A `4.0 dm^(3)` flask containing `N_(2) at 4` bar was connected to a `6.0 dm^(3)` flask containing helium at `6` bar , and the gases were allowed to mix isothermally. The total pressure of the resulting mixture will be

Options

A. 10.0 bar
B. 5.2 bar
C. 1.6 bar
D. 5.0 bar

Answer

B

Explanation

At constant temperature
`P_(1)V_(1) + P_(2) V_(2) = P_(3) (V_(1) + V_(2))`
`(4.0 "bar") (4.0 dm^(3)) + (6.0 "bar") (6.0 dm^(3))`
`= P_(3) (4.0 dm^(3) + 6.0 dm^(3))`
`:. P_(3) = (16 + 36)/(10) = (52)/(10) = 5.2` Bar