# 13.3 The ideal gas law  (Page 4/11)

 Page 4 / 11

## 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.

what is the dimension of strain
Is there a formula for time of free fall given that the body has initial velocity? In other words, formula for time that takes a downward-shot projectile to hit the ground. Thanks!
hi
Agboro
hiii
Chandan
Hi
Sahim
hi
Jeff
hey
Priscilla
sup guys
Bile
Hy
Kulsum
What is unit of watt?
Kulsum
watt is the unit of power
Rahul
p=f.v
Rahul
watt can also be expressed as Nm/s
Rahul
what s i unit of mass
Maxamed
SI unit of mass is Kg(kilogram).
Robel
what is formula of distance
Maxamed
Formula for for the falling body with initial velocity is:v^2=v(initial)^2+2*g*h
Mateo
i can't understand
Maxamed
we can't do this calculation without knowing the height of the initial position of the particle
Chathu
sorry but no more in science
Imoreh
2 forces whose resultant is 100N, are at right angle to each other .if one of them makes an angle of 30 degree with the resultant determine it's magnitude
50 N... (50 *1.732)N
Sahim
Plz cheak the ans and give reply..
Sahim
Is earth is an inertial frame?
The abacus (plural abaci or abacuses), also called a counting frame, is a calculating tool that was in use in Europe, China and Russia, centuries before the adoption of the written Hindu–Arabic numeral system
Sahim
thanks
Irungu
Most welcome
Sahim
Hey.. I've a question.
Is earth inertia frame?
Sahim
only the center
Shii
What is an abucus?
Irungu
what would be the correct interrogation "what is time?" or "how much has your watch ticked?"
a load of 20N on a wire of cross sectional area 8×10^-7m produces an extension of 10.4m. calculate the young modules of the material of the wire is of length 5m
Young's modulus = stress/strain strain = extension/length (x/l) stress = force/area (F/A) stress/strain is F l/A x
El
so solve it
Ebenezer
Ebenezer
two bodies x and y start from rest and move with uniform acceleration of a and 4a respectively. if the bodies cover the same distance in terms of tx and ty what is the ratio of tx to ty
what is cesium atoms?
The atoms which form the element Cesium are known as Cesium atoms.
Naman
A material that combines with and removes trace gases from vacuum tubes.
Shankar
what is difference between entropy and heat capacity
Varun
Heat capacity can be defined as the amount of thermal energy required to warm the sample by 1°C. entropy is the disorder of the system. heat capacity is high when the disorder is high.
Chathu
I want learn physics
sir how to understanding clearly
Vinodhini
try to imagine everything you study in 3d
revolutionary
pls give me one title
Vinodhini
displacement acceleration how understand
Vinodhini
vernier caliper usage practically
Vinodhini
karthik sir is there
Vinodhini
what are the solution to all the exercise..?
What is realm
The quantum realm, also called the quantum scale, is a term of art inphysics referring to scales where quantum mechanical effects become important when studied as an isolated system. Typically, this means distances of 100 nanometers (10−9meters) or less or at very low temperature.
revolutionary
How to understand physics
i like physics very much
Vinodhini
i want know physics practically where used in daily life
Vinodhini
I want to teach physics very interesting to studentd
Vinodhini
how can you build interest in physics
Prince
Austin
understanding difficult
Vinodhini
vinodhini mam, physics is used in our day to day life in all events..... everything happening around us can be explained in the base of physics..... saying simple stories happening in our daily life and relating it to physics and questioning students about how or why its happening like that can make
revolutionary
revolutionary
anything send about physics daily life
Vinodhini
How to understand easily
Vinodhini
revolutionary
even when you see this message in your phone...it works accord to a physics principle. you touch screen works based on physics, your internet works based on physics, etc....... check out google and search for it
revolutionary
what is mean by Newtonian principle of Relativity? definition and explanation with example
what is art physics
I've been trying to download a good and comprehensive textbook for physics, pls can somebody help me out?
Olanrewaju
try COLLEGE PHYSICS!! I think it will give you an edge.
Lawal
smith