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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.

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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" "." } {}

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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.

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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