Question

A balloon contains 14.0 L of air at 760 torr, What will be the volume of the balloon when it is taken to a depth
of 10 f, in a swimming pool? Assume that the temperature of the air and water are equal. (density: Hg 13.6
g/ml)
a) 110
b) 11.3
c) 10
d) 10.81​

Answer 1

Answer:

D 10.89

Explanation:

Taking P1V1=P2V2

P. pressure

V. volume

P1. 760

V1. 14

10ft 3.048m

density of water. 1000

g. 9.8

so. P2 1000 ×9.8×3.048

(u will get se value in Pascal)

In torr 977

so V2= P1V1 ÷P2

760 ×14÷977=10.89

Thank U

Answer 2

Answer:

In an application of P1V1=P2V2 , where P is pressure and V is volume.

we know that 760 torr= 1 atm,

P1= 1 atm and V1= 14.0 L.

Given that we increase our pressure by submerging in a pool (of water) down 10 ft, so we have h=10 ft.

P2= 1 atm + (Density of water)(Gravitational Constant)(Depth)

= 1+ (1000 kg/m^3)(9.81 m/s^2)(10 ft)(12 in / 1 ft)(2.54 cm/ 1 in)(1 m/ 100 cm)(1 atm/ 101,325 Pa) = 1.295 atm

P1V1=P2V2 ;

 P1V1/P2 = V2 ;

=(1 atm)(14.0L)/(1.295 atm)= 10.8 L