Answer
(a) increase
(b) $308.4K$
Work Step by Step
(a) As the work is done on the system, the internal energy of the system increases when no heat enters or leaves the system. Thus, the temperature of the gas increases.
(b) We can find the final temperature as follows:
$T_f=(\frac{P_f}{P_i})^{\frac{1}{\gamma}}T_i$
We plug in the known values to obtain:
$T_f=(\frac{140KPa}{110KPa})^{\frac{2}{5}}(280K)$
$T_f=308.4K$
