The average velocity at T K of CO2 gas is 9-0 - 104 cm sec-1- The value of T is-

Average velocity =√(8RTπ M) 9 × 10^4 cm/sec = 9 × 10^4100 m/sec =9 × 10^2 m/sec ∴ 9 × 10^2 = √(8 × 8.314× T_13.14 × 44 × 10^ 3) nbsp; … (A) nbsp; nbs ...

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Standard XII

Chemistry

Maxwell's Distribution of Speeds

The average v...

Question

The average velocity at $T K$ of $C O_{2}$ gas is $9.0 \times 10^{4} {cm sec}^{- 1}$ . The value of $T$ is______.

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Solution

Average velocity $= \sqrt{\frac{8 R T}{π M}}$
and most probable velocity $= \sqrt{\frac{2 R T}{M}}$
Given - for $C O_{2}$
Average velocity at $T_{1} =$ Most probable velocity at $T_{2}$
$9 \times 10^{4} cm/sec = \frac{9 \times 10^{4}}{100} m/sec$
$= 9 \times 10^{2} m/sec$

$∴ 9 \times 10^{2} = \sqrt{\frac{8 \times 8.314 \times T_{1}}{3.14 \times 44 \times 10^{- 3}}} \dots (A)$ [Average velocity at $T_{1} K$ ]

On solving, $T_{1} = 1682.5 K$

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