Room in a Mole — phryzby

Q. A mole of hydrogen gas fills about 22.4 litres at room pressure and 0 °C. Take a hydrogen atom to be a ball of radius 0.5 Å. How many times the room taken by the atoms themselves is the room the gas fills? expects 7.1 × 10⁴

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The web of concepts

given read from the words equation asked

The equations it used

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Check an answer

Q. A mole $n = 1$ mol of hydrogen gas fills about 22.4 litres $V_{m} = 0.0224$ m³ at room pressure and 0 °C. Take a hydrogen atom to be a ball of radius 0.5 Å $r = 0.5$ Å. How many times the room taken by the atoms themselves is the room the gas fills?

The playground

Outputs

times

R_{V}

The solution

Solved with Dimensions

Given

$r = 0.5$ Å from “0.5 Å”
$n = 1$ mol from “A mole”
$V_{m} = 0.0224$ m³ from “22.4 litres”

Find $R_{V}$ .

Working

$r = 0.5$ Å $= 5 \times 10^{−11}$ m
entities in moles $N = n \cdot N_{A}$
$N = 1 \cdot 6.02214076 \times 10^{23}$
$N = 6.02214076 \times 10^{23}$
volume of a sphere $V = \frac{4 \cdot π \cdot r^{3}}{3}$
$V = \frac{4 \cdot 3.14159265 \cdot {(5 \times 10^{−11})}^{3}}{3}$
$V = 5.23598776 \times 10^{−31}$ m³
atoms side by side $V_{N} = N \cdot V$
$V_{N} = 6.02214076 \times 10^{23} \cdot 5.23598776 \times 10^{−31}$
$V_{N} = 3.15318553 \times 10^{−7}$ m³
molar over atomic volume $R_{V} = \frac{V_{m}}{V_{N}}$
$R_{V} = \frac{0.0224}{3.15318553 \times 10^{−7}}$
$R_{V} = 71039.2706$
$R_{V} \approx 7.1 \times 10^{4}$ to 2 significant figures

Conclusion

A gas is almost all empty space, and that is why it can be squeezed and a solid cannot.

The Ruby

1require_relative "../physics"23question :room_in_a_mole do4  kind :numerical5  describes "A mole of hydrogen gas fills about 22.4 litres at room pressure and 0 °C. Take a hydrogen atom to be a ball of radius 0.5 Å. %{times}?"6  asks times: "How many times the room taken by the atoms themselves is the room the gas fills"7end

1require_relative "../questions/room_in_a_mole"2require_relative "../science/units_and_measurement/dimensions"34solve :room_in_a_mole, using: Dimensions do5  deduce_from "0.5 Å", r: "0.5 Å"6  deduce_from "A mole", n: "1 mol"7  deduce_from "22.4 litres", V_m: "0.0224 m3"8  take times: :R_V9  answer find_value(:times, figures: 2)1011  conclusion "A gas is almost all empty space, and that is why it can be squeezed and a solid " \12             "cannot."13end

1require_relative "../solutions/room_in_a_mole"23play_with :room_in_a_mole do4  title "Room in a Mole"5end

The laws

quantity	written	units	within	from
speed of light	$c$	m/s	—	Units
avogadro constant	$N_{A}$	mol⁻¹	—	Units
length	$l$	m	0 … ∞	Units
mass	$m$	kg	—	Units
time	$t$	s	0 … ∞	Units
electric current	$I$	A	—	Units
temperature	$T$	K	—	Units
amount of substance	$n$	mol	—	Units
luminous intensity	$I_{v}$	cd	—	Units
arc length	$s$	m	—	Units
radius	$r$	m	0 … ∞	Units
plane angle	$θ$	—	—	Units
area	$A$	m²	—	Units
solid angle	$Ω$	—	—	Units
light time	$t_{light}$	s	—	Units
light distance	$d_{light}$	m	—	Units
number of entities	$N$	—	—	Units
baseline	$B$	m	—	Units
distance away	$D$	m	—	Units
parallax angle	$θ_{p}$	—	—	Units
pi	$π$	—	0 … ∞	Dimensions
gravitational constant	$G$	N m² kg⁻²	—	Dimensions
volume	$V$	m³	—	Dimensions
speed	$v$	m/s	—	Dimensions
acceleration	$a$	m/s²	—	Dimensions
force	$F$	N	—	Dimensions
density	$ρ$	kg/m³	—	Dimensions
work	$W$	J	—	Dimensions
surface area	$S$	m²	—	Dimensions
height	$h$	m	—	Dimensions
breadth	$b$	m	—	Dimensions
cylinder surface area	$S_{c}$	m²	—	Dimensions
volume of a slab	$V_{s}$	m³	—	Dimensions
relative density	$ρ_{rel}$	—	—	Dimensions
density of water	$ρ_{w}$	kg/m³	—	Dimensions
linear magnification	$m_{l}$	—	—	Dimensions
image length	$l_{i}$	m	—	Dimensions
object length	$l_{o}$	m	—	Dimensions
image area	$A_{i}$	m²	—	Dimensions
object area	$A_{o}$	m²	—	Dimensions
volume of them all	$V_{N}$	m³	—	Dimensions
molar volume	$V_{m}$	m³	—	Dimensions
room to spare	$R_{V}$	—	—	Dimensions

θ = \frac{s}{r}

arc over radius

Ω = \frac{A}{r^{2}}

area over radius squared

d_{light} = c \cdot t_{light}

distance at light speed

N = n \cdot N_{A}

entities in moles

θ_{p} = \frac{B}{D}

baseline over distance

V = l^{3}

volume of a cube

v = \frac{l}{t}

average speed

a = \frac{v}{t}

average acceleration

F = m \cdot a

newtons second law

ρ = \frac{m}{V}

mass per volume

W = F \cdot l

force through a distance

S = 6 \cdot l^{2}

surface of a cube

V = \frac{4 \cdot π \cdot r^{3}}{3}

volume of a sphere

S_{c} = 2 \cdot π \cdot r \cdot (r + h)

surface of a cylinder

A = l \cdot b

area of a rectangle

V_{s} = l \cdot b \cdot h

volume from three sides

ρ_{rel} = \frac{ρ}{ρ_{w}}

density against water

m_{l} = \frac{l_{i}}{l_{o}}

lengths magnified

{m_{l}}^{2} = \frac{A_{i}}{A_{o}}

areas magnified

V_{N} = N \cdot V

atoms side by side

R_{V} = \frac{V_{m}}{V_{N}}

molar over atomic volume