At what temperature, hydrogen molecules will have same average speed as nitrogen molecules at $47^{0}$ C?

32.2

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50

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22.86

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14.84

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Solution

The correct option is B $22.86$ K
Average speed for gaseous molecule.
$C_{a v g} = \sqrt{\frac{S R T}{π M}}$
For $N_{2}$ , $(C_{a v g})_{N_{2}}$ = $\sqrt{\frac{S R T_{M_{2}}}{π M_{N_{2}}}}$
For $H_{2}$ , $(C_{a v g})_{H_{2}}$ = $\sqrt{\frac{S R T_{H_{2}}}{π M_{H_{2}}}}$
as $(C_{a v g})_{M_{2}}$ = $(C_{a v g})_{H_{2}}$ (given)
$\sqrt{\frac{T_{M_{2}}}{M_{M_{2}}}} = \sqrt{\frac{T_{H_{2}}}{M_{H_{2}}}}$
$\frac{320 K}{28}$ = $\frac{x}{2}$
Hence temperature will be $22.86$ K

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