13.3 The ideal gas law  (Page 4/11)

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Calculating moles per cubic meter and liters per mole

Calculate: (a) the number of moles in $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}$ of gas at STP, and (b) the number of liters of gas per mole.

Strategy and Solution

(a) We are asked to find the number of moles per cubic meter, and we know from [link] that the number of molecules per cubic meter at STP is $2\text{.}\text{68}×{\text{10}}^{\text{25}}$ . The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let $n$ stand for the number of moles,

$n\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}=\frac{N\phantom{\rule{0.25em}{0ex}}{\text{molecules/m}}^{3}}{6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{molecules/mol}}=\frac{2\text{.}\text{68}×{\text{10}}^{\text{25}}\phantom{\rule{0.25em}{0ex}}{\text{molecules/m}}^{3}}{6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{molecules/mol}}=\text{44}\text{.}5\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}\text{.}$

(b) Using the value obtained for the number of moles in a cubic meter, and converting cubic meters to liters, we obtain

$\frac{\left({\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{L/m}}^{3}\right)}{44\text{.}5\phantom{\rule{0.25em}{0ex}}{\text{mol/m}}^{3}}=\text{22}\text{.}5\phantom{\rule{0.25em}{0ex}}\text{L/mol}\text{.}$

Discussion

This value is very close to the accepted value of 22.4 L/mol. The slight difference is due to rounding errors caused by using three-digit input. Again this number is the same for all gases. In other words, it is independent of the gas.

The (average) molar weight of air (approximately 80% ${\text{N}}_{2}$ and 20% ${\text{O}}_{2}$ is $M=\text{28}\text{.}8\phantom{\rule{0.25em}{0ex}}\text{g}\text{.}$ Thus the mass of one cubic meter of air is 1.28 kg. If a living room has dimensions $5\phantom{\rule{0.25em}{0ex}}\text{m}×\text{5 m}×\text{3 m,}$ the mass of air inside the room is 96 kg, which is the typical mass of a human.

The density of air at standard conditions $\left(P=1\phantom{\rule{0.25em}{0ex}}\text{atm}$ and $T=\text{20}\text{º}\text{C}\right)$ is $1\text{.}\text{28}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$ . At what pressure is the density $0\text{.}{\text{64 kg/m}}^{3}$ if the temperature and number of molecules are kept constant?

The best way to approach this question is to think about what is happening. If the density drops to half its original value and no molecules are lost, then the volume must double. If we look at the equation $\text{PV}=\text{NkT}$ , we see that when the temperature is constant, the pressure is inversely proportional to volume. Therefore, if the volume doubles, the pressure must drop to half its original value, and ${P}_{\text{f}}=0\text{.}\text{50}\phantom{\rule{0.25em}{0ex}}\text{atm}\text{.}$

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

$\text{PV}=\frac{N}{{N}_{\text{A}}}{N}_{\text{A}}\text{kT}\text{.}$

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

$R={N}_{\text{A}}k=\left(6\text{.}\text{02}×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{-1}\right)\left(1\text{.}\text{38}×{\text{10}}^{-\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\right)=8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J}/\text{mol}\cdot \text{K}.$

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.

Calculating number of moles: gas in a bike tire

How many moles of gas are in a bike tire with a volume of $2\text{.}\text{00}×{\text{10}}^{–3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\left(2\text{.}\text{00 L}\right),$ 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.

$\begin{array}{lll}P& =& 7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\\ V& =& 2\text{.}\text{00}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\\ T& =& \text{18}\text{.}0\text{º}\text{C}=\text{291 K}\\ R& =& 8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}\end{array}$

2. Rearrange the equation to solve for $n$ and substitute known values.

$\begin{array}{lll}n& =& \frac{\text{PV}}{\text{RT}}=\frac{\left(7\text{.}\text{00}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{Pa}\right)\left(2\text{.}00×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{3}\right)}{\left(8\text{.}\text{31}\phantom{\rule{0.25em}{0ex}}\text{J/mol}\cdot \text{K}\right)\left(\text{291}\phantom{\rule{0.25em}{0ex}}\text{K}\right)}\\ \text{}& =& \text{0}\text{.}\text{579}\phantom{\rule{0.25em}{0ex}}\text{mol}\end{array}$

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 [link] , but we would get the same answer if we used the final values.

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