a rigid container makes the pressure increase.
There are a number of chemical reactions that require ammonia. In order to carry out a reaction efficiently, we need to know how much ammonia we have for stoichiometric purposes. Using gas laws, we can determine the number of moles present in the tank if we know the volume, temperature, and pressure of the system.
Ideal Gas Law
The combined gas law shows that the pressure of a gas is inversely proportional to volume and directly proportional to temperature. Avogadro's Law shows that volume or pressure is directly proportional to the number of moles of gas. Putting these laws together gives us the following equation:
\[\frac{P_1 \times V_1}{T_1 \times n_1} = \frac{P_2 \times V_2}{T_2 \times n_2}\]
As with the other gas laws, we can also say that \(\frac{\left( P \times V \right)}{\left( T \times n \right)}\) is equal to a constant. The constant can be evaluated provided that the gas being described is considered to be ideal.
The ideal gas law is a single equation which relates the pressure, volume, temperature, and number of moles of an ideal gas. If we substitute in the variable \(R\) for the constant, the equation becomes:
\[\frac{P \times V}{T \times n} = R\]
The ideal gas law is conveniently rearranged to look this way, with the multiplication signs omitted:
\[PV = nRT\]
The variable \(R\) in the equation is called the ideal gas constant.
Evaluating the Ideal Gas Constant
The value of \(R\), the ideal gas constant, depends on the units chosen for pressure, temperature, and volume in the ideal gas equation. It is necessary to use Kelvin for the temperature and it is conventional to use the SI unit of liters for the volume. However, pressure is commonly measured in one of three units: \(\text{kPa}\), \(\text{atm}\), or \(\text{mm} \: \ce{Hg}\). Therefore, \(R\) can have three different values.
We will demonstrate how \(R\) is calculated when the pressure is measured in \(\text{kPa}\). Recall that the volume of \(1.00 \: \text{mol}\) of any gas at STP is measured to be \(22.414 \: \text{L}\). We can substitute \(101.325 \: \text{kPa}\) for pressure, \(22.414 \: \text{L}\) for volume, and \(273.15 \: \text{K}\) for temperature into the ideal gas equation and solve for \(R\).
\[R = \frac{PV}{nT} = \frac{101.325 \: \text{kPa} \times 22.414 \: \text{L}}{1.000 \: \text{mol} \times 273.15 \: \text{K}} = 8.314 \: \text{kPa} \cdot \text{L/K} \cdot \text{mol}\]
This is the value of \(R\) that is to be used in the ideal gas equation when the pressure is given in \(\text{kPa}\). The table below shows a summary of this and the other possible values of \(R\). It is important to choose the correct value of \(R\) to use for a given problem.
| Unit of \(P\) | Unit of \(V\) | Unit of \(n\) | Unit of \(T\) | Value and Unit of \(R\) |
|---|---|---|---|---|
| \(\text{kPa}\) | \(\text{L}\) | \(\text{mol}\) | \(\text{K}\) | \(8.314 \: \text{J/K} \cdot \text{mol}\) |
| \(\text{atm}\) | \(\text{L}\) | \(\text{mol}\) | \(\text{K}\) | \(0.08206 \: \text{L} \cdot \text{atm/K} \cdot \text{mol}\) |
| \(\text{mm} \: \ce{Hg}\) | \(\text{L}\) | \(\text{mol}\) | \(\text{K}\) | \(62.36 \: \text{L} \cdot \text{mm} \: \ce{Hg}/\text{K} \cdot \text{mol}\) |
Notice that the unit for \(R\) when the pressure is in \(\text{kPa}\) has been changed to \(\text{J/K} \cdot \text{mol}\). A kilopascal multiplied by a liter is equal to the SI unit for energy, a joule \(\left( \text{J} \right)\).
What volume is occupied by \(3.760 \: \text{g}\) of oxygen gas at a pressure of \(88.4 \: \text{kPa}\) and a temperature of \(19^\text{o} \text{C}\)? Assume the oxygen is an ideal gas.
Solution
Step 1: List the known quantities and plan the problem.
Known
- \(P = 88.4 \: \text{kPa}\)
- \(T = 19^\text{o} \text{C} = 292 \: \text{K}\)
- Mass \(\ce{O_2} = 3.760 \: \text{g}\)
- \(\ce{O_2} = 32.00 \: \text{g/mol}\)
- \(R = 8.314 \: \text{J/K} \cdot \text{mol}\)
Unknown
In order to use the ideal gas law, the number of moles of \(\ce{O_2}\) \(\left( n \right)\) must be found from the given mass and the molar mass. Then, use \(PV = nRT\) to solve for the volume of oxygen.
Step 2: Solve.
\[3.760 \: \text{g} \times \frac{1 \: \text{mol} \: \ce{O_2}}{32.00 \: \text{g} \: \ce{O_2}} = 0.1175 \: \text{mol} \: \ce{O_2}\]
Rearrange the ideal gas law and solve for \(V\).
\[V = \frac{nRT}{P} = \frac{0.1175 \: \text{mol} \times 8.314 \: \text{J/K} \cdot \text{mol} \times 292 \: \text{K}}{88.4 \: \text{kPa}} = 3.23 \: \text{L} \: \ce{O_2}\]
Step 3: Think about your result.
The number of moles of oxygen is far less than one mole, so the volume should be fairly small compared to molar volume \(\left( 22.4 \: \text{L/mol} \right)\) since the pressure and temperature are reasonably close to standard. The result has three significant figures because of the values for \(T\) and \(P\). Since a joule \(\left( \text{J} \right) = \text{kPa} \cdot \text{L}\), the units cancel out correctly, leaving a volume in liters.
Key Takeaway
- The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the ideal gas law, PV = nRT. The proportionality constant, R, is called the gas constant and has the value 0.08206 (L·atm)/(K·mol), 8.3145 J/(K·mol), or 1.9872 cal/(K·mol), depending on the units used.
- The ideal gas law can be used to calculate any of the four properties if the other three are known.