Oxygen Drawn from a Cylinder

solved Moles

Q. An oxygen cylinder $V_{l} = V_{f}$ of volume 30 L $V_{f} = 30$ L has an initial gauge pressure of 15 atm $P_{f} = 15$ atm and a temperature of 27 °C $T_{f} = 300$ K. After some oxygen $M_{w} = M_{f}$ is withdrawn from the cylinder $n_{w} = n_{f} - n_{l}$ , the gauge pressure drops to 11 atm $P_{l} = 11$ atm and its temperature drops to 17 °C $T_{l} = 290$ K. The molecular mass of O2 is 32 g/mol $M_{f} = 32$ g/mol. Estimate the mass of oxygen taken out of the cylinder, in kilograms?

Inputs

volume

V_{f}

30 L

initial pressure

P_{f}

15 atm

final pressure

P_{l}

11 atm

molar mass

M_{f}

32 g/mol

Outputs

mass

m_{w}

How it's solved

Solved with Moles

Given

$V_{f} = 30$ L
$P_{f} = 15$ atm
$P_{l} = 11$ atm
$M_{f} = 32$ g/mol
$T_{f} = 300$ K from “a temperature of 27 °C”
$T_{l} = 290$ K from “drops to 17 °C”
$V_{l} = V_{f}$ from “An oxygen cylinder”
$M_{w} = M_{f}$ from “some oxygen”
$n_{w} = n_{f} - n_{l}$ from “withdrawn from the cylinder”

Find $m_{w}$ .

Working

$V_{f} = 30$ L $= 0.03$ m³
$P_{f} = 15$ atm $= 1.519875 \times 10^{6}$ Pa
$P_{l} = 11$ atm $= 1.114575 \times 10^{6}$ Pa
$M_{f} = 32$ g/mol $= 0.032$ kg/mol
ideal gas law(full) $P_{f} \cdot V_{f} = n_{f} \cdot R \cdot T_{f}$
$1.519875 \times 10^{6} \cdot 0.03 = n_{f} \cdot 8.31446262 \cdot 300$
$n_{f} = 18.2798946$ mol
An oxygen cylinder $V_{l} = V_{f}$
$V_{l} = 0.03$
$V_{l} = 0.03$ m³
ideal gas law(left) $P_{l} \cdot V_{l} = n_{l} \cdot R \cdot T_{l}$
$1.114575 \times 10^{6} \cdot 0.03 = n_{l} \cdot 8.31446262 \cdot 290$
$n_{l} = 13.8675062$ mol
withdrawn from the cylinder $n_{w} = n_{f} - n_{l}$
$n_{w} = 18.2798946 - 13.8675062$
$n_{w} = 4.41238834$ mol
some oxygen $M_{w} = M_{f}$
$M_{w} = 0.032$
$M_{w} = 0.032$ kg/mol
mass of a gas(withdrawn) $m_{w} = n_{w} \cdot M_{w}$
$m_{w} = 4.41238834 \cdot 0.032$
$m_{w} = 0.1412$ kg
$m_{w} \approx 0.14$ kgto 2 significant figures

Conclusion

The cylinder keeps its volume, so the ideal gas law counts the moles in it before and after; the moles missing, times the mass of a mole, went out with the oxygen drawn.

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 :oxygen_drawn_from_a_cylinder do4  kind :numerical5  describes "An oxygen cylinder of volume %{volume} has an initial gauge pressure of %{initial_pressure} and a " \6            "temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to " \7            "%{final_pressure} and its temperature drops to 17 °C. The molecular mass of O2 is %{molar_mass}. %{mass}, " \8            "in kilograms?"9  given volume: "30 L"10  given initial_pressure: "15 atm"11  given final_pressure: "11 atm"12  given molar_mass: "32 g/mol"13  asks mass: "Estimate the mass of oxygen taken out of the cylinder"14end

