Question

The root mean square velocity of the molecules in a sample of helium is $\frac{5}{7}$ th that molecules in a sample of hydrogen. If the temperature of hydrogen gas is $0^{o}C$ that of helium sample is near about :-

• A. $0^{o}C$
• B. $5^{0}K$
• C. ${273}^{o}C$
• D. ${100}^{o}K$

Answer 1

Answer: (B)

$5^{0}K$

We know that,

$v_{rms}=\sqrt{3RT/M}$

Now, the root mean square velocity of helium

$v_{rms}\left(He\right)=\sqrt{\frac{3R\times T}{M}}....\left(I\right)$

Now, the root mean square velocity of hydrogen

$v_{rms}\left(H\right)=\sqrt{\frac{3R\times 273}{M}}.....\left(II\right)$

Now, from equation (I) and (II)

$\frac{v_{rms}\left(He\right)}{v_{rms}\left(H\right)}=\frac{\sqrt{\frac{3R\times T}{M}}}{\sqrt{\frac{3R\times 273}{M}}}$

$\frac{5}{7}=\sqrt{\frac{T\times 2}{273\times 4}}$

$\frac{25}{49}=\frac{T\times 2}{273\times 4}$

$T=\frac{273\times 25\times 4}{49\times 2}$

$T={278}^{0}C$

$T=278-273$

$T=5^{0}K$

Hence, this is the required solution