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The $\mathrm{KE}$ of $\mathrm{N}$ molecule of $\mathrm{O}_2$ is x joules of $-123^{\circ} \mathrm{C}$ . Anothen sample of $\mathrm{O}_2$ at $27^{\circ} \mathrm{C}$ has a $\mathrm{kE}$ of $\mathrm{2 x}$ joules. The latters sample contain-

Option: 1
$\mathrm{N}$

Option: 2
$2 \mathrm{~N}$

Option: 3
$\frac{ \mathrm{N}}{ 2}$

Option: 4
$\frac{ \mathrm{N} }{4}$

Answers (1)

best_answer

KE of asr molecule = $\mathrm{\frac{3}{2} k T}$

$\mathrm{T=-123+273=+150 K }$

$\mathrm{\frac{3}{2} K \times 150 \times N=x }$

from (i) and (i), N=N )

$\mathrm{At ~27^{\circ} \mathrm{C}=(27+273) \mathrm{K}=300 \mathrm{K} }$

$\mathrm{\frac{3}{2} K \times 300 \times N^{\prime}=2 n }$

Sn both the cases N molecules are present.

Posted by

Ritika Kankaria

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