RMS Speed Doubled From the Ice Point

Q. The temperature at which the RMS velocity of the gas molecules would become twice of its value at 0 °C is expects A

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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

The playground

The solution

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 Ruby

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

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

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

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