Neon and Oxygen by Density

solved IdealGas

Q. A vessel contains two $V_{n} = V_{o}$ , $T_{n} = T_{o}$ non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2 $P_{n} = 1.5 \cdot P_{o}$ . The atomic mass of neon is 20.2 g/mol $M_{n} = 20.2$ g/mol and the molecular mass of oxygen is 32.0 g/mol $M_{o} = 32.0$ g/mol. What is the ratio of the mass densities of neon to oxygen in the vessel?

Inputs

neon

M_{n}

20.2 g/mol

oxygen

M_{o}

32 g/mol

Outputs

ratio?

How it's solved

Solved with IdealGas

Given

$M_{n} = 20.2$ g/mol
$M_{o} = 32.0$ g/mol
$P_{n} = 1.5 \cdot P_{o}$ from “partial pressures is 3:2”
$V_{n} = V_{o}$ from “A vessel contains two”
$T_{n} = T_{o}$ from “A vessel contains two”

Find $\frac{ρ_{n}}{ρ_{o}}$ .

Working

ideal gas law(oxygen) $P_{o} \cdot V_{o} = n_{o} \cdot R \cdot T_{o}$
$n_{o} = \frac{P_{o} \cdot V_{o}}{8.31446262 \cdot T_{o}}$
ideal gas law(neon) $P_{n} \cdot V_{n} = n_{n} \cdot R \cdot T_{n}$
$n_{n} = \frac{1.5 \cdot P_{o} \cdot V_{o}}{8.31446262 \cdot T_{o}}$
density of a gas(oxygen) $ρ_{o} = \frac{n_{o} \cdot M_{o}}{V_{o}}$
$ρ_{o} = \frac{n_{o} \cdot 0.032}{V_{o}}$
density of a gas(neon) $ρ_{n} = \frac{n_{n} \cdot M_{n}}{V_{n}}$
$ρ_{n} = \frac{n_{n} \cdot 0.0202}{V_{o}}$
$\frac{ρ_{n}}{ρ_{o}} = 0.946875$
$M_{n} = 20.2$ g/mol $= 0.0202$ kg/mol
$M_{o} = 32.0$ g/mol $= 0.032$ kg/mol

Conclusion

Neon has more molecules in the vessel, but each is lighter, so its greater number nearly cancels its lightness.

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 :neon_and_oxygen_by_density do4  kind :numerical5  describes "A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their " \6            "partial pressures is 3:2. The atomic mass of neon is %{neon} and the molecular mass of oxygen is " \7            "%{oxygen}. %{ratio} of neon to oxygen in the vessel?"8  given neon: "20.2 g/mol"9  given oxygen: "32.0 g/mol"10  asks ratio: "What is the ratio of the mass densities"11end

solution.rb

1require_relative "../questions/neon_and_oxygen_by_density"2require_relative "../science/kinetic_theory/ideal_gas"34solve :neon_and_oxygen_by_density, using: IdealGas do5  objects :neon, :oxygen67  take neon: neon.molar_mass8  take oxygen: oxygen.molar_mass910  deduce_from "partial pressures is 3:2", neon.pressure => 1.5 * oxygen.pressure11  deduce_from "A vessel contains two", neon.volume => oxygen.volume, neon.temperature => oxygen.temperature1213  answer find_relation(neon.density / oxygen.density), figures: 31415  conclusion "Neon has more molecules in the vessel, but each is lighter, so its greater number nearly cancels " \16             "its lightness."17end

playground.rb

1require_relative "../solutions/neon_and_oxygen_by_density"23play_with :neon_and_oxygen_by_density do4  title "Neon and Oxygen by Density"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 neon	$P_{n}$	Pa	0 … ∞
volume of neon	$V_{n}$	m³	0 … ∞
amount of neon	$n_{n}$	mol	—
temperature of neon	$T_{n}$	K	0 … 1e+09
pressure of oxygen	$P_{o}$	Pa	0 … ∞
volume of oxygen	$V_{o}$	m³	0 … ∞
amount of oxygen	$n_{o}$	mol	—
temperature of oxygen	$T_{o}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of neon	$ρ_{n}$	kg/m³	0 … ∞
molar mass of neon	$M_{n}$	kg/mol	—
density of oxygen	$ρ_{o}$	kg/m³	0 … ∞
molar mass of oxygen	$M_{o}$	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_{n} \cdot V_{n} = n_{n} \cdot R \cdot T_{n}

ideal gas law(neon)

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

ideal gas law(oxygen)

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

density of a gas(ground)

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

density of a gas(neon)

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

density of a gas(oxygen)

P_{n} = 1.5 \cdot P_{o}

partial pressures is 3:2

V_{n} = V_{o}

A vessel contains two (volume of neon)

T_{n} = T_{o}

A vessel contains two (temperature of neon)