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How many air molecules are in a 14.0×12.0×10.0 ft room? assume atmospheric pressure of 1.00 atm, a room temperature of 20.0 ∘c, and ideal behavior. volume conversion:there are 28.2 liters in one cubic foot?

Respuesta :

skyluke89
First, let's find the total volume of the room (and of the gas):
[tex]V=(14.0 ft\cdot 12.0 ft \cdot 10.0 ft)\cdot 28.0 \frac{L}{ft^3}=47040 L [/tex]

Then we can use the ideal gas law to find the number of moles of the gas, n:
[tex]pV=nRT[/tex]
where 
p= 1.00 atm is the pressure
V= 47040 L is the volume of the gas
n is the number of moles 
[tex]R=0.082 atm L K^{-1} mol^{-1}[/tex] is the gas constant
[tex]T=20.0^{\circ}=293 K[/tex] is the temperature

Using these data and re-arranging the formula, we find
[tex]n= \frac{pV}{RT}=1957.9 mol [/tex]

Then we know that 1 mol of gas contains [tex]N_A = 6.022 \cdot 10^{23} [/tex] molecules (Avogadro number), so the total number of molecules of the air in the problem is
[tex]N= n N_A = (1957.9 mol)(6.022 \cdot 10^{23} mol^{-1})=1.18 \cdot 10^{27}[/tex] molecules.
ardni313

There are 1.19.10²⁷ air molecules in the room

Further explanation

In general, the gas equation can be written

[tex] \large {\boxed {\bold {PV = nRT}}} [/tex]

where

P = pressure, atm, N / m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P = atm, v = liter), or 8.314 J / mol K (P = Pa or N / m2, v = m³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / m

m = mass

M = relative molecular mass

room volume: 14x12x10 = 1680 ft³

T = 20 °C = 20 +273 = 293 K

1 ft³ = 28.2 liters, then

1680 ft³ = 47376 liters

[tex]\rm n=\dfrac{PV}{RT}\\\\n=\dfrac{1\times 47376}{0.082\times 293}\\\\n=1971.864[/tex]

[tex]\rm n=\dfrac{N}{No}\\\\N=1971.864\times 6.02.10^{23}\\\\N=1.19.10^{27}\:molecules[/tex]

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