Question

A balloon contains $14.0L$ of air at $760torr$. What will be the volume of the balloon when it is taken to a depth of $10ft$. in a swimming pool? Assume that the temperature of the air and water are equal.

(Density: $Hg=13.6g/mL.)$

• A. $11.0$
• B. $11.3$
• C. $10$
• D. $10.8$

Answer 1

The correct option is D $10.8$

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