Question

"a balloon contains 2.0 l of air at 101.5 kpa. you squeeze the balloon to a volume of 0.25 l. what is the pressure of air inside the balloon?"

Answer 1

In solving this problem, we assume that the temperature remains constant

At constant temperature, the volume of a gas is inversely proportional to its pressure. This means, as the pressure is increased, the volume decreases and vice versa.

We can, therefore, use the formula

p1v1 = p2v2

Where;
p1 = initial pressure
v1 = initial volume
p2 = final pressure
v2 = final volume

p1 = 101.5 kpa
v1 = 2.0 L
p2 = ?
v2 = 0.25 L

101.5 x 2 = p2 x 0.25

p2 = (101.5 x 2)/0.25 = 812

∴ the pressure of air inside the balloon = 812 kpa

Answer 2

Explanation:

According to Boyle’s law, the volume of a gas at constant temperature varies inversely with the pressure exerted on it. In an application of

-->P1V1=P2V2,

where P is pressure and V is volume.

The balloons containing 14L of air at 760torr. So,

P1=760torr and

V1=14 L

The height of the balloon is equal to, h=10ft which is converted to meter-

10ft=10x0.3048=3.048

And the Density of water (ρ) is 1000, acceleration due to gravity is,

9.8m/s2

By using the equation,

p=ρgh,for the value of

so,p2=1000x9.8x3.048+101325=984torr



So, the volume of the balloon when it is taken to a depth of 10ft.

v2=p1V1/P2

V2=760x14/984=10.8L