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(C) For an ideal gas,the equation of state is $PV = \mu RT$. Since the mass is fixed,$\mu$ is constant. Thus,$T \propto PV$.From the graph,we can choo

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$2$

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(C) For an ideal gas,the equation of state is $PV = \mu RT$. Since the mass is fixed,$\mu$ is constant. Thus,$T \propto PV$.From the graph,we can choose points on the isotherms to find the product $PV$ for each temperature.For isotherm $T_1$,at $V = 1 	ext{ m}^3$,$P = 1 	imes 10^5 	ext{ Pa}$. So,$T_1 \propto (1 	imes 1) = 1$.For isotherm $T_2$,at $V = 2 	ext{ m}^3$,$P = 2 	imes 10^5 	ext{ Pa}$. So,$T_2 \propto (2 	imes 2) = 4$.Therefore,the ratio of temperatures is $\frac{T_2}{T_1} = \frac{4}{1} = 4$.The $r.m.s.$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Thus,the ratio of $r.m.s.$ speeds is $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}} = \sqrt{4} = 2$.

The following graph shows two isotherms for a fixed mass of an ideal gas. The ratio of $r.m.s.$ speed of the molecules at temperatures $T_2$ and $T_1$ is

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