An air bubble of volume 15 cc is formed at a depth of 50 m in a lake. If the temperature of the bubble while rising remains constant then what will be the volume of the bubble at the surface?(Given: g=10 m/s2 & atm. pressure = 1.0 * 100000 Pa)
To find the volume of the bubble at the surface, we need to consider the effect of pressure on the volume of gas.
According to Boyle's Law, the volume of a gas is inversely proportional to the pressure exerted on it, when the temperature remains constant. Mathematically, it can be represented as:
V1/P1 = V2/P2
Where:
V1 = initial volume of the gas (15 cc)
P1 = initial pressure on the gas (at depth in the lake)
V2 = final volume of the gas (to be determined)
P2 = final pressure on the gas (at the surface of the lake)
Given:
Pressure at the surface (P2) = atmospheric pressure = 1.0 * 100000 Pa
Acceleration due to gravity (g) = 10 m/s^2
To determine the pressure at depth in the lake (P1), we can use the hydrostatic pressure formula:
P1 = P2 + ρgh
Where:
ρ = density of water = 1000 kg/m^3
g = acceleration due to gravity
h = depth of the bubble in meters
Substituting the given values, we get:
P1 = 1.0 * 100000 + (1000 * 10 * 50)
= 1.0 * 100000 + 500000
= 1.5 * 1000000 Pa
Now, let's calculate the final volume of the bubble (V2), assuming the temperature remains constant:
V1/P1 = V2/P2
15/1.5 * 1000000 = V2/1.0 * 100000
Simplifying the equation:
15 * 1.0 * 100000 = V2 * 1.5 * 1000000
Dividing both sides by 1.5 * 1000000:
V2 = (15 * 1.0 * 100000) / (1.5 * 1000000)
V2 = 1500000 / 1500000
V2 = 1 cc
Therefore, the volume of the bubble at the surface will be 1 cc.
p1•V1 = p2•V2,
ρ•g•h• V1 = p2•V2,
V2 = ρ•g•h• V1/ p2,
ρ = 1000 kg/m^3 is the density of water,
p is the atm. pressure
i am not getting the answer which is 90cc
take p1= ro*g*h + p atmospheric
then u will get it:
(10^3*10*50+10^5)15=10^5*v
v=90cc
njoy...:)