The following graph shows two isotherms for a | vedclass

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(C) For an ideal gas,the equation of state is $PV = nRT$. Since the mass is fixed,$n$ is constant,so $T \propto PV$.For isotherm $T_1$,we can choose a

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$2$

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(C) For an ideal gas,the equation of state is $PV = nRT$. Since the mass is fixed,$n$ is constant,so $T \propto PV$.For isotherm $T_1$,we can choose a point on the curve,e.g.,$P = 4 	imes 10^5 	ext{ Pa}$ and $V = 1 	ext{ m}^3$. Thus,$T_1 \propto (4 	imes 10^5) 	imes 1 = 4 	imes 10^5$.For isotherm $T_2$,we can choose a point on the curve,e.g.,$P = 1 	imes 10^5 	ext{ Pa}$ and $V = 1 	ext{ m}^3$. Thus,$T_2 \propto (1 	imes 10^5) 	imes 1 = 1 	imes 10^5$.Therefore,the ratio of temperatures is $\frac{T_1}{T_2} = \frac{4 	imes 10^5}{1 	imes 10^5} = 4$.The $r.m.s.$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.The ratio of $r.m.s.$ speeds at $T_1$ and $T_2$ is $\frac{v_{rms1}}{v_{rms2}} = \sqrt{\frac{T_1}{T_2}} = \sqrt{4} = 2$.

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 $T_1$ and $T_2$ is

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