<< Chapter < Page Chapter >> Page >

Calculating moles per cubic meter and liters per mole

Calculate: (a) the number of moles in 1 . 00 m 3 size 12{1 "." "00"" m" rSup { size 8{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 . 68 × 10 25 size 12{2 "." "68"´"10" rSup { size 8{"25"} } } {} . The number of moles can be found by dividing the number of molecules by Avogadro’s number. We let n size 12{n} {} stand for the number of moles,

n mol/m 3 = N molecules/m 3 6 . 02 × 10 23 molecules/mol = 2 . 68 × 10 25 molecules/m 3 6 . 02 × 10 23 molecules/mol = 44 . 5 mol/m 3 . size 12{n`"mol/m" rSup { size 8{3} } = { {N`"molecules/m" rSup { size 8{3} } } over {6 "." "02" times "10" rSup { size 8{"23"} } `"molecules/mol"} } = { {2 "." "68" times "10" rSup { size 8{"25"} } `"molecules/m" rSup { size 8{3} } } over {6 "." "02" times "10" rSup { size 8{"23"} } `"molecules/mol"} } ="44" "." 5`"mol/m" rSup { size 8{3} } "." } {}

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

10 3 L/m 3 44 . 5 mol/m 3 = 22 . 5 L/mol . size 12{ { { left ("10" rSup { size 8{3} } `"L/m" rSup { size 8{3} } right )} over {44 "." 5`"mol/m" rSup { size 8{3} } } } ="22" "." 5`"L/mol" "." } {}

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% N 2 size 12{N rSub { size 8{2} } } {} and 20% O 2 size 12{O rSub { size 8{2} } } {} is M = 28 . 8 g . size 12{M="28" "." 8" g" "." } {} Thus the mass of one cubic meter of air is 1.28 kg. If a living room has dimensions 5 m × 5 m × 3 m, size 12{5" m" times "5 m" times "3 m,"} {} the mass of air inside the room is 96 kg, which is the typical mass of a human.

Got questions? Get instant answers now!

The density of air at standard conditions ( P = 1 atm size 12{ \( P=1" atm"} {} and T = 20 º C ) size 12{T="20"°C \) } {} is 1 . 28 kg/m 3 size 12{1 "." "28"" kg/m" rSup { size 8{3} } } {} . At what pressure is the density 0 . 64 kg/m 3 size 12{0 "." "64 kg/m" rSup { size 8{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 PV = NkT size 12{ ital "PV"= ital "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 f = 0 . 50 atm . size 12{P rSub { size 8{f} } =0 "." "50"" atm" "." } {}

Got questions? Get instant answers now!

The ideal gas law restated using moles

A very common expression of the ideal gas law uses the number of moles, n size 12{n} {} , rather than the number of atoms and molecules, N size 12{N} {} . We start from the ideal gas law,

PV = NkT, size 12{ ital "PV"= ital "NkT"} {}

and multiply and divide the equation by Avogadro’s number N A size 12{N rSub { size 8{A} } } {} . This gives

PV = N N A N A kT . size 12{ ital "PV"= { {N} over {N rSub { size 8{A} } } } N rSub { size 8{A} } ital "kT" "." } {}

Note that n = N / N A size 12{n=N/N rSub { size 8{A} } } {} is the number of moles. We define the universal gas constant R = N A k size 12{R=N rSub { size 8{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

PV = nRT . size 12{ ital "PV"= ital "nRT"} {}

The numerical value of R size 12{R} {} in SI units is

R = N A k = 6 . 02 × 10 23 mol 1 1 . 38 × 10 23 J/K = 8 . 31 J / mol K . size 12{R=N rSub { size 8{A} } k= left (6 "." "02" times "10" rSup { size 8{"23"} } `"mol" rSup { size 8{ - 1} } right ) left (1 "." "38" times "10" rSup { size 8{ - "23"} } `"J/K" right )=8 "." "31"`J/"mol" cdot K} {}

In other units,

R = 1 . 99 cal/mol K R = 0 . 0821 L atm/mol K . alignl { stack { size 12{R=1 "." "99"" cal/mol" cdot K} {} #size 12{R"=0" "." "0821 L" cdot "atm/mol" cdot K "." } {} } } {}

