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# The root mean square velocity of the molecules in a sample of helium is ${{\left( \dfrac{5}{7} \right)}^{th}}$ that molecules in a sample of hydrogen. If the temperature of hydrogen gas is $0{}^\circ C$, that of helium sample is about:A. $5.6{}^\circ C$B. $4K$ C. $213{}^\circ C$ D. $100{}^\circ C$

Last updated date: 20th Jun 2024
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Hint: We are given the relation between root mean square velocity of a helium sample and hydrogen sample and also the temperature of the hydrogen sample. We do know the equation for root mean square velocity. By using this equation we can find the root mean square velocity of hydrogen and helium and later we can relate them using the given relation. By substituting the known values in the root mean square velocity equation we will get the temperature of the helium sample.

Formula used:
$\left( {{v}_{rms}} \right)=\sqrt{\dfrac{3RT}{M}}$

In the question we are given the relation between root mean square velocity of a helium sample and hydrogen sample.
It says that,
${{\left( {{v}_{rms}} \right)}_{He}}=\dfrac{5}{7}{{\left( {{v}_{rms}} \right)}_{H}}$, were ‘${{\left( {{v}_{rms}} \right)}_{He}}$’ is the root mean square velocity of the helium sample and ‘${{\left( {{v}_{rms}} \right)}_{H}}$’ is the root mean square velocity of the hydrogen sample.
We know that the equation for root mean square velocity is given as,
$\left( {{v}_{rms}} \right)=\sqrt{\dfrac{3RT}{M}}$, were ‘R’ is the universal gas constant, ‘T’ is the temperature and ‘M’ is the mass.
Now we can write the root mean square velocity of the given helium sample as,
${{\left( {{v}_{rms}} \right)}_{He}}=\sqrt{\dfrac{3R{{T}_{He}}}{{{M}_{He}}}}$
We know that mass of helium, ${{M}_{He}}=4kg$. Therefore,
$\Rightarrow {{\left( {{v}_{rms}} \right)}_{He}}=\sqrt{\dfrac{3R{{T}_{He}}}{4}}$
Now let us write the root mean square velocity equation of the given hydrogen sample.
${{\left( {{v}_{rms}} \right)}_{H}}=\sqrt{\dfrac{3R{{T}_{H}}}{{{M}_{H}}}}$
In the question we are given the temperature of the hydrogen sample, ${{T}_{H}}=0{}^\circ C=273K$ and we know that the mass of hydrogen, ${{M}_{H}}=2kg$. Thus we get,
${{\left( {{v}_{rms}} \right)}_{H}}=\sqrt{\dfrac{3R\times 273}{2}}$
Since we are given ${{\left( {{v}_{rms}} \right)}_{He}}=\dfrac{5}{7}{{\left( {{v}_{rms}} \right)}_{H}}$, we can write
$\Rightarrow \sqrt{\dfrac{3R{{T}_{He}}}{4}}=\left( \dfrac{5}{7} \right)\sqrt{\dfrac{3R\times 273}{2}}$
By solving this we will get,
$\Rightarrow \left( \dfrac{\left( \sqrt{\dfrac{3R{{T}_{He}}}{4}} \right)}{\left( \sqrt{\dfrac{3R\times 273}{2}} \right)} \right)=\left( \dfrac{5}{7} \right)$
$\Rightarrow \sqrt{\dfrac{\left( \dfrac{{{T}_{He}}}{4} \right)}{\left( \dfrac{273}{2} \right)}}=\left( \dfrac{5}{7} \right)$
$\Rightarrow \sqrt{\dfrac{\left( \dfrac{{{T}_{He}}}{2} \right)}{273}}=\left( \dfrac{5}{7} \right)$
By squaring both sides, we get
$\Rightarrow \dfrac{{{T}_{He}}}{2\times 273}=\dfrac{25}{49}$
From the above equation we get the temperature of the helium sample as,
$\Rightarrow {{T}_{He}}=\dfrac{25\times 2\times 273}{49}$
$\Rightarrow {{T}_{He}}=278.6K$
By converting this into Celsius, we get
$\Rightarrow {{T}_{He}}=5.6{}^\circ C$
Therefore the temperature of the helium sample is $5.6{}^\circ C$.

So, the correct answer is “Option A”.

Note: Root mean square velocity, also known as RMS velocity is the square root of the mean squares of the velocities of each individual gas molecule in a sample.
Here we take the mass of hydrogen as 2kg. This is because hydrogen never exists as an atom. It will always exist as ${{\text{H}}_{\text{2}}}$ molecule. Hence the mass is 2 kg.