For an ideal gas the density of the gas is $\rho_0$ when temperature and pressure of the gas are $T_0$ and…
For an ideal gas the density of the gas is $\rho_0$ when temperature and pressure of the gas are $T_0$ and $\mathrm{P}_0$ respectively. When the temperature of the gas is $2 \mathrm{~T}_0$, its pressure will be $3 \mathrm{P}_0$. The new density will be
$\frac{3}{2} \rho_0$
$\frac{4}{3} \rho_0$
$\frac{3}{4} \rho_0$
$\frac{2}{3} \rho_0$
Solution
Density $\propto \mathrm{P} / \mathrm{T}$
So, $\frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{P}_2}{\mathrm{~T}_2} \times \frac{\mathrm{T}_1}{\mathrm{P}_{\mathrm{t}}}$
$\therefore \quad$ The new density is:
$\begin{aligned}
& \mathrm{d}_2=\rho_0 \times \frac{3 \mathrm{P}_0}{2 \mathrm{~T}_0} \times \frac{\mathrm{T}_0}{\mathrm{P}_0} \\
& \mathrm{~d}_2=\frac{3}{2} \rho_0
\end{aligned}$