Question

collapsible balloon is inflated to a volume of 10 lit at a pressure of 1atm .when the balloon is immersed in a lake to its bottom,its volume is reduced to 1.25 lit.Assuming 1 atm pressure is equivalent to 10 m column of H2O and no change in temperature,then the depth of the lake will be?​

Answer 1

Answer:

The depth of the lake is 80 m.

Explanation:

Initial pressure of the inflated balloon = $P_1= 1atm$

Initial volume of the inflated balloon = $V_1=10 L$

Final pressure of the inflated balloon = $P_2= ?$

Final volume of the inflated balloon = $V_2=1.25 L$

Applying Boyle's law:

$P_1V_1=P_2V_2$ (constant temperature)

$P_2=\frac{P_1V_1}{V_2}=\frac{1 atm \times 10 L}{1.25 L}$

$P_2=8 atm$

The pressure on the balloon when immersed in lake to its bottom is 8 atm.

1 atm pressure = 10 m column of water

Then 8 atm pressure corresponds to : 8 × 10 m = 80 m

The depth of the lake is 80 m.