The pressure Vs temperature graph of an ideal | vedclass

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(C) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass,$M$ is the molar mass,and $\rho = \frac{m}{V}$ is the density.Thus,$P

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$\frac{5\rho _0}{6}$

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(C) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass,$M$ is the molar mass,and $ho = \frac{m}{V}$ is the density.Thus,$P = \frac{ho RT}{M}$,which implies $ho = \frac{PM}{RT}$.At point $A$: $P_A = 3P_0$,$T_A = 2T_0$,and $ho_A = ho_0 = \frac{(3P_0)M}{R(2T_0)} = \frac{3P_0 M}{2RT_0}$.At point $B$: $P_B = 5P_0$,$T_B = 4T_0$,and $ho_B = \frac{(5P_0)M}{R(4T_0)} = \frac{5P_0 M}{4RT_0}$.Taking the ratio: $\frac{ho_B}{ho_0} = \frac{5P_0 M / 4RT_0}{3P_0 M / 2RT_0} = \frac{5}{4} 	imes \frac{2}{3} = \frac{10}{12} = \frac{5}{6}$.Therefore,$ho_B = \frac{5}{6}ho_0$.

The pressure $Vs$ temperature graph of an ideal gas is shown in the figure. If the density of the gas at point $A$ is $\rho _0$,what will be the density at point $B$?

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