• # question_answer Pressure versus temperature graph of an ideal gas is as shown in the 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)  $2{{\rho }_{0}}$

$\rho =\frac{PM}{RT}$or $\rho \propto \frac{P}{T}$or ${{\left( \frac{P}{T} \right)}_{A}}=\frac{{{P}_{0}}}{{{T}_{o}}}$and ${{\left( \frac{P}{T} \right)}_{B}}=\frac{3}{2}\frac{{{P}_{0}}}{{{T}_{o}}}$ $\therefore$${{\left( \frac{P}{T} \right)}_{B}}=\frac{3}{2}{{\left( \frac{P}{T} \right)}_{A}}\Rightarrow {{\rho }_{B}}=\frac{3}{2}{{\rho }_{A}}=\frac{3}{2}{{\rho }_{0}}$ Hence, the correction option is .