The pressure P _ 1 and density d _ 1 of diatomic gas (γ = (7)/(5))...

For adiabatic process, $P{V}^{\gamma} =$ constant

We know, density $d = \dfrac{mass}{Volume}$.

Then, we have $\dfrac{P}{{d}^{\gamma}} =$ constant

Given here, $\dfrac{{d}_{2}}{{d}_{1}} = 32$ and $\gamma = \dfrac{7}{5}$.

The ratio $\dfrac{{P}_{1}}{{P}_{2}} = {\left(\dfrac{{d}_{1}}{{d}_{2}}\right)}^{\gamma} = \left(\dfrac{1}{32}\right)\dfrac{7}{5} = \dfrac{1}{128}$

The ratio of temperature using ideal gas equation is $\dfrac{{T}_{1}}{{T}_{2}} = \dfrac{{P}_{1}{V}_{1}}{{P}_{2}{V}_{2}} = \dfrac{1}{128} \times 32 = \dfrac{1}{4}$

Thus, ${T}_{2} = 4{T}_{1}$.