Question

Calculate the rms speed of hydrogen gas molecule at 27°C. Given that the molar mass for hydrogen molecules is 2 g/mol, and R = 8.31 J mol-1 K-1

Answer 1

GIVEN :–

• Temperature of hydrogen molecule = 27°C = 300 k

• Molar mass for hydrogen molecules is 2 g/mol = 2/1000 kg/mol = 0.002 kg/mol

• R = 8.31 J mol-1 K-1

TO FIND :–

• rms speed of hydrogen molecules = ?

SOLUTION :–

• We know that –

$\\ \: \: \longrightarrow \: { \boxed{ \sf V_{rms} = \sqrt{ \dfrac{3RT}{M} } }}$

• So that –

$\\ \: \: \implies \: \sf V_{rms} = \sqrt{ \dfrac{3(8.31)(300)}{(0.002)}}$

$\\ \: \: \implies \: \sf V_{rms} = \sqrt{ \dfrac{3(831)(3)(1000)}{(2)}}$

$\\ \: \: \implies \: \sf V_{rms} = \sqrt{ \dfrac{831 \times 9 \times 1000}{2}}$

$\\ \: \: \implies \: \sf V_{rms} = \sqrt{ 3739500}$

$\\ \: \: \implies \large{ \boxed{ \sf V_{rms} = 1,933.78 \: \dfrac{m}{s}}}$

Answer 2

We know that:-

$\begin{lgathered} { \sf V_{rms} = \sqrt{ \dfrac{3RT}{M} } } \\\end{lgathered}$

Solution:-

$\begin{lgathered} ⇝ \sf V_{rms} = \sqrt{ \dfrac{3(8.31)(300)}{(0.002)}} \\\end{lgathered}$

$\begin{lgathered} ⇝ \sf V_{rms} = \sqrt{ \dfrac{3(831)(3)(1000)}{(2)}} \\\end{lgathered}$

$\begin{lgathered} ⇝ \sf V_{rms} = \sqrt{ \dfrac{831 \times 9 \times 1000}{2}} \\\end{lgathered}$

$\begin{lgathered} ⇝ \sf V_{rms} = \sqrt{ 3739500} \\\end{lgathered}$

$\begin{lgathered} ⇝ {\red{ \sf V_{rms} = 1,933.78 \: \dfrac{m}{s}}} \\\end{lgathered}$