Answer
(a) 9.5 mL
(b) 950 mL
(c) 65.1 mL
(d) 7220 mL or 7.22 L
(e) $6.56\times10^{6}\,mL$ or 6560 L
Work Step by Step
$V_{1}=95.0\,mL$
$P_{1}=0.500\,atm$
Temperature and amount of gas is constant.
Boyle's law states that
$P_{1}V_{1}=P_{2}V_{2}$ (unchanging T and n)
$\implies V_{2}=\frac{P_{1}V_{1}}{P_{2}}$
(a) When $P_{2}=5.00\,atm$,
$V_{2}=\frac{0.500\,atm\times 95.0\,mL}{5.00\,atm}=9.5\,mL$
(b) When $P_{2}=0.0500\,atm$,
$V_{2}=\frac{0.500\,atm\times 95.0\,mL}{0.0500\,atm}=950\,mL$
(c) When $P_{2}=555\,torr=555\,torr\times\frac{1\,atm}{760\,torr}=0.730\,atm$
$V_{2}=\frac{0.500\,atm\times 95.0\,mL}{0.730\,atm}=65.1\,mL$
(d) When $P_{2}=5.00\,torr=5.00\,torr\times\frac{1\,atm}{760\,torr}=0.00658\,atm$
$V_{2}=\frac{0.500\,atm\times 95.0\,mL}{0.00658\,atm}=7220\,mL=7.22\,L$
(e) When $P_{2}=5.5\times10^{-2}\,torr=0.055\,torr\times\frac{1\,atm}{760\,torr}=7.237\times10^{-5}\,atm$
$V_{2}=\frac{0.500\,atm\times 95.0\,mL}{7.237\times10^{-5}\,atm}=6.56\times10^{6}\,mL=6560\,L$
