Question

the ratio of most probable speed of helium at 100 k to that of 02 at 400 k is?​

Answer 1

Answer:

Underoot 2 ratio 1 is the correct answer

Explanation:

Answer 2

Given:

The ratio of the most probable speed of helium at 100 k to that of 02 at 400 k is?​

To find:

The ratio of the most probable speeds

Solution:

To solve the given problem we will use the following formula of the most probable speed of a gas:

$\boxed{\boxed{\bold{V_m_p = \sqrt{\frac{2RT}{M} } }}}$

where R = gas constant = 8.314 kg m²/mol K s², T = temperature (K), M = molecular mass (kg/mol)

For Helium gas:

T = 100 K

M = 4 g/mol = 0.004 kg/mol

∴ Most probable speed of He, $V_H_e = \sqrt{\frac{2R \times 100}{0.004} } = \sqrt{\frac{2R \times 100\times 10^3}{4} }$ . . . (1)

For Oxygen gas:

T = 400 K

M = 32 g/mol = 0.032 kg/mol

∴ Most probable speed of O₂, $V_O_ 2 = \sqrt{\frac{2R \times 400}{0.032} } = \sqrt{\frac{2R \times 400\times 10^3}{32} }$ . . . (2)

Now,

The ratio of the most probable speed of Helium to that of O₂ is,

= $\frac{V_H_e (at \:100 K)}{V_O_2 (at \:400 K)}$

on substituting from (1) and (2), we get

= $\frac{\sqrt{\frac{2R \times 100\times 10^3}{4} }}{ \sqrt{\frac{2R \times 400\times 10^3}{32} }}$

= $\frac{\sqrt{2R} \times \sqrt{\frac{100}{4}} \times \sqrt{ 10^3} }{ \sqrt{2R} \times \sqrt{\frac{400}{32}} \times \sqrt{ 10^3}}$

On cancelling the similar terms from numerator and denominator, we get

= $\frac{ \sqrt{\frac{100}{4}} }{ \sqrt{\frac{400}{32}}}$

= $\frac{ \sqrt{25} }{ \sqrt{\frac{25}{2}}}$

= $\frac{ \sqrt{25} \times \sqrt{2} }{ \sqrt{25}}$

= $\bold{\sqrt{2} : 1}$

Thus, the ratio of the most probable speed of helium at 100 k to that of 0₂ at 400 k is → $\underline{\bold{\sqrt{2} : 1}}$.

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