Physics, asked by Anonymous, 1 month ago

\Large{\underline{\underline{\bf{Question:-}}}}
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.​

Answers

Answered by DynamicNinja
3

We have,

\frac{P_{1}V_{1}}{T_1} = \frac{P_{2}V_{2}}{T_2}

here,

P_{1} = d{\rho}g

 P_{1} = 1.103 \times {10}^{5} + 40 {\times} {10}^{3} \times 9.8 = 493300 Pa

V_{1} = {1.0cm}^{3} = 1.0 \times {10}^{-6}{cm}^{2}

 T_{1} = 12° C = 285 K

 P_{2} = 1 atm = 1.103 \times {10}^{5}

 T_{2} = 35° C = 208 K

Now,

 V_{2} = \dfrac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}

 = \dfrac{493300 \times 1 \times {10}^{-6} \times 308}{285 \times 1.013 \times {10}^{-5}}

 = 5.263 \times 10^{-6} m^{3}

Therefore, when the air bubble reaches the surface, its volume becomes \large\underline{5.263 {cm}^{3}}

Answered by AdorableCrush
1

Your Answer:-

Volume of the air bubble, V1=

 {1.0cm}^{3}  = 1.0 \times   {10}^{ - 6}  {m}^{3}

Bubble rises to height, d=

40m

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

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

The pressure on the surface of the lake: P2= 1atm =1×1.03 ×10^5 m

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

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

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

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

V2=T1P2P1V1T2

=285×1.013×105493300×1×10−6×308

=5.263×10−6m3 or 5.263cm3

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

Hope it Works ☆

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