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

Graph is a straight line not passing through origin.

Ideal gas: PV= nRT $=\frac{\mathrm{mRT}}{\mathrm{M}}\Rightarrow \mathrm{P}=\frac{\mathrm{\rho RT}}{\mathrm{M}}$.....(1)
$\text{From graph : }\mathrm{P}=\left[\frac{2{\mathrm{P}}_0}{{\mathrm{T}}_0}\right]\mathrm{T}-{\mathrm{P}}_0$....(2)
 

At A; P=P₀; T = T₀ ;  $\mathrm{\rho }={\mathrm{\rho }}_0$

Using $\frac{\mathrm{\rho RT}}{\mathrm{M}}=\left[\frac{2{\mathrm{P}}_0}{{\mathrm{T}}_0}\right]\mathrm{T}-{\mathrm{P}}_0$

$\mathrm{At} \mathrm{A} ; \frac{{\mathrm{\rho }}_0{\mathrm{RT}}_0}{\mathrm{M}}=\left[\frac{2{\mathrm{P}}_0}{{\mathrm{T}}_0}\right]{\mathrm{T}}_0-{\mathrm{P}}_0$

$\Rightarrow \frac{{\mathrm{\rho }}_0{\mathrm{RT}}_0}{\mathrm{M}}={\mathrm{P}}_0$

$\Rightarrow \frac{\mathrm{R}}{\mathrm{M}}=\frac{{\mathrm{P}}_0}{{\mathrm{\rho }}_0{\mathrm{T}}_0}$

At B; P = 3P₀; T = 2T₀

$\mathrm{At} \mathrm{B} ; \frac{\mathrm{\rho R}2{\mathrm{T}}_0}{\mathrm{M}}=\left[\frac{2{\mathrm{P}}_0}{{\mathrm{T}}_0}\right]2{\mathrm{T}}_0-{\mathrm{P}}_0$

$\Rightarrow \mathrm{\rho }=\frac{3}{2}{\mathrm{\rho }}_0$