Question

(5 Points)

a) At ground level, the pressure of the helium in a balloon is 1x105
Pa. The volume occupied by the helium is 9.6m The balloon is
released and it rises quickly through the atmosphere. Calculate
the pressure of the helium when it occupies a volume of 12m3.
(3 Marks)
b) A box is 15m below the surface of the sea. The density of sea-
water is 1020 kg/m.
Calculate the pressure on the box due to the sea-water.
(2 Marks)

Answer 1

1.
`P_(2)` = 8 x
`10^(4)` Pa

2. P = 1.5 x
`10^(5)` N/
`m^(2)`

Step-by-step explanation:

1. From Boyles' law;

`P_(1)`
`V_(1)` =
`P_(2)`
`V_(2)`

`P_(1)` = 1 x
`10^(5)` Pa

`V_(1)` = 9.6
`m^(3)`

`V_(2)` = 12
`m^(3)`

Thus,

1 x
`10^(5)` x 9.6 =
`P_(2)` x 12

`P_(2)` =
`(100000 x 9.6)/(12)`

= 80000

`P_(2)` = 8 x
`10^(4)` Pa

2. Pressure, P = ρhg

where: ρ is the density of the fluid, h is the height/ depth and g is the acceleration due to gravity (9.8 m/
`s^(2)`).

Thus,

P = 1020 x 15 x 9.8

= 149940

P = 1.5 x
`10^(5)` N/
`m^(2)`