Hydrogen to Match Oxygen's PV/T

solved Moles

Q. Plots of PV/T versus P are drawn for 1.00 × 10⁻³ kg $m_{o} = 1.00 \times 10^{−3}$ kg of oxygen gas at two different temperatures. The molecular mass of H2 is 2.02 g/mol $M_{h} = 2.02$ g/mol and of O2 is 32.0 g/mol $M_{o} = 32.0$ g/mol. If we obtained similar plots for hydrogen, what mass of hydrogen yields the same value of PV/T $n_{h} = n_{o}$ (for low pressure high temperature region of the plot), in kilograms?

Inputs

oxygen mass

m_{o}

0.001 kg

hydrogen

M_{h}

2.02 g/mol

oxygen

M_{o}

32 g/mol

Outputs

mass

m_{h}

How it's solved

Solved with Moles

Given

$m_{o} = 1.00 \times 10^{−3}$ kg
$M_{h} = 2.02$ g/mol
$M_{o} = 32.0$ g/mol
$n_{h} = n_{o}$ from “the same value of PV/T”

Find $m_{h}$ .

Working

$M_{h} = 2.02$ g/mol $= 0.00202$ kg/mol
$M_{o} = 32.0$ g/mol $= 0.032$ kg/mol
mass of a gas(oxygen) $m_{o} = n_{o} \cdot M_{o}$
$0.001 = n_{o} \cdot 0.032$
$n_{o} = 0.03125$ mol
the same value of PV/T $n_{h} = n_{o}$
$n_{h} = 0.03125$
$n_{h} = 0.03125$ mol
mass of a gas(hydrogen) $m_{h} = n_{h} \cdot M_{h}$
$m_{h} = 0.03125 \cdot 0.00202$
$m_{h} = 6.312 \times 10^{−5}$ kg
$m_{h} \approx 6.3 \times 10^{−5}$ kgto 2 significant figures

Conclusion

Where a gas behaves ideally, PV/T is the number of moles times the gas constant, so the same value needs the same number of moles, and hydrogen's moles are far lighter.

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 :hydrogen_to_match_oxygen_s_pv_over_t do4  kind :numerical5  describes "Plots of PV/T versus P are drawn for %{oxygen_mass} of oxygen gas at two different temperatures. The " \6            "molecular mass of H2 is %{hydrogen} and of O2 is %{oxygen}. If we obtained similar plots for hydrogen, " \7            "%{mass} yields the same value of PV/T (for low pressure high temperature region of the plot), in " \8            "kilograms?"9  given oxygen_mass: "1.00e-3 kg"10  given hydrogen: "2.02 g/mol"11  given oxygen: "32.0 g/mol"12  asks mass: "what mass of hydrogen"13end

solution.rb

1require_relative "../questions/hydrogen_to_match_oxygen_s_pv_over_t"2require_relative "../science/kinetic_theory/moles"34solve :hydrogen_to_match_oxygen_s_pv_over_t, using: Moles do5  objects :oxygen, :hydrogen67  take oxygen_mass: oxygen.mass8  take hydrogen: hydrogen.molar_mass9  take oxygen: oxygen.molar_mass1011  deduce_from "the same value of PV/T", hydrogen.amount => oxygen.amount1213  mass = hydrogen.mass14  answer mass:, in: "kg", figures: 21516  conclusion "Where a gas behaves ideally, PV/T is the number of moles times the gas constant, so the same value " \17             "needs the same number of moles, and hydrogen's moles are far lighter."18end

playground.rb

1require_relative "../solutions/hydrogen_to_match_oxygen_s_pv_over_t"23play_with :hydrogen_to_match_oxygen_s_pv_over_t do4  title "Hydrogen to Match Oxygen's PV/T"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 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
pressure of hydrogen	$P_{h}$	Pa	0 … ∞
volume of hydrogen	$V_{h}$	m³	0 … ∞
amount of hydrogen	$n_{h}$	mol	—
temperature of hydrogen	$T_{h}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of oxygen	$ρ_{o}$	kg/m³	0 … ∞
molar mass of oxygen	$M_{o}$	kg/mol	—
density of hydrogen	$ρ_{h}$	kg/m³	0 … ∞
molar mass of hydrogen	$M_{h}$	kg/mol	—
mass of ground	$m_{g}$	kg	0 … ∞
mass of oxygen	$m_{o}$	kg	0 … ∞
mass of hydrogen	$m_{h}$	kg	0 … ∞
molecules of ground	$N_{g}$	—	—
molecules of oxygen	$N_{o}$	—	—
molecules of hydrogen	$N_{h}$	—	—

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

ideal gas law(oxygen)

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

ideal gas law(hydrogen)

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

density of a gas(ground)

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

density of a gas(oxygen)

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

density of a gas(hydrogen)

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

mass of a gas(ground)

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

mass of a gas(oxygen)

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

mass of a gas(hydrogen)

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

molecules in a gas(ground)

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

molecules in a gas(oxygen)

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

molecules in a gas(hydrogen)

n_{h} = n_{o}

the same value of PV/T