The pressure versus temperature graph of an i | vedclass

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(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.Density $\rho = \frac{m}{V} = \frac{PM}{RT}

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

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(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.Density $\rho = \frac{m}{V} = \frac{PM}{RT}$.Since $M$ and $R$ are constants,$\rho \propto \frac{P}{T}$.At point $A$,the coordinates are $(T_0, P_0)$,so $\left(\frac{P}{T}\right)_A = \frac{P_0}{T_0}$.At point $B$,the coordinates are $(2T_0, 3P_0)$,so $\left(\frac{P}{T}\right)_B = \frac{3P_0}{2T_0} = \frac{3}{2} \left(\frac{P_0}{T_0}\right)$.Since $\rho \propto \frac{P}{T}$,we have $\frac{\rho_B}{\rho_A} = \frac{(P/T)_B}{(P/T)_A} = \frac{3}{2}$.Therefore,$\rho_B = \frac{3}{2} \rho_A = \frac{3}{2} \rho_0$.

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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