Two gases bulbs `A` and `B` are connected by a tube having a stopcock. Bulb `A` has a volume of `100 mL` and contains `H_(2)` gas . After opening the gas from `A` to the evacuated bulb `B` , the pressure falls down by `40%`. The volume `(mL) of `B` must be

Options

A. 75
B. 150
C. 125
D. 200

Answer

B

Explanation

Applying Boyle's law
`P_(A) V_(B) = (0.40 P_(A)) (V_(A) + V_(B))`
`P_(A) xx 100 = 0.40 P_(A) (100 + V_(B))`
or `100 = 0.4 (100 + V_(B)) " or " 100 + V_(B) = 250`
`:. V_(B) = 150 mL`