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(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.Substituting $V = \frac{m}{\rho}$,we get $P(\frac{m}{\rho}) = \frac{m}{M}RT$.This s

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$\frac{3}{2}{\rho _0}$

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(B) From the ideal gas equation,$PV = nRT$,where $n = \frac{m}{M}$.Substituting $V = \frac{m}{\rho}$,we get $P(\frac{m}{\rho}) = \frac{m}{M}RT$.This simplifies to $P = \frac{\rho RT}{M}$,or $\rho = \frac{PM}{RT}$.Since $M$ and $R$ are constants,$\rho \propto \frac{P}{T}$.From the graph,at point $A$,the coordinates are $(T_0, P_0)$,so $(\frac{P}{T})_A = \frac{P_0}{T_0}$.At point $B$,the coordinates are $(2T_0, 3P_0)$,so $(\frac{P}{T})_B = \frac{3P_0}{2T_0} = \frac{3}{2}(\frac{P_0}{T_0})$.Therefore,the ratio of densities is $\frac{\rho_B}{\rho_A} = \frac{(P/T)_B}{(P/T)_A} = \frac{3}{2}$.Thus,$\rho_B = \frac{3}{2}\rho_A = \frac{3}{2}\rho_0$.

The pressure versus temperature graph of an ideal gas is as shown in the figure. The density of the gas at point $A$ is ${\rho _0}$. The density at point $B$ will be

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