solution.rb

1require_relative "../questions/oxygen_drawn_from_a_cylinder"2require_relative "../science/kinetic_theory/moles"34solve :oxygen_drawn_from_a_cylinder, using: Moles do5  objects :full, :left, :withdrawn67  take volume: full.volume8  take initial_pressure: full.pressure9  take final_pressure: left.pressure10  take molar_mass: full.molar_mass1112  deduce_from "a temperature of 27 °C", full.temperature => "300 K"13  deduce_from "drops to 17 °C", left.temperature => "290 K"14  deduce_from "An oxygen cylinder", left.volume => full.volume15  deduce_from "some oxygen", withdrawn.molar_mass => full.molar_mass16  deduce_from "withdrawn from the cylinder", withdrawn.amount => full.amount - left.amount1718  mass = withdrawn.mass19  answer mass:, in: "kg", figures: 22021  conclusion "The cylinder keeps its volume, so the ideal gas law counts the moles in it before and after; the " \22             "moles missing, times the mass of a mole, went out with the oxygen drawn."23end

playground.rb

1require_relative "../solutions/oxygen_drawn_from_a_cylinder"23play_with :oxygen_drawn_from_a_cylinder do4  title "Oxygen Drawn from a Cylinder"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
avogadro constant	$N_{A}$	mol⁻¹	—	Moles
mass	$m$	kg	0 … ∞	Moles
molecules	$N$	—	—	Moles
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 full	$P_{f}$	Pa	0 … ∞
volume of full	$V_{f}$	m³	0 … ∞
amount of full	$n_{f}$	mol	—
temperature of full	$T_{f}$	K	0 … 1e+09
pressure of left	$P_{l}$	Pa	0 … ∞
volume of left	$V_{l}$	m³	0 … ∞
amount of left	$n_{l}$	mol	—
temperature of left	$T_{l}$	K	0 … 1e+09
pressure of withdrawn	$P_{w}$	Pa	0 … ∞
volume of withdrawn	$V_{w}$	m³	0 … ∞
amount of withdrawn	$n_{w}$	mol	—
temperature of withdrawn	$T_{w}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of full	$ρ_{f}$	kg/m³	0 … ∞
molar mass of full	$M_{f}$	kg/mol	—
density of left	$ρ_{l}$	kg/m³	0 … ∞
molar mass of left	$M_{l}$	kg/mol	—
density of withdrawn	$ρ_{w}$	kg/m³	0 … ∞
molar mass of withdrawn	$M_{w}$	kg/mol	—
mass of ground	$m_{g}$	kg	0 … ∞
mass of full	$m_{f}$	kg	0 … ∞
mass of left	$m_{l}$	kg	0 … ∞
mass of withdrawn	$m_{w}$	kg	0 … ∞
molecules of ground	$N_{g}$	—	—
molecules of full	$N_{f}$	—	—
molecules of left	$N_{l}$	—	—
molecules of withdrawn	$N_{w}$	—	—

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

m_{g} = n_{g} \cdot M_{g}

mass of a gas

N_{g} = n_{g} \cdot N_{A}

molecules in a gas

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

ideal gas law(ground)

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

ideal gas law(full)

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

ideal gas law(left)

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

ideal gas law(withdrawn)

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

density of a gas(ground)

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

density of a gas(full)

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

density of a gas(left)

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

density of a gas(withdrawn)

m_{g} = n_{g} \cdot M_{g}

mass of a gas(ground)

m_{f} = n_{f} \cdot M_{f}

mass of a gas(full)

m_{l} = n_{l} \cdot M_{l}

mass of a gas(left)

m_{w} = n_{w} \cdot M_{w}

mass of a gas(withdrawn)

N_{g} = n_{g} \cdot N_{A}

molecules in a gas(ground)

N_{f} = n_{f} \cdot N_{A}

molecules in a gas(full)

N_{l} = n_{l} \cdot N_{A}

molecules in a gas(left)

N_{w} = n_{w} \cdot N_{A}

molecules in a gas(withdrawn)

V_{l} = V_{f}

An oxygen cylinder

M_{w} = M_{f}

some oxygen

n_{w} = n_{f} - n_{l}

withdrawn from the cylinder