Room in a Mole — phryzby

solved Dimensions

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?

Outputs

times

R_{V}

How it's solved

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 web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

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

solution.rb

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

playground.rb

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

The laws it drew on

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