Question

The variation of density $\left(\rho \right)$ with temperature (T) of an ideal gas of molecular mass (M) at constant pressure is plotted as shown in figure.
The pressure of the gas at state P is $\frac{R{T}^2}{\sqrt{K}M}$
The value of K is

Answer 1

We know that $\rho =\frac{PM}{RT}\Rightarrow \frac{d\rho }{dt}=\frac{-PM}{R{T}^2}$
At point P, the slope is tan ${150}^{\circ }$
$\therefore \frac{d\rho }{dt}=tan{150}^{\circ }=\frac{-1}{\sqrt{3}}$
Since the pressure at point P is $\frac{R{T}^2}{\sqrt{K}M}$
then $\frac{d\rho }{dt}=-\frac{R{T}^2}{\sqrt{K}M}\times \frac{M}{R{T}^2}=\frac{-1}{\sqrt{K}}$
$\therefore \frac{-1}{\sqrt{K}}=\frac{-1}{\sqrt{3}}\Rightarrow K=3$