Molar Volume at STP

solved IdealGas

Q. Molar volume is the volume occupied by 1 mol $n = 1$ mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure $P = 1$ atm, 0 °C $T = 273.15$ K). What is the molar volume, in litres?

Inputs

amount

n

1 mol

Outputs

volume

V

How it's solved

Solved with IdealGas

Given

$n = 1$ mol
$P = 1$ atm from “1 atmospheric pressure”
$T = 273.15$ K from “0 °C”

Find $V$ .

Working

$P = 1$ atm $= 101325$ Pa
ideal gas law(gas) $P \cdot V = n \cdot R \cdot T$
$101325 \cdot V = 1 \cdot 8.31446262 \cdot 273.15$
$V = 22.41$ L
$V = 0.0224139695$ m³ $= 22.4139695$ L
$V \approx 22.4$ Lto 3 significant figures

Conclusion

The ideal gas law has no room for what the molecules are, so a mole of any gas at the same pressure and temperature takes up the same volume.

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 :molar_volume_at_stp do4  kind :numerical5  describes "Molar volume is the volume occupied by %{amount} of any (ideal) gas at standard temperature and pressure " \6            "(STP : 1 atmospheric pressure, 0 °C). %{volume}, in litres?"7  given amount: "1 mol"8  asks volume: "What is the molar volume"9end

solution.rb

1require_relative "../questions/molar_volume_at_stp"2require_relative "../science/kinetic_theory/ideal_gas"34solve :molar_volume_at_stp, using: IdealGas do5  objects :gas67  take amount: gas.amount89  deduce_from "1 atmospheric pressure", gas.pressure => "1 atm"10  deduce_from "0 °C", gas.temperature => "273.15 K"1112  volume = gas.volume13  answer volume:, in: "L", figures: 31415  conclusion "The ideal gas law has no room for what the molecules are, so a mole of any gas at the same pressure " \16             "and temperature takes up the same volume."17end

playground.rb

1require_relative "../solutions/molar_volume_at_stp"23play_with :molar_volume_at_stp do4  title "Molar Volume at STP"5end

The laws it drew on

quantity	written	units	within	from
gas constant	$R$	J mol⁻¹ K⁻¹	—	IdealGas
pressure	$P$	Pa	0 … ∞	IdealGas
volume	$V$	m³	0 … ∞	IdealGas
amount	$n$	mol	—	IdealGas
temperature	$T$	K	0 … 1e+09	IdealGas
molar mass	$M$	kg/mol	—	IdealGas
density	$ρ$	kg/m³	0 … ∞	IdealGas
pressure of ground	$P_{g}$	Pa	0 … ∞
volume of ground	$V_{g}$	m³	0 … ∞
amount of ground	$n_{g}$	mol	—
temperature of ground	$T_{g}$	K	0 … 1e+09
pressure of gas	$P$	Pa	0 … ∞
volume of gas	$V$	m³	0 … ∞
amount of gas	$n$	mol	—
temperature of gas	$T$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of gas	$ρ$	kg/m³	0 … ∞
molar mass of gas	$M$	kg/mol	—

P_{g} \cdot V_{g} = n_{g} \cdot R \cdot T_{g}

ideal gas law

ρ_{g} = \frac{n_{g} \cdot M_{g}}{V_{g}}

density of a gas

P_{g} \cdot V_{g} = n_{g} \cdot R \cdot T_{g}

ideal gas law(ground)

P \cdot V = n \cdot R \cdot T

ideal gas law(gas)

ρ_{g} = \frac{n_{g} \cdot M_{g}}{V_{g}}

density of a gas(ground)

ρ = \frac{n \cdot M}{V}

density of a gas(gas)