Neon and Oxygen in a Vessel

Q. A vessel contains two non-reactive gases: neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. What is the ratio of the number of molecules of neon to oxygen in the vessel? expects 1.5

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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 $P_{n} = 1.5 \cdot P_{o}$ . What is the ratio of the number of molecules of neon to oxygen in the vessel?

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Outputs

ratio?

The solution

Solved with IdealGas

Given

$P_{n} = 1.5 \cdot P_{o}$ 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 $\frac{n_{n}}{n_{o}}$ .

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) $P_{n} \cdot V_{n} = n_{n} \cdot R \cdot T_{n}$
$n_{n} = \frac{1.5 \cdot P_{o} \cdot V_{o}}{8.31446262 \cdot T_{o}}$
$\frac{n_{n}}{n_{o}} = 1.5$

Conclusion

Sharing one vessel at one temperature, each gas presses in proportion to how many molecules it has, whatever those molecules are.

The Ruby

1require_relative "../physics"23question :neon_and_oxygen_in_a_vessel 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. %{ratio} of neon to oxygen in the vessel?"7  asks ratio: "What is the ratio of the number of molecules"8end

1require_relative "../questions/neon_and_oxygen_in_a_vessel"2require_relative "../science/kinetic_theory/ideal_gas"34solve :neon_and_oxygen_in_a_vessel, using: IdealGas do5  objects :neon, :oxygen67  deduce_from "partial pressures is 3:2", neon.pressure => 1.5 * oxygen.pressure8  deduce_from "A vessel contains two", neon.volume => oxygen.volume, neon.temperature => oxygen.temperature910  answer find_relation(neon.amount / oxygen.amount), figures: 21112  conclusion "Sharing one vessel at one temperature, each gas presses in proportion to how many molecules it " \13             "has, whatever those molecules are."14end

1require_relative "../solutions/neon_and_oxygen_in_a_vessel"23play_with :neon_and_oxygen_in_a_vessel do4  title "Neon and Oxygen in a Vessel"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)