Air Shared With an Evacuated Vessel

Q. A vessel containing 0.1 m³ of air at 76 cm of Hg is connected to an evacuated vessel of capacity 0.09 m³. The resultant air pressure is expects C

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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

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The solution

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 Ruby

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

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

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

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