To increase the rms speed of gas molecules by | vedclass

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(B) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this relation,we can see that $v_{

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$56.25$

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(B) The root mean square (rms) speed of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this relation,we can see that $v_{rms} \propto \sqrt{T}$.Let the initial rms speed be $v_1$ and the initial temperature be $T_1$. Let the final rms speed be $v_2$ and the final temperature be $T_2$.Given that the rms speed increases by $25 \%$,we have $v_2 = v_1 + 0.25v_1 = 1.25v_1$.Since $v \propto \sqrt{T}$,we have $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values,we get $1.25 = \sqrt{\frac{T_2}{T_1}}$.Squaring both sides,we get $(1.25)^2 = \frac{T_2}{T_1}$,which gives $1.5625 = \frac{T_2}{T_1}$.This implies $T_2 = 1.5625 T_1$.The percentage increase in temperature is given by $\frac{T_2 - T_1}{T_1} 	imes 100 \%$.Percentage increase $= (1.5625 - 1) 	imes 100 \% = 0.5625 	imes 100 \% = 56.25 \%$.

To increase the rms speed of gas molecules by $25 \%$,the percentage increase in absolute temperature of the gas is to be

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