. 04. The RMS velocity of Nitrogen molecules at N.T.P. is?

1.	. 04. The RMS velocity of Nitrogen molecules at N.T.P. is?
Answer» ANSWER: The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s Given: Here, N.T.P specifies STANDARD conditions. Therefore, Temperature will be 273 K. Molar mass of Nitrogen = 28 grams. Explanation: $\rule{300}{1.5}$ From the Formula, $\large \bigstar\;{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T\;}{M}}}}$ $\bold{Here}\begin{cases}\text{R Denotes Universal gas constant} \\ \text{T Denotes Temperature} \\ \text{M Denotes Molar mass (Kg)} \end{cases}$ Now, $\large{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T}{M}}}}$ Substituting the VALUES, $\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{3 \times 8.314 \times 273}{0.028}}}$ As, R = 8.314 J/mol.K Mass of Nitrogen = 28 grams = 0.028 Kg. $\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{24.942 \times 273}{0.028}}}$ $\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{6809.166}{0.028}}}$ $\longmapsto\large{\tt V_{RMS} = \sqrt{243184.5}}$ $\longmapsto\large{\underline{\boxed{\red{\tt V_{RMS} = 493.137 \; m/s}}}}$ ∴ The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s (approx). $\boxed{\begin{minipage}{7.5cm} $ \bigstar \underline{\text{Extra Formulas}} \\\\ \tt \star \;V_{(average)} = \sqrt{\dfrac{8 \; R \; T}{\pi \; M}} \\\\ \tt \star \;V_{(mp)} = \sqrt{\dfrac{2 \; R \; T}{ M}} \\\\ \star \; \text{Ratio of Three velocities,} \\ \longmapsto V_{(RMS)} : V_{(mp)} : V_{(Average)} = \sqrt{3} : \sqrt{2} : \sqrt{\dfrac{8}{\pi}} \\\\ \rule{200}{1} \\ \text{Note} \\ \bullet\; V_{(mp)} = \text{Most probable Speed} \\ \bullet\;V_{(average)} = \text{Average Speed} $ \end{minipage}}$ $\rule{300}{1.5}$

Answer»

ANSWER:

The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s

Given:

Here, N.T.P specifies STANDARD conditions.

Therefore, Temperature will be 273 K.
Molar mass of Nitrogen = 28 grams.

Explanation:

$\rule{300}{1.5}$

From the Formula,

$\large \bigstar\;{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T\;}{M}}}}$

$\bold{Here}\begin{cases}\text{R Denotes Universal gas constant} \\ \text{T Denotes Temperature} \\ \text{M Denotes Molar mass (Kg)} \end{cases}$

Now,

$\large{\boxed{\tt V_{RMS} = \sqrt{\dfrac{3\;R\;T}{M}}}}$

Substituting the VALUES,

$\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{3 \times 8.314 \times 273}{0.028}}}$

As,

R = 8.314 J/mol.K
Mass of Nitrogen = 28 grams = 0.028 Kg.

$\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{24.942 \times 273}{0.028}}}$

$\longmapsto\large{\tt V_{RMS} = \sqrt{\dfrac{6809.166}{0.028}}}$

$\longmapsto\large{\tt V_{RMS} = \sqrt{243184.5}}$

$\longmapsto\large{\underline{\boxed{\red{\tt V_{RMS} = 493.137 \; m/s}}}}$

∴ The R.M.S velocity of Nitrogen molecules at N.T.P. is 493 m/s (approx).

$\boxed{\begin{minipage}{7.5cm} $ \bigstar \underline{\text{Extra Formulas}} \\\\ \tt \star \;V_{(average)} = \sqrt{\dfrac{8 \; R \; T}{\pi \; M}} \\\\ \tt \star \;V_{(mp)} = \sqrt{\dfrac{2 \; R \; T}{ M}} \\\\ \star \; \text{Ratio of Three velocities,} \\ \longmapsto V_{(RMS)} : V_{(mp)} : V_{(Average)} = \sqrt{3} : \sqrt{2} : \sqrt{\dfrac{8}{\pi}} \\\\ \rule{200}{1} \\ \text{Note} \\ \bullet\; V_{(mp)} = \text{Most probable Speed} \\ \bullet\;V_{(average)} = \text{Average Speed} $ \end{minipage}}$

$\rule{300}{1.5}$