The Ideal Gas Law Restated Using Moles
The Ideal Gas Law Restated Using Moles
A very common expression of the ideal gas law uses the number of moles, \(n\), rather than the number of atoms and molecules, \(N\). We start from the ideal gas law,
\(\text{PV}=\text{NkT,}\)
and multiply and divide the equation by Avogadro’s number \({N}_{\text{A}}\). This gives
\(\text{PV}=\cfrac{N}{{N}_{\text{A}}}{N}_{\text{A}}\text{kT}\text{.}\)
Note that \(n=N/{N}_{\text{A}}\) is the number of moles. We define the universal gas constant \(R={N}_{\text{A}}k\), and obtain the ideal gas law in terms of moles.
Ideal Gas Law (in terms of moles)
The ideal gas law (in terms of moles) is
\(\text{PV}=\text{nRT}.\)
The numerical value of \(R\) in SI units is
\(R={N}_{\text{A}}k=(6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{-1})(1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K})=8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J}/\text{mol}\cdot \text{K}.\)
In other units,
\(\begin{array}{lll}R& =& 1\text{.}\text{99}\phantom{\rule{0.25em}{0ex}}\text{cal/mol}\cdot \text{K}\\ R& =& \text{0}\text{.}\text{0821 L}\cdot \text{atm/mol}\cdot \text{K}\text{.}\end{array}\)
You can use whichever value of \(R\) is most convenient for a particular problem.
Example: Calculating Number of Moles: Gas in a Bike Tire
How many moles of gas are in a bike tire with a volume of \(2\text{.}\text{00}×{\text{10}}^{–3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}(2\text{.}\text{00 L}),\) a pressure of \(7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\) (a gauge pressure of just under \(\text{90}\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{lb/in}}^{2}\)), and at a temperature of \(\text{18}\text{.}0\text{º}\text{C}\)?
Strategy
Identify the knowns and unknowns, and choose an equation to solve for the unknown. In this case, we solve the ideal gas law, \(\text{PV}=\text{nRT}\), for the number of moles \(n\).
Solution
1. Identify the knowns.
\(\begin{array}{lll}P& =& 7\text{.}\text{00}\times{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\\ V& =& 2\text{.}\text{00}\times{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\\ T& =& \text{18}\text{.}0\text{º}\text{C}=\text{291 K}\\ R& =& 8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}\end{array}\)
2. Rearrange the equation to solve for \(n\) and substitute known values.
\(\begin{array}{lll}n& =& \cfrac{\text{PV}}{\text{RT}}=\cfrac{\left(7\text{.}\text{00}\times{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\right)\left(2\text{.}00\times{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\right)}{\left(8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}\right)\left(\text{291}\phantom{\rule{0.25em}{0ex}}\text{K}\right)}\\ \text{}& =& \text{0}\text{.}\text{579}\phantom{\rule{0.25em}{0ex}}\text{mol}\end{array}\)
Discussion
The most convenient choice for \(R\) in this case is \(8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K,}\) because our known quantities are in SI units. The pressure and temperature are obtained from the initial conditions in this example, but we would get the same answer if we used the final values.
The ideal gas law can be considered to be another manifestation of the law of conservation of energy. Work done on a gas results in an increase in its energy, increasing pressure and/or temperature, or decreasing volume. This increased energy can also be viewed as increased internal kinetic energy, given the gas’s atoms and molecules.
This lesson is part of:
Temperature, Kinetic Theory, and Gas Laws