Physics - Heat & Thermo Question with Solution | TestHub

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

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

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

Given here, $\frac{d_2}{ d_1}=32$ and $\gamma =\frac{7}{5}$.

The ratio $\frac{P_1}{P_2}={\left(\frac{d_1}{ d_2}\right)}^{\gamma }={\left(\frac{1}{32}\right)}^{\frac{7}{5}}=\frac{1}{128}$

The ratio of temperature using ideal gas equation is $\frac{T_1}{ T_2}=\frac{P_1 V_1}{P_2 V_2}=\frac{1}{128}\times 32=\frac{1}{4}$

Thus, $T_2=4T_1$.