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ChemistryIdeal GasMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasynumerical

The root mean square speed of $N_{2}$ molecules in sample at temperature T is 'x'. If the temperature is doubled, then nitrogen molecules dissociate into atoms, the root mean square speed of nitrogen atoms becomes n times of 'x' find the value of n here?

Answer:

2.00

Solution:

$v_{r m s} = \sqrt{\frac{3 R T}{m}}$
$x = \sqrt{\frac{3 R T}{28}}$ …(i)
$v_{r m s} = \sqrt{\frac{3 R \times 2 T}{14}}$ …(ii)
Dividing equation (i) by equation (ii)
$\frac{x}{v_{r m s}} = \sqrt{\frac{1}{4}}$
So, $v_{r m s} = 2 x$
So, n = 2.

Stream:NTA_ABHYASSubject:ChemistryTopic:Ideal GasSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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