The root mean square velocity of the molecule | vedclass

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Question

(A) The root mean square velocity is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.This implies that $v_{rms} \propto \sqrt{\frac{T}{M}}$.Giv

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$0^{\circ}C$

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(A) The root mean square velocity is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.This implies that $v_{rms} \propto \sqrt{\frac{T}{M}}$.Given: $\frac{v_{He}}{v_{H_2}} = \frac{5}{7}$,$T_{H_2} = 0^{\circ}C = 273 \, K$,$M_{He} = 4 \, g/mol$,and $M_{H_2} = 2 \, g/mol$.Using the ratio: $\frac{v_{He}}{v_{H_2}} = \sqrt{\frac{T_{He}}{M_{He}} 	imes \frac{M_{H_2}}{T_{H_2}}} = \frac{5}{7}$.Squaring both sides: $\frac{T_{He}}{4} 	imes \frac{2}{273} = \frac{25}{49}$.$T_{He} = \frac{25}{49} 	imes \frac{4 	imes 273}{2} = \frac{25 	imes 2 	imes 273}{49} = \frac{50 	imes 273}{49} \approx 278.57 \, K$.Converting to Celsius: $278.57 - 273 = 5.57^{\circ}C$,which is approximately $0^{\circ}C$ given the options provided.

The root mean square velocity of the molecules in a sample of helium is $(5/7)^{th}$ that of the molecules in a sample of hydrogen. If the temperature of the hydrogen sample is $0^{\circ}C$,then the temperature of the helium sample is about:

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