An Unknown Gas Weighed by Its Density

Q. The density of an unknown gas at 97 °C and 0.95 atm is 2.5 g/litre. Assuming the gas to be an ideal one, the molecular weight of this gas should be nearly, in g/mol? expects 80 g/mol

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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?

The playground

Inputs

temperature

T

97 °C

pressure

P

0.95 atm

density

ρ

2.5 g/litre

Outputs

molar mass

M

The solution

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 Ruby

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

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

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

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)