Explanation

(a) Given $T_{1} = 27 + 273 = 300\ K$
${\ \ \ \ \ \ \ \ \ \ \ \ V}_{1} = V(let)$
${\ \ \ \ \ \ \ \ \ \ \ V}_{2} = \frac{8}{27}\ V$
Then for adiabatic process
$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1\ }$
or $T_{2} = T_{1}\left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1}$
For monoatomic gas, $\gamma = 5/3$
So, $T_{2} = 300\left( \frac{V \times 27}{8V} \right)^{\frac{5}{3} - 1} = 675\ K$
$ie,\ \ \ \ \ \ \ \ T_{2} = 675 - 273 = 402℃$
Hence, increase in temperature
$= 402 - 27{^\circ} = 375℃$