Warming Helium in a Cylinder

solved SpecificHeatOfGases

Q. A cylinder of fixed capacity 44.8 L $V = 44.8$ L contains helium $f = 3$ gas at standard temperature and pressure $T = 273$ K, $P = 1$ atm. Taking R as 8.31 J mol⁻¹ K⁻¹ $R = 8.31$ J mol⁻¹ K⁻¹, what is the amount of heat needed to raise the temperature of the gas in the cylinder by 15.0 °C $Δ T = 15.0$ °C, in joules?

Inputs

volume

V

44.8 L

constant

R

8.31 J mol⁻¹ K⁻¹

rise

Δ T

15 °C

Outputs

heat

Q

How it's solved

Solved with SpecificHeatOfGases

Given

$V = 44.8$ L
$R = 8.31$ J mol⁻¹ K⁻¹
$Δ T = 15.0$ °C
$T = 273$ K from “standard temperature and pressure”
$P = 1$ atm from “standard temperature and pressure”
$f = 3$ from “helium”

Find $Q$ .

Working

$V = 44.8$ L $= 0.0448$ m³
$Δ T = 15.0$ °C $= 15$ K
$P = 1$ atm $= 101325$ Pa
ideal gas law(helium) $P \cdot V = n \cdot R \cdot T$
$101325 \cdot 0.0448 = n \cdot 8.31 \cdot 273$
$n = 2.00092567$ mol
energy shared among the degrees of freedom(helium) $C_{v} = \frac{f \cdot R}{2}$
$C_{v} = \frac{3 \cdot 8.31}{2}$
$C_{v} = 12.465$ J mol⁻¹ K⁻¹
heat to warm a gas at fixed volume(helium) $Q = n \cdot C_{v} \cdot Δ T$
$Q = 2.00092567 \cdot 12.465 \cdot 15$
$Q = 374.1$ J
$Q \approx 374$ Jto 3 significant figures

Conclusion

At standard conditions the cylinder holds two moles; helium's atoms only move, three ways, so each mole takes three halves of R for every kelvin.

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 :warming_helium_in_a_cylinder do4  kind :numerical5  describes "A cylinder of fixed capacity %{volume} contains helium gas at standard temperature and pressure. " \6            "Taking R as %{constant}, %{heat} to raise the temperature of the gas in the cylinder by %{rise}, " \7            "in joules?"8  given volume: "44.8 L"9  given constant: "8.31 J mol-1 K-1"10  given rise: "15.0 °C"11  asks heat: "what is the amount of heat needed"12end

solution.rb

1require_relative "../questions/warming_helium_in_a_cylinder"2require_relative "../science/kinetic_theory/specific_heat_of_gases"34solve :warming_helium_in_a_cylinder, using: SpecificHeatOfGases do5  objects :helium67  take volume: helium.volume8  take constant: R9  take rise: helium.temperature_rise1011  deduce_from "standard temperature and pressure", helium.temperature => "273 K", helium.pressure => "1 atm"12  deduce_from "helium", helium.degrees_of_freedom => "3"1314  heat = helium.heat_at_fixed_volume15  answer heat:, in: "J", figures: 31617  conclusion "At standard conditions the cylinder holds two moles; helium's atoms only move, three ways, " \18             "so each mole takes three halves of R for every kelvin."19end

playground.rb

1require_relative "../solutions/warming_helium_in_a_cylinder"23play_with :warming_helium_in_a_cylinder do4  title "Warming Helium in a Cylinder"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
degrees of freedom	$f$	—	—	SpecificHeatOfGases
molar heat at fixed volume	$C_{v}$	J mol⁻¹ K⁻¹	—	SpecificHeatOfGases
temperature rise	$Δ T$	K	—	SpecificHeatOfGases
heat at fixed volume	$Q$	J	—	SpecificHeatOfGases
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 helium	$P$	Pa	0 … ∞
volume of helium	$V$	m³	0 … ∞
amount of helium	$n$	mol	—
temperature of helium	$T$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of helium	$ρ$	kg/m³	0 … ∞
molar mass of helium	$M$	kg/mol	—
molar heat at fixed volume of ground	$C_{v,g}$	J mol⁻¹ K⁻¹	—
degrees of freedom of ground	$f_{g}$	—	—
molar heat at fixed volume of helium	$C_{v}$	J mol⁻¹ K⁻¹	—
degrees of freedom of helium	$f$	—	—
heat at fixed volume of ground	$Q_{g}$	J	—
temperature rise of ground	$Δ T_{g}$	K	—
heat at fixed volume of helium	$Q$	J	—
temperature rise of helium	$Δ T$	K	—

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

C_{v,g} = \frac{f_{g} \cdot R}{2}

energy shared among the degrees of freedom

Q_{g} = n_{g} \cdot C_{v,g} \cdot Δ T_{g}

heat to warm a gas at fixed volume

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(helium)

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(helium)

C_{v,g} = \frac{f_{g} \cdot R}{2}

energy shared among the degrees of freedom(ground)

C_{v} = \frac{f \cdot R}{2}

energy shared among the degrees of freedom(helium)

Q_{g} = n_{g} \cdot C_{v,g} \cdot Δ T_{g}

heat to warm a gas at fixed volume(ground)

Q = n \cdot C_{v} \cdot Δ T

heat to warm a gas at fixed volume(helium)