You can use whichever value of R size 12{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 . 00 × 10 3 m 3 ( 2 . 00 L ) , size 12{2 "." "00"´"10" rSup { size 8{ +- 3} } " m" rSup { size 8{3} } \( 2 "." "00 L" \) ,} {} a pressure of 7 . 00 × 10 5 Pa size 12{7 "." "00"´"10" rSup { size 8{5} } " Pa"} {} (a gauge pressure of just under 90 . 0 lb/in 2 size 12{"90" "." 0" lb/in" rSup { size 8{2} } } {} ), and at a temperature of 18 . 0 º C size 12{"18" "." 0°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, PV = nRT size 12{ ital "PV"= ital "nRT"} {} , for the number of moles n size 12{n} {} .

Solution

1. Identify the knowns.

P = 7 . 00 × 10 5 Pa V = 2 . 00 × 10 3 m 3 T = 18 . 0 º C = 291 K R = 8 . 31 J/mol K alignl { stack { size 12{P=7 "." "00" times "10" rSup { size 8{5} } " Pa"} {} #V=2 "." "00" times "10" rSup { size 8{ - 3} } " m" rSup { size 8{3} } {} # T="18" "." 0°C="291 K" {} #R=8 "." "31"" J/mol" cdot K {} } } {}

2. Rearrange the equation to solve for n size 12{n} {} and substitute known values.

n = PV RT = 7 . 00 × 10 5 Pa 2 . 00 × 10 3 m 3 8 . 31 J/mol K 291 K = 0 . 579 mol alignl { stack { size 12{n= { { ital "PV"} over { ital "RT"} } = { { left (7 "." "00" times "10" rSup { size 8{5} } `"Pa" right ) left (2 "." 00 times "10" rSup { size 8{ - 3} } `m rSup { size 8{3} } right )} over { left (8 "." "31"`"J/mol" cdot K right ) left ("291"" K" right )} } } {} #" "=" 0" "." "579"`"mol" {} } } {}

Discussion

The most convenient choice for R size 12{R} {} in this case is 8 . 31 J/mol K, size 12{8 "." "31"" J/mol" cdot "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.

Got questions? Get instant answers now!

Questions & Answers

what is thermodynamics
wana Reply
thermodynamic is a branch of physics that teaches on the relationship about heat and anyother form of energy
Emmanuel
if l cary box and stop is ther any work
Tamirat Reply
no that because u have moved no distance. for work to be performed a force needs to be applied and a distance needs to be moved
Emmanuel
Different between fundamental unit and derived unit
Alimi Reply
fundamental unit are independent quantities that do not depend on any other unit while derived unit are quantities that depend on two or more units for it definition
Emmanuel
what is nuclear fission
Sadik Reply
hello
Shawty
are you there
Shawty
miss your absence here...
Shawty
what is a vector
Temitayo Reply
vectors are quantities that have numerical value or magnitude and direction.
Muhammad
what is regelation
Oladele
vector is any quantity that has magnitude and direction
Emmanuel
Physics is a physical science that deals with the study of matter in relation to energy
Divine Reply
Hi
Jimoh
hello
Salaudeen
hello
Sadik
Yes
Maxamuud
hi everyone
Muhammad
what is physics
Rhema Reply
physics is a physical science that deals with the study of matter in relation to energy
Osayuwa
a15kg powerexerted by the foresafter 3second
Firdos Reply
what is displacement
Xolani Reply
movement in a direction
Jason
hello
Hosea
Hey
Smart
haider
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
anas Reply
Hi
saeid
hi
Yimam
Hi
Jimoh
An object made of several thin conducting layers separated by insulation may not be affected by magnetic damping because the eddy current produced in each layer due to induction will be very small and the opposing magnetic flux produced by the eddy currents will be very small
Muhammad
What is thê principle behind movement of thê taps control
Oluwakayode Reply
while
Hosea
what is atomic mass
thomas Reply
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Jesuovie Reply
Explain why it is difficult to have an ideal machine in real life situations.
Isaac Reply
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
It is difficult to have an ideal machine in real life situation because in ideal machines all the input energy should be converted to output energy . But , some part of energy is always lost in overcoming friction and input energy is always greater than output energy . Hence , no machine is ideal.
Muhammad
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
Zaharadeen Reply
Practice Key Terms 4

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask