Question

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

• A. $\frac{3}{4}{\rho }_0$
• B. $\frac{3}{2}{\rho }_0$
• C. $\frac{4}{3}{\rho }_0$
• D. $2{\rho }_0$

Answer 1

The correct option is B $\frac{3}{2}{\rho }_0$

Ideal gas equation $PV=nRT$

Also $\frac{n}{V}=\frac{d}{M}$ where $d$ and $M$ are the density and molar mass of the gas respectively

$\therefore$ $P=\frac{d}{M}RT$ $⟹d=\frac{PM}{RT}$

At point A : $d_A=\frac{P_{A}M}{R{T}_A}$

$⟹{\rho }_o=\frac{P_{o}M}{R{T}_o}$ .........(1)

At point B : $d_B=\frac{P_{B}M}{R{T}_B}$

$⟹d_B=\frac{\left(3P_o\right)M}{R\left(2T_o\right)}=\frac{3}{2}{\rho }_o$ (from (1))