Answer
(a) $75\,atm=57000\,Torr$
$75\,atm=76\,bar$
(b) $580\,Torr=0.76\,atm$
$580\,Torr=770\,millibars$
(c) $5.2\,atm=5.3\times10^{5}\,Pa=530\,kPa$
(d) $920\,Torr=1.2\times10^{5}\,Pa$
$920\,Torr=1.2\,atm$
Work Step by Step
We find:
(a) $75\,atm=75\,atm\times\frac{760\,Torr}{1\,atm}=57000\,Torr$
$75\,atm=75\,atm\times\frac{1.01325\,bar}{1\,atm}=76\,bar$
(b) $580\,Torr=580\,Torr\times\frac{1\,atm}{760\,Torr}=0.76\,atm$
$580\,Torr=580\,Torr\times\frac{1\,atm}{760\,Torr}\times\frac{1.01325\,bar}{1\,atm}\times\frac{10^{3}\,millibar}{1\,bar}$
$=770\,millibars$
(c) $5.2\,atm=5.2\,atm\times\frac{1.01325\times10^{5}\,Pa}{1\,atm}=5.3\times10^{5}\,Pa$
$=5.3\times10^{5}\,Pa\times\frac{100\,kPa}{1\times10^{5}\,Pa}=530\,kPa$
(d) $920\,Torr=920\,Torr\times\frac{1.01325\times10^{5}\,Pa}{760\,Torr}=1.2\times10^{5}\,Pa$
$920\,Torr=920\,Torr\times\frac{1\,atm}{760\,Torr}=1.2\,atm$
