The given diagram shows isotherms for a fixed | vedclass

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(B) The root mean square (r.m.s.) speed of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Since $V_{rms} \propto \sqrt{T}$,the ratio is $\

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$2$

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(B) The root mean square (r.m.s.) speed of gas molecules is given by $V_{rms} = \sqrt{\frac{3RT}{M}}$.Since $V_{rms} \propto \sqrt{T}$,the ratio is $\frac{V_{rms,2}}{V_{rms,1}} = \sqrt{\frac{T_2}{T_1}}$.From the ideal gas equation $PV = nRT$,we have $T = \frac{PV}{nR}$.For the isotherm at $T_2$,choosing a point $(V=2 	ext{ m}^3, P=2 	imes 10^5 	ext{ Pa})$,we get $T_2 \propto P_2V_2 = (2 	imes 10^5) 	imes 2 = 4 	imes 10^5$.For the isotherm at $T_1$,choosing a point $(V=1 	ext{ m}^3, P=1 	imes 10^5 	ext{ Pa})$,we get $T_1 \propto P_1V_1 = (1 	imes 10^5) 	imes 1 = 1 	imes 10^5$.Therefore,$\frac{T_2}{T_1} = \frac{4 	imes 10^5}{1 	imes 10^5} = 4$.The ratio of the r.m.s. speeds is $\sqrt{\frac{T_2}{T_1}} = \sqrt{4} = 2$.

The given diagram shows isotherms for a fixed mass of an ideal gas at temperatures $T_1$ and $T_2$. What is the value of the ratio $\frac{\text{r.m.s. speed of the molecules at temperature } T_2}{\text{r.m.s. speed of the molecules at temperature } T_1}$?

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