RMS Speed Doubled From the Ice Point

solved RmsSpeed

Q. The temperature at which the RMS velocity of the gas molecules $M_{h} = M_{g}$ would become twice of its value $v_{rms,h} = 2 \cdot v_{rms,g}$ at 0 °C $T_{g} = 0$ °C is

How it's solved

Solved with RmsSpeed

Given

$T_{g} = 0$ °C
$M_{h} = M_{g}$ from “the gas molecules”
$v_{rms,h} = 2 \cdot v_{rms,g}$ from “would become twice of its value”

Assumed

$M_{g} = 28$ g/mol

Find $T_{h}$ .

Working

$T_{g} = 0$ °C $= 273.15$ K
$M_{g} = 28$ g/mol $= 0.028$ kg/mol
rms speed of a gas(gas) $v_{rms,g} = {(\frac{3 \cdot R \cdot T_{g}}{M_{g}})}^{0.5}$
$v_{rms,g} = {(\frac{3 \cdot 8.31446262 \cdot 273.15}{0.028})}^{0.5}$
$v_{rms,g} = 493.286587$ m/s
would become twice of its value $v_{rms,h} = 2 \cdot v_{rms,g}$
$v_{rms,h} = 2 \cdot 493.286587$
$v_{rms,h} = 986.573174$ m/s
the gas molecules $M_{h} = M_{g}$
$M_{h} = 0.028$
$M_{h} = 0.028$ kg/mol
rms speed of a gas(hot) $v_{rms,h} = {(\frac{3 \cdot R \cdot T_{h}}{M_{h}})}^{0.5}$
$986.573174 = {(\frac{3 \cdot 8.31446262 \cdot T_{h}}{0.028})}^{0.5}$
$T_{h} = 819.4$ °C
$T_{h} = 1092.6$ K $= 819.45$ °C

Conclusion

The rms speed goes as the square root of the absolute temperature, so doubling it takes four times 273 K, which is 1092 K or 819 °C.

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 :rms_speed_doubled_from_the_ice_point do4  kind :scq5  describes "The temperature at which the RMS velocity of the gas molecules would become twice of its value at " \6            "%{cold} is"7  given cold: "0 °C"89  option "819 °C"10  option "1092 °C"11  option "1100 °C"12  option "1400 °C"13end

solution.rb

1require_relative "../questions/rms_speed_doubled_from_the_ice_point"2require_relative "../science/kinetic_theory/rms_speed"34solve :rms_speed_doubled_from_the_ice_point, using: RmsSpeed do5  objects :gas, :hot67  take cold: gas.temperature89  deduce_from "the gas molecules", hot.molar_mass => gas.molar_mass10  deduce_from "would become twice of its value", hot.rms_speed => 2 * gas.rms_speed11  assume gas.molar_mass => "28 g/mol"1213  answer find_value(hot.temperature, in: "°C")1415  conclusion "The rms speed goes as the square root of the absolute temperature, so doubling it takes four times " \16             "273 K, which is 1092 K or 819 °C."17end

playground.rb

1require_relative "../solutions/rms_speed_doubled_from_the_ice_point"23play_with :rms_speed_doubled_from_the_ice_point do4  title "RMS Speed Doubled From the Ice Point"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
avogadro constant	$N_{A}$	mol⁻¹	—	RmsSpeed
rms speed	$v_{rms}$	m/s	—	RmsSpeed
average thermal energy	$E_{avg}$	J	—	RmsSpeed
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	$P_{g}$	Pa	0 … ∞
volume of gas	$V_{g}$	m³	0 … ∞
amount of gas	$n_{g}$	mol	—
temperature of gas	$T_{g}$	K	0 … 1e+09
pressure of hot	$P_{h}$	Pa	0 … ∞
volume of hot	$V_{h}$	m³	0 … ∞
amount of hot	$n_{h}$	mol	—
temperature of hot	$T_{h}$	K	0 … 1e+09
density of ground	$ρ_{g}$	kg/m³	0 … ∞
molar mass of ground	$M_{g}$	kg/mol	—
density of gas	$ρ_{g}$	kg/m³	0 … ∞
molar mass of gas	$M_{g}$	kg/mol	—
density of hot	$ρ_{h}$	kg/m³	0 … ∞
molar mass of hot	$M_{h}$	kg/mol	—
rms speed of ground	$v_{rms,g}$	m/s	—
rms speed of gas	$v_{rms,g}$	m/s	—
rms speed of hot	$v_{rms,h}$	m/s	—
average thermal energy of ground	$E_{avg,g}$	J	—
average thermal energy of gas	$E_{avg,g}$	J	—
average thermal energy of hot	$E_{avg,h}$	J	—

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

v_{rms,g} = {(\frac{3 \cdot R \cdot T_{g}}{M_{g}})}^{0.5}

rms speed of a gas

E_{avg,g} = \frac{3 \cdot R \cdot T_{g}}{2 \cdot N_{A}}

three halves k t for each molecule

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

ideal gas law(ground)

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

ideal gas law(gas)

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

ideal gas law(hot)

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

density of a gas(ground)

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

density of a gas(gas)

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

density of a gas(hot)

v_{rms,g} = {(\frac{3 \cdot R \cdot T_{g}}{M_{g}})}^{0.5}

rms speed of a gas(ground)

v_{rms,g} = {(\frac{3 \cdot R \cdot T_{g}}{M_{g}})}^{0.5}

rms speed of a gas(gas)

v_{rms,h} = {(\frac{3 \cdot R \cdot T_{h}}{M_{h}})}^{0.5}

rms speed of a gas(hot)

E_{avg,g} = \frac{3 \cdot R \cdot T_{g}}{2 \cdot N_{A}}

three halves k t for each molecule(ground)

E_{avg,g} = \frac{3 \cdot R \cdot T_{g}}{2 \cdot N_{A}}

three halves k t for each molecule(gas)

E_{avg,h} = \frac{3 \cdot R \cdot T_{h}}{2 \cdot N_{A}}

three halves k t for each molecule(hot)

M_{h} = M_{g}

the gas molecules

v_{rms,h} = 2 \cdot v_{rms,g}

would become twice of its value