Answer
(a) $0.062m^3$
(b) $0.083m^3$
Work Step by Step
(a) We can find the required volume as follows:
$P_iV_i^{\gamma}=P_fV_f^{\gamma}$
This can be rearranged as:
$V_f=(\frac{P_i}{P_f})V_i$
We plug in the known values to obtain:
$V_f=(\frac{105KPa}{145KPa})^{\frac{3}{5}}(0.075m^3)$
$V_f=0.062m^3$
(b) We know that
$V_f=(\frac{T_f}{T_i})^{\frac{1}{1-\gamma}}V_i$
We plug in the known values to obtain:
$V_f=(\frac{295K}{317K})^{\frac{1}{1-\frac{5}{3}}}(0.075m^3)$
$V_f=0.083m^3$