A bubble of gas released at the bottom of a lake increases 8 times its original volume when it reaches the surface . Assuming that atmospheric pressure exerted by a column of water ,10m height the depth of lake is ?

Question

Answer

Throughout the entire process, the number of moles of the gas bubble remains same. So, we can apply the ideal gas equation PV = constant.

Let us assume that the bubble is h meters below the surface of the water, and P₀ is the atmospheric pressure.

Pressure only due to water at a depth of h meters would be equal to ρgh. Also, atmospheric pressure would be acting on the bubble. So, the total pressure on the gas bubble at that depth = P₀ + ρgh = ρg*10 + ρgh = ρg(h+10).

At the surface of water, only atmospheric pressure would be acting on the bubble i.e. P = P₀ = ρg*10

As PV = constant, ρg(h+10) * V= ρg10 * 8V → h+10 = 80 → h = 70 meters.

Answer

PV(underwater) =P'V'(surface) (x+10)ρgV=8V×10ρg , x+10=80 , x=70, so ans is 70m

Answer

Pressure and Volume Relationship

According to Boyle's Law, there is an inverse relationship between the pressure and volume of a gas at constant temperature. This means that as the volume of a gas increases, the pressure it exerts decreases, and vice versa.

Bubble Expansion

In this scenario, a bubble of gas is released at the bottom of a lake and rises to the surface. As the bubble rises, it experiences a decrease in pressure due to the column of water above it. This decrease in pressure causes the bubble to expand.

Volume Increase

The problem states that the bubble increases in volume by 8 times its original size when it reaches the surface. This means that the final volume of the bubble is 9 times its original volume (original volume + 8 times original volume). Let's denote the original volume as V and the final volume as 9V.

Pressure and Volume Relationship

We can use Boyle's Law to relate the initial and final volumes and the corresponding pressures. Boyle's Law states that P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume.

In this case, since the temperature remains constant, we can write the equation as P1 * V = P2 * (9V).

Atmospheric Pressure

The problem also provides information about the atmospheric pressure exerted by a column of water with a height of 10m. The pressure exerted by a column of fluid is given by the equation P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the column.

Given the height of the column as 10m, we can substitute the values into the equation to find the atmospheric pressure.

Calculation

Let's assume the initial pressure of the bubble is P1 and the atmospheric pressure is Pa.

According to the information given in the problem, the final pressure when the bubble reaches the surface is Pa + ρgh.
Using Boyle's Law, we can set up the equation P1 * V = (Pa + ρgh) * (9V).

Simplifying the equation, we get P1 = 9(Pa + ρgh) - Equation 1.

To find the depth of the lake, we need to equate the pressure at the bottom of the lake to the atmospheric pressure plus the pressure exerted by the column of water.

P1 = Pa + ρgh - Equation 2.

By equating the equations 1 and 2, we can solve for h, the depth of the lake.

Conclusion

By setting up the equations using Boyle's Law and the pressure equation for a column of fluid, we can determine the depth of the lake based on the expansion of the gas bubble.

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