A gas is adiabatically compressed till its pressure becomes 64 times to its initial pressure. What is the final temperature if initial temperature is  ${27}^{0}C$  and $\gamma$  is 3/2,

$adiabatic relation of pressure and temperature is P^{1-\gamma }{T}^{\gamma }=cons\tan t;$

$if initial pressure is P_1;$

$final pressure P_2=64P_1;$

$initial temperature T_1=27+273=300K;$

$final temperature=T_2=?$

$specific heat ratio=\frac{3}{2}$

$P^{1-\gamma }{T}^{\gamma }=cons\tan t$

$T^{\gamma }=\frac{cons\tan t}{P^{1-\gamma }}$

$T={\left(\frac{cons\tan t}{P^{1-\gamma }}\right)}^{\frac{1}{{ }^{\gamma }}}=\frac{cons\tan t}{P^{{\displaystyle \frac{1-\gamma }{\gamma }}}}=cons\tan t\left(P^{\frac{{ }^{\gamma }-1}{{ }^{\gamma }}}\right)$

$\frac{{ }^{\gamma }-1}{{ }^{\gamma }}=\frac{\frac{3}{2}-1}{\frac{3}{2}}=\frac{{\displaystyle \frac{3-2}{2}}}{\frac{3}{2}}=\frac{1}{3}----\left(1\right)$

$$\begin{gathered} \frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{{ }^{\frac{{ }^{\gamma }-1}{{ }^{\gamma }}}} \\ \frac{T_2}{300}={\left(\frac{64P_1}{P_1}\right)}^{{ }^{\frac{{ }^{\gamma }-1}{{ }^{\gamma }}}} \end{gathered}$$

$\begin{array}{l}{T}_2=300{\left(4^3\right)}^{\frac{1}{3}} from\left(1\right)\\ T_2=300\times 4\\ =1200K\end{array}$