Form the available data : `V_(1) = 600 mL , V_(2) = 640 mL`
`P_(1) = 760 mm` , `P_(2) = ?`
`T_(1) = (25+273) = 298 K` , `T_(2) = (10+273) = 283 K`
According to gas equation, `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `P_(2) = (P_(1)V_(1)T_(2))/(V_(2)T_(1))`
`P_(2) = ((760 mm) xx (600 mL) xx (283 K))/((640 mL) xx (298 K)) = 676.6 mm` of Hg