The average speed at temperature $T_{1}$ and most probable speed at $T_{2}$ of $C O_{2}$ gas is $2.8282 m / s e c$ . Calculate the values of $T_{1}$ and $T_{2}$ .

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Solution

$v_{a v g} = \sqrt{\frac{8 R T}{π M}}$
$= \sqrt{\frac{8 \times 0.0821 T_{1}}{π \times 44}} = 2.8282$
$T_{1} = 1682.5 K$ ,

$v_{m p s} = \sqrt{\frac{2 R T_{2}}{M}}$
$⟹ T_{2} = 2143.4 K$

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