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(D) From the ideal gas equation,$PV = nRT = \frac{m}{M} RT$,where $m$ is the mass and $M$ is the molar mass.Since density $\varrho = \frac{m}{V}$,we c

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$4 P_0$

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(D) From the ideal gas equation,$PV = nRT = \frac{m}{M} RT$,where $m$ is the mass and $M$ is the molar mass.Since density $\varrho = \frac{m}{V}$,we can write $P = \frac{\varrho RT}{M}$,which implies $P \propto \varrho T$.At point $A$: $P_A = P_0$,$T_A = T_0$,and $\varrho_A = \varrho_0$. Thus,$P_0 = k \cdot \varrho_0 T_0$ (where $k = \frac{R}{M}$).At point $B$: $P_B = Y$,$T_B = 3T_0$,and $\varrho_B = \frac{4}{3} \varrho_0$.Taking the ratio of the two equations:$\frac{P_B}{P_A} = \frac{\varrho_B T_B}{\varrho_A T_A}$$\frac{Y}{P_0} = \frac{(\frac{4}{3} \varrho_0) (3T_0)}{\varrho_0 T_0}$$\frac{Y}{P_0} = \frac{4}{3} 	imes 3 = 4$Therefore,$Y = 4 P_0$.

For a gas,as shown in the $P-T$ graph,the densities at points $A$ and $B$ are $\varrho_0$ and $\frac{4}{3} \varrho_0$ respectively. The value of $Y$ on the pressure $(P)$ axis is

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