Hydrogen to Match Oxygen's PV/T

Q. Plots of PV/T versus P are drawn for 1.00 × 10⁻³ kg of oxygen gas at two different temperatures. The molecular mass of H2 is 2.02 g/mol and of O2 is 32.0 g/mol. If we obtained similar plots for hydrogen, what mass of hydrogen yields the same value of PV/T (for low pressure high temperature region of the plot), in kilograms? expects 6.3 × 10⁻⁵ kg

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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?

The playground

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}

The solution

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 Ruby

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

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

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

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