Question

At \(300 K\) the rms speed of oxygen molecules is \(\sqrt {\frac {\alpha+5}{\alpha}}\) times to that of its average speed in the gas. Then, the value of \(\alpha\) will be
(used \(\pi=\frac {22}{7}\))

• A. 28
• B. 24
• C. 32
• D. 27

Answer 1

The Correct Option is A

 The given problem asks us to find the value of \(\alpha\) based on the relationship between the root mean square (RMS) speed and the average speed of oxygen molecules at a temperature of \(300 K\).

To solve this problem, we need to use the known formulae for RMS speed and average speed of gas molecules:

• The RMS speed, \( v_{\text{rms}} \), is given by: \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), where \( k \) is the Boltzmann constant, \( T \) is the temperature, and \( m \) is the mass of a molecule.
• The average speed, \( v_{\text{avg}} \), is given by: \(v_{\text{avg}} = \sqrt{\frac{8kT}{\pi m}}\).

The problem states that:

\(v_{\text{rms}} = \sqrt {\frac {\alpha+5}{\alpha}} \times v_{\text{avg}}\)

Substitute the expressions for \( v_{\text{rms}} \) and \( v_{\text{avg}} \):

\(\sqrt{\frac{3kT}{m}} = \sqrt {\frac {\alpha+5}{\alpha}} \times \sqrt{\frac{8kT}{\pi m}}\)

Square both sides to eliminate the square roots:

\(\frac{3kT}{m} = \frac {\alpha+5}{\alpha} \times \frac{8kT}{\pi m}\)

Cancel out common terms: \( kT/m \) from both sides:

\(3 = \frac {\alpha+5}{\alpha} \times \frac{8}{\pi}\)

Now, rearrange the equation to solve for \( \alpha \):

\(3 = \frac {8(\alpha + 5)}{\pi \alpha}\)

Using \( \pi = \frac{22}{7} \), substitute it into the equation:

\(3 = \frac {8(\alpha + 5)}{\frac{22}{7} \alpha}\)

Cross-multiply to simplify:

\(3 \times \frac{22}{7} \alpha = 8(\alpha + 5)\)

Further simplification gives:

\(\frac{66 \alpha}{7} = 8\alpha + 40\)

Multiply through by 7 to clear the fraction:

\(66 \alpha = 56 \alpha + 280\)

Subtract \( 56 \alpha \) from both sides:

\(10 \alpha = 280\)

Divide by 10 to find \( \alpha \):

\(\alpha = 28\)

Thus, the correct value of \( \alpha \) is \(28\), making the correct option

28

.