Explanation

(a) At STP,
22.4 L of any gas is 1 mol,
$\therefore\ \ \ \ 5.6\ L = \frac{5.6}{22.4} = \frac{1}{4}mol = n$
In adiabatic process,
$TV^{\gamma - 1} = constant$
$\therefore\ \ \ \ T_{2}V_{2}^{\gamma - 1\ } = T_{1}V_{1}^{\gamma - 1}$
or $T_{2} = T_{1}\left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1}$
$\gamma = \frac{C_{P}}{C_{V}} = \frac{5}{3}$ for monoatomic He gas
$\therefore\ \ \ T_{2} = T_{1}\ \left( \frac{5.6}{0.7} \right)^{\frac{5}{3} - 1} = 4T_{1}$
Further in adiabatic process,
Q=0
$\therefore W + \mathrm{\Delta}U = 0$
or $W = - \mathrm{\Delta}U = - nC_{V}\mathrm{\Delta}T$
$= - n\ \left( \frac{R}{\gamma - 1} \right)(T_{2} - T_{1})$
$= - \frac{1}{4}\left( \frac{R}{\frac{5}{3} - 1} \right)(4T_{1} - T_{1})$
$= - \frac{9}{8}RT_{1}$