At 600 K,the root mean square (rms) speed of | vedclass

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(B) The root mean square speed is given by $U_{rms} = \sqrt{\frac{3RT}{M_X}}$.The most probable speed is given by $U_{mp} = \sqrt{\frac{2RT}{M_Y}}$.Gi

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$4$

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(B) The root mean square speed is given by $U_{rms} = \sqrt{\frac{3RT}{M_X}}$.The most probable speed is given by $U_{mp} = \sqrt{\frac{2RT}{M_Y}}$.Given that $(U_{rms})_{X, 600} = (U_{mp})_{Y, 90}$,we have:$\sqrt{\frac{3 	imes R 	imes 600}{40}} = \sqrt{\frac{2 	imes R 	imes 90}{M_Y}}$.Squaring both sides:$\frac{3 	imes 600}{40} = \frac{2 	imes 90}{M_Y}$.$\frac{1800}{40} = \frac{180}{M_Y}$.$45 = \frac{180}{M_Y}$.$M_Y = \frac{180}{45} = 4 \ g \ mol^{-1}$.

At $600 \ K$,the root mean square (rms) speed of gas $X$ (molar mass $= 40 \ g \ mol^{-1}$) is equal to the most probable speed of gas $Y$ at $90 \ K$. The molar mass of the gas $Y$ is $....... \ g \ mol^{-1}$. (Nearest integer)

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