At $${27^ \circ }C$$  a gas is compressed suddenly such that its pressure becomes $$\left( {\frac{1}{8}} \right)$$ of original pressure. Final temperature will be $$\left( {\gamma = \frac{5}{3}} \right)$$

Solution :
The adiabatic relation between $$p$$ and $$V$$ for a perfect gas is $$p{V^\gamma } = k\left( {a\,{\text{constant}}} \right)\,......\left( {\text{i}} \right)$$
Again from standard gas equation
$$\eqalign{ & pV = nRT \cr & \Rightarrow V = \frac{{RT}}{p} \cr} $$
Putting in Eq. (i), we get
$$p\frac{{{R^\gamma }{T^\gamma }}}{{{p^\gamma }}} = k$$
or $${p^{1 - \gamma }}{T^\gamma } = \frac{k}{{{R^\gamma }}} = {\text{another}}\,{\text{constant}}$$
i.e. $${p^{1 - \gamma }}{T^\gamma } = {\text{constant}}$$
Comparing two different situations,
$$p_1^1{ - ^\gamma }T_1^\gamma = p_2^1{ - ^\gamma }T_2^\gamma $$
Here, $${p_2} = \left( {\frac{1}{8}} \right){p_1}$$
$$\eqalign{ & {T_1} = {27^ \circ }C = 273 + 27 = 300\,K \cr & {T_2} = ?,\gamma = \frac{5}{3} \cr & \therefore {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^\gamma } = {\left( {\frac{{{p_1}}}{{{p_2}}}} \right)^{1 - \gamma }} \cr} $$
or $${\left( {\frac{{{T_2}}}{{300}}} \right)^{\frac{5}{3}}} = {\left( 8 \right)^{1 - \frac{5}{3}}} = {\left( 8 \right)^{ - \frac{2}{3}}}$$
$$\eqalign{ & \Rightarrow {T_2} = 130.6\,K \cr & \therefore {T_2} = - {142^ \circ }C \cr} $$