A quantity of gas is collected in a gradutated tube over the mercury. The volume of the gas at `20^(@)C` is `50.0 mL` and the level of the mercury in the tube is `100 mm` above the outside mercury level. The barometer reads `750 mm`. Volume at `STP` is

Options

A. 40 mL
B. 42 mL
C. 39.8 mL
D. 60 mL

Answer

C

Explanation

`T_(1) = 20 + 273 = 293 K, " " T_(2) = 273 K`
`V_(1) = 50 mL. " " V_(2) = ?`
`P_(1) = 750 - 100 = 650 mm Hg, P_(2) = 760 mm Hg`
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))`
`V_(2) = (P_(1) V_(1)T_(2))/(T_(1)P_(2)) = (50 xx 650 xx 273)/(293 xx 760)`
`= 39.8 mL`