The temperature of an ideal gas is increased | vedclass

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(B) The root mean square ($R$.$M$.$S$.) velocity of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this relation,we can s

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$2x$

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(B) The root mean square ($R$.$M$.$S$.) velocity of an ideal gas is given by the formula: $v_{rms} = \sqrt{\frac{3RT}{M}}$.From this relation,we can see that $v_{rms} \propto \sqrt{T}$.Let $v_1$ be the $R$.$M$.$S$. velocity at $T_1 = 100 \ K$ and $v_2$ be the $R$.$M$.$S$. velocity at $T_2 = 400 \ K$.Given $v_1 = x$.Using the proportionality: $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $\frac{v_2}{x} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2$.Therefore,$v_2 = 2x$.

The temperature of an ideal gas is increased from $100 \ K$ to $400 \ K$. If '$x$' is the $R$.$M$.$S$. velocity of its molecules at $100 \ K$,what does it become at $400 \ K$?

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