Answer
(a) 18.2 atm
(b) 0.274 L
Work Step by Step
(a) $V_{1}=35.0\,L$
$P_{1}=59.4\,torr$
$V_{2}=0.150\,L$
Temperature and amount of gas is constant.
Boyle's law states that
$P_{1}V_{1}=P_{2}V_{2}$ (unchanging T and n)
$\implies P_{2}=\frac{P_{1}V_{1}}{V_{2}}=\frac{59.4\,torr\times35.0\,L}{0.150\,L}$
$=13860\,torr\times\frac{1\,atm}{760\,torr}=18.2\,atm$
(b) $V_{1}=35.0\,L$
$P_{1}=59.4\,torr$
$P_{2}=10.0\,atm= 7600\,torr$
$P_{1}V_{1}=P_{2}V_{2}$ (unchanging T and n)
$\implies V_{2}=\frac{P_{1}V_{1}}{P_{2}}=\frac{59.4\,torr\times35.0\,L}{7600\,torr}$
$=0.274\,L$
