Home
Class 12
CHEMISTRY
At constant temperature if an air bubble...

At constant temperature if an air bubble present at the bottom of a lake at 8 atm pressure and with radius 0.1 cm rises to the surface then its new radius will become

A

0.4 cm

B

0.3 cm

C

0.2 cm

D

0.1 cm

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the new radius of an air bubble that rises from the bottom of a lake to the surface, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is held constant. ### Step-by-Step Solution: 1. **Identify Initial Conditions**: - Initial pressure (P1) = 8 atm (at the bottom of the lake) - Initial radius (r1) = 0.1 cm 2. **Calculate Initial Volume (V1)**: - The volume of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] - Therefore, the initial volume (V1) can be calculated as: \[ V1 = \frac{4}{3} \pi (0.1)^3 = \frac{4}{3} \pi (0.001) = \frac{4\pi}{3000} \text{ cm}^3 \] 3. **Identify Final Conditions**: - Final pressure (P2) = 1 atm (at the surface of the lake) 4. **Apply Boyle's Law**: - According to Boyle's Law: \[ P1 \cdot V1 = P2 \cdot V2 \] - Rearranging gives: \[ V2 = \frac{P1 \cdot V1}{P2} \] 5. **Substitute Values**: - Substitute P1, V1, and P2 into the equation: \[ V2 = \frac{8 \cdot V1}{1} = 8 \cdot V1 \] 6. **Calculate New Volume (V2)**: - Since we know that V1 = \(\frac{4}{3} \pi (0.1)^3\): \[ V2 = 8 \cdot \left(\frac{4}{3} \pi (0.1)^3\right) = \frac{32}{3} \pi (0.1)^3 \] 7. **Relate Volume to Radius**: - The volume of the new bubble (V2) can also be expressed in terms of the new radius (r2): \[ V2 = \frac{4}{3} \pi r2^3 \] 8. **Set the Two Volume Equations Equal**: - Set the two expressions for V2 equal: \[ \frac{32}{3} \pi (0.1)^3 = \frac{4}{3} \pi r2^3 \] 9. **Cancel Common Terms**: - Cancel \(\frac{4}{3} \pi\) from both sides: \[ 32 \cdot (0.1)^3 = r2^3 \] 10. **Calculate r2**: - Calculate \(r2^3\): \[ r2^3 = 32 \cdot 0.001 = 0.032 \] - Now take the cube root to find r2: \[ r2 = \sqrt[3]{0.032} \approx 0.2 \text{ cm} \] ### Final Answer: The new radius of the air bubble when it rises to the surface is approximately **0.2 cm**.
Promotional Banner

Similar Questions

Explore conceptually related problems

As an air bubble comes from the bottom of a lake to the top, its radius "_____"

Assertion. A bubble comes from the bottom of a lake to the top. Reason. Its radius increases.

When an air bubble rise from the bottom of a deep lake to a point just below the water surface, the pressure of air inside the bubble

Radius of an air bubble at the bottom of the lake is r and it becomes 2 r when the air bubbles rises to the top surface of the lake. If P cm of water be the atmospheric pressure, then the depth of the lake is

The radius of an air bubble is increasing at the rate of 0.5 cm/sec. At what rate is the volume of the bubble increasing when the radius is 1 cm?

As an air bubble rises from the bottom of a lake to the surface, its volume is doubled. Find the depth of the lake. Take atmospheric pressure = 76 cm of Hg.

The temperature of an air bubble while rising from bottom to surface of a lake remains constant but its diameter is doubled if the pressure on the surface is equal to h meter of mercury column and relative density of mercury is then the depth of lake in metre is -

A spherical air bubble is rising from the depth of a lake when pressure is P atm and temperature is T K. The percentage increase in its radius when it comes to the surface of lake will be: (Assume temperature and pressure at the surface to be respectively 2 TK and (P)/(4) )