Answer
a) $P_{water}=1.706kPa$
b) $P_{air}=0.9366kPa$
Work Step by Step
a) $P_{water}=P_{sat}$
From table A-4 at $T=15^{\circ}C$
$P_{water}=1.706kPa$
b) From table A-4 at $T=20^{\circ}C$
$P_{sat}=2.339kPa$
$P_{air}=\phi P_{sat}=0.4*2.339kPa=0.9366kPa$
