To what temperature should the hydrogen at $327^{o} C$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?

- 123^{o} C

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123^{o} C

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- 100^{o} C

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0^{o} C

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Solution

The correct option is A $- 123^{o} C$
$v_{r m s} \propto \sqrt{\frac{3 R T}{M}}$
$\Rightarrow T \propto v_{r m s}^{2}$
$\Rightarrow \frac{T_{2}}{T_{1}} = {[\frac{v_{2}}{v_{1}}]}^{2} = \frac{1}{4}$
$\Rightarrow T_{2} = \frac{T_{1}}{4} = \frac{(273 + 327)}{4}$
$= 150 K = - 123^{o} C$

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To what temperature should the hydrogen at 327oC be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?

The correct option is A −123oCvrms∝√3RTM⇒T∝v2rms⇒T2T1=[v2v1]2=14⇒T2=T14=(273+327)4=150K=−123oC

To what temperature should the hydrogen at $327^{o} C$ be cooled at constant pressure, so that the root mean square velocity of its molecules becomes half of its previous value?

The correct option is A $- 123^{o} C$
$v_{r m s} \propto \sqrt{\frac{3 R T}{M}}$
$\Rightarrow T \propto v_{r m s}^{2}$
$\Rightarrow \frac{T_{2}}{T_{1}} = {[\frac{v_{2}}{v_{1}}]}^{2} = \frac{1}{4}$
$\Rightarrow T_{2} = \frac{T_{1}}{4} = \frac{(273 + 327)}{4}$
$= 150 K = - 123^{o} C$