Neon and Oxygen by Density

Q. A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. The atomic mass of neon is 20.2 g/mol and the molecular mass of oxygen is 32.0 g/mol. What is the ratio of the mass densities of neon to oxygen in the vessel? expects 0.947

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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?

The playground

Inputs

neon

M_{n}

20.2 g/mol

oxygen

M_{o}

32 g/mol

Outputs

ratio?

The solution

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 Ruby

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

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

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

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)