An Unknown Gas Weighed by Its Density

solved IdealGas

Q. The density of an unknown gas at 97 °C $T = 97$ °C and 0.95 atm $P = 0.95$ atm is 2.5 g/litre $ρ = 2.5$ g/litre. Assuming the gas to be an ideal one, the molecular weight of this gas should be nearly, in g/mol?

Inputs

temperature

T

97 °C

pressure

P

0.95 atm

density

ρ

2.5 g/litre

Outputs

molar mass

M

How it's solved

Solved with IdealGas

Given

$T = 97$ °C
$P = 0.95$ atm
$ρ = 2.5$ g/litre

Assumed

$V = 1$ L

Find $M$ .

Working

$T = 97$ °C $= 370.15$ K
$P = 0.95$ atm $= 96258.75$ Pa
$ρ = 2.5$ g/litre $= 2.5$ kg/m³
$V = 1$ L $= 0.001$ m³
ideal gas law(gas) $P \cdot V = n \cdot R \cdot T$
$96258.75 \cdot 0.001 = n \cdot 8.31446262 \cdot 370.15$
$n = 0.0312772297$ mol
density of a gas(gas) $ρ = \frac{n \cdot M}{V}$
$2.5 = \frac{0.0312772297 \cdot M}{0.001}$
$M = 79.93$ g/mol
$M = 0.0799303528$ kg/mol $= 79.9303528$ g/mol
$M \approx 80$ g/molto 2 significant figures

Conclusion

A litre of any ideal gas at this temperature and pressure holds the same number of moles, so its mass over that number of moles is the molar mass; 97 °C is 370 K on the absolute scale the gas law needs.

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 :an_unknown_gas_weighed_by_its_density do4  kind :numerical5  describes "The density of an unknown gas at %{temperature} and %{pressure} is %{density}. Assuming the gas to be " \6            "an ideal one, %{molar_mass}, in g/mol?"7  given temperature: "97 °C"8  given pressure: "0.95 atm"9  given density: "2.5 g/litre"10  asks molar_mass: "the molecular weight of this gas should be nearly"11end

solution.rb

1require_relative "../questions/an_unknown_gas_weighed_by_its_density"2require_relative "../science/kinetic_theory/ideal_gas"34solve :an_unknown_gas_weighed_by_its_density, using: IdealGas do5  objects :gas67  take temperature: gas.temperature, pressure: gas.pressure, density: gas.density89  assume gas.volume => "1 L"1011  molar_mass = gas.molar_mass12  answer molar_mass:, in: "g/mol", figures: 21314  conclusion "A litre of any ideal gas at this temperature and pressure holds the same number of moles, so its " \15             "mass over that number of moles is the molar mass; 97 °C is 370 K on the absolute scale the gas law " \16             "needs."17end

playground.rb

1require_relative "../solutions/an_unknown_gas_weighed_by_its_density"23play_with :an_unknown_gas_weighed_by_its_density do4  title "An Unknown Gas Weighed by Its Density"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
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 gas	$P$	Pa	0 … ∞
volume of gas	$V$	m³	0 … ∞
amount of gas	$n$	mol	—
temperature of gas	$T$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of gas	$ρ$	kg/m³	0 … ∞
molar mass of gas	$M$	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 \cdot V = n \cdot R \cdot T

ideal gas law(gas)

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

density of a gas(ground)

ρ = \frac{n \cdot M}{V}

density of a gas(gas)