What will be the pressure of the gas mixture when `0.5 L` of `H_(2)` at `0.8` bar `2.0 L` of oxygen at `0.7` bar are introduced in a `1L` vessel at `27^(@)C` ?

Options

A. 1.8 bar
B. 2.8 bar
C. 3.0 bar
D. 5 bar

Answer

A

Explanation

Partial pressure of hydrogen gas
`V_(1)=0.5 L " " V_(2)=1.0 L`
`P_(1)=0.8 "bar" " " P_(2) = ?`
Applying Boyle's law (constant T and n)
`P_(1)V_(1)=P_(2)V_(2)` or `P_(2)=(P_(1)V_(1))/(V_(2))`
`therefore P_(2)=((0.8 "bar")xx(0.5L))/((1.0L))=0.40` bar
Partial pressure of oxygen gas,
`V_(1)=2.0 L, " " V_(2)=1.0 L`
`P_(1)=0.7 "bar" " " P_(2) = ?`
or `P_(2)=(P_(1)V_(1))/(V_(2))=((0.7" bar")xx(2.0L))/((1.0L))=1.40` bar
Pressure of the gas mixture,
`P_("mix")=P_(H_(2))+P_(O_(2))=0.40+1.40=1.80` bar