Air Shared With an Evacuated Vessel

solved IdealGas

Q. A vessel containing 0.1 m³ $V_{a} = 0.1$ m³ of air at 76 cm of Hg $P_{a} = 76$ cm of Hg is connected to an evacuated vessel $V_{s} = V_{a} + V_{e}$ , $n_{s} = n_{a}$ of capacity 0.09 m³ $V_{e} = 0.09$ m³. The resultant air pressure is

How it's solved

Solved with IdealGas

Given

$V_{a} = 0.1$ m³
$P_{a} = 76$ cm of Hg
$V_{e} = 0.09$ m³
$V_{s} = V_{a} + V_{e}$ from “is connected to an evacuated vessel”
$n_{s} = n_{a}$ from “is connected to an evacuated vessel”
$T_{s} = T_{a}$ from “assumed”

Assumed

$T_{a} = 300$ K

Find $P_{s}$ .

Working

$P_{a} = 76$ cm of Hg $= 101325.014$ Pa
is connected to an evacuated vessel (volume of spread) $V_{s} = V_{a} + V_{e}$
$V_{s} = 0.1 + 0.09$
$V_{s} = 0.19$ m³
ideal gas law(air) $P_{a} \cdot V_{a} = n_{a} \cdot R \cdot T_{a}$
$101325.014 \cdot 0.1 = n_{a} \cdot 8.31446262 \cdot 300$
$n_{a} = 4.06219937$ mol
is connected to an evacuated vessel (amount of spread) $n_{s} = n_{a}$
$n_{s} = 4.06219937$
$n_{s} = 4.06219937$ mol
assumed $T_{s} = T_{a}$
$T_{s} = 300$
$T_{s} = 300$ K
ideal gas law(spread) $P_{s} \cdot V_{s} = n_{s} \cdot R \cdot T_{s}$
$P_{s} \cdot 0.19 = 4.06219937 \cdot 8.31446262 \cdot 300$
$P_{s} = 40$ cm of Hg
$P_{s} = 53328.955$ Pa $= 40$ cm of Hg

Conclusion

The same air at the same temperature now fills 0.19 m³ instead of 0.1 m³, so its pressure falls in the same proportion, from 76 to 40 cm of mercury.

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 :air_shared_with_an_evacuated_vessel do4  kind :scq5  describes "A vessel containing %{volume} of air at %{pressure} is connected to an evacuated vessel of capacity " \6            "%{capacity}. The resultant air pressure is"7  given volume: "0.1 m3"8  given pressure: "76 cm of Hg"9  given capacity: "0.09 m3"1011  option "20 cm of Hg"12  option "30 cm of Hg"13  option "40 cm of Hg"14  option "60 cm of Hg"15end

solution.rb

1require_relative "../questions/air_shared_with_an_evacuated_vessel"2require_relative "../science/kinetic_theory/ideal_gas"34solve :air_shared_with_an_evacuated_vessel, using: IdealGas do5  objects :air, :spread, :empty67  take volume: air.volume, pressure: air.pressure, capacity: empty.volume89  deduce_from "is connected to an evacuated vessel", spread.volume => air.volume + empty.volume,10                                                     spread.amount => air.amount11  assume air.temperature => "300 K", spread.temperature => air.temperature1213  answer find_value(spread.pressure, in: "cm of Hg")1415  conclusion "The same air at the same temperature now fills 0.19 m³ instead of 0.1 m³, so its pressure falls in " \16             "the same proportion, from 76 to 40 cm of mercury."17end

playground.rb

1require_relative "../solutions/air_shared_with_an_evacuated_vessel"23play_with :air_shared_with_an_evacuated_vessel do4  title "Air Shared With an Evacuated Vessel"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 air	$P_{a}$	Pa	0 … ∞
volume of air	$V_{a}$	m³	0 … ∞
amount of air	$n_{a}$	mol	—
temperature of air	$T_{a}$	K	0 … 1e+09
pressure of spread	$P_{s}$	Pa	0 … ∞
volume of spread	$V_{s}$	m³	0 … ∞
amount of spread	$n_{s}$	mol	—
temperature of spread	$T_{s}$	K	0 … 1e+09
pressure of empty	$P_{e}$	Pa	0 … ∞
volume of empty	$V_{e}$	m³	0 … ∞
amount of empty	$n_{e}$	mol	—
temperature of empty	$T_{e}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of air	$ρ_{a}$	kg/m³	0 … ∞
molar mass of air	$M_{a}$	kg/mol	—
density of spread	$ρ_{s}$	kg/m³	0 … ∞
molar mass of spread	$M_{s}$	kg/mol	—
density of empty	$ρ_{e}$	kg/m³	0 … ∞
molar mass of empty	$M_{e}$	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_{a} \cdot V_{a} = n_{a} \cdot R \cdot T_{a}

ideal gas law(air)

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

ideal gas law(spread)

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

ideal gas law(empty)

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

density of a gas(ground)

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

density of a gas(air)

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

density of a gas(spread)

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

density of a gas(empty)

V_{s} = V_{a} + V_{e}

is connected to an evacuated vessel (volume of spread)

n_{s} = n_{a}

is connected to an evacuated vessel (amount of spread)

T_{s} = T_{a}

assumed