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An Air Bubble of Volume 1.0 Cm3 Rises from the Bottom of a Lake 40 M Deep at a Temperature of 12 °C. to What Volume Does It Grow When It Reaches the Surface, Which is at a Temperature of 35 °C? - Physics

An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

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Solution 1

Volume of the air bubble, V1 = 1.0 cm3 = 1.0 × 10–6 m3

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T1 = 12°C = 285 K

Temperature at the surface of the lake, T2 = 35°C = 308 K

The pressure on the surface of the lake:

P2 = 1 atm = 1 ×1.013 × 105 Pa

The pressure at the depth of 40 m:

P1 = 1 atm + g

Where,

ρ is the density of water = 103 kg/m3

g is the acceleration due to gravity = 9.8 m/s2

P1 = 1.013 × 105 + 40 × 103 × 9.8 = 493300 Pa

We have : `(P_1V_1)/T_1 = (P_2V_2)/T_2`

Where, V2 is the volume of the air bubble when it reaches the surface

`V_2 = (P_1V_1T_2)/(T_1P_2)`

`= ((493300)(1.0 xx 10^(-6))308)/(285 xx 1.013 xx 10^5)`

= 5.263 × 10–6 m3 or 5.263 cm3

Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm3

Solution 2

Volume of the bubble inside,  `V_1 = 1.0 cm^3 = 1xx10^(-6) m^3`

Pressure on the bubble, `P_1` = Pressure of water + Atmospheric pressure

= `pgh + 1.01 xx 10^5 = 1000 xx 9.8 xx 40 + 1.01 xx 10^5`

`=3.92 xx 10^5 + 1.01 xx 10^5 = 4.93 xx 10^5` Pa

Temperature, `T_1 = 12 ^@C = 273 + 12 = 285 K`

Also, pressure outside the lake, `P_2 = 1.01 xx 10^5 n m^(-2)`

Temperature, `T_2 = 35 ^@C = 273 + 35 = 308` K, Volume `V_2` = ?

Now  `(P_1V_1)/T_1 = (P_2V_2)/T_2`

`:. V_2 = (P_1V_1)/T_1. T_2/P_2  = (4.93xx10^5xx1xx10^(-6)xx 308)/(285 xx 1.01 xx 10^5) = 5.3 xx 10^(-6) m^(-3)`

  Is there an error in this question or solution?
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APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 13 Kinetic Theory
Q 5 | Page 334
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