In an adiabatic process, the density of a diatomic gas becomes 32 n times its initial value. The final…

Given,

Final density of diatomic gas is ${\rho }_f=32n{\rho }_i$

Final pressure is $P_f=n{P}_i$

In adiabatic process, equation of state is given by,

$P{V}^{\gamma }=\text{constant} ...\left(1\right)$

Where, $V$ is the volume, $P$ is the pressure and $\gamma =\frac{7}{5}$ is the ratio of specific heat of diatomic gas

$P{\left(\frac{m}{\rho }\right)}^{\gamma }=\text{constant} ...\left(2\right)$, As $V=\frac{m}{\rho }$

and mass is constant, then the pressure is given by

$\therefore P\propto {\rho }^{\gamma }$

$\Rightarrow \frac{P_f}{P_i}={\left(\frac{{\rho }_f}{{\rho }_i}\right)}^{\gamma }$

$\Rightarrow \frac{n{P}_i}{P_i}={\left(\frac{32{\rho }_i}{{\rho }_i}\right)}^{{}^{7/5}}=(32{)}^{7/5}$

$\therefore \mathrm{n}=2^7=128$