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Q12.5: NCERT Physics - Part 2 KINETIC THEORY 12.5 Solution: An air bubble of volume 1.0 cm_ rises from the bottom of a lake …

An air bubble of volume 1.0 cm_ 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?

Step-by-Step Verified Solution

**Answer:** Final volume = 5.3×10−65.3 \times 10^{-6} m_ =5.3= 5.3 cm_ **Step-by-Step Explanation:** Initial volume V1=1.0V_1 = 1.0 cm_ =1.0×10−6= 1.0 \times 10^{-6} m_. Initial temperature T1=12∘C=285T_1 = 12^\circ\text{C} = 285 K. Initial pressure at depth 40 m:
P1=Patm+ρgh=1.013×105+1000×9.8×40=4.93×105 PaP_1 = P_{atm} + \rho g h = 1.013 \times 10^5 + 1000 \times 9.8 \times 40 = 4.93 \times 10^5 \text{ Pa}
Final temperature T2=35∘C=308T_2 = 35^\circ\text{C} = 308 K. Final pressure P2=Patm=1.013×105P_2 = P_{atm} = 1.013 \times 10^5 Pa. Using combined gas law:
P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}
V2=V1×P1P2×T2T1=1.0×10−6×4.93×1051.013×105×308285=5.3×10−6 m3=5.3 cm3V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} = 1.0 \times 10^{-6} \times \frac{4.93 \times 10^5}{1.013 \times 10^5} \times \frac{308}{285} = 5.3 \times 10^{-6} \text{ m}^3 = 5.3 \text{ cm}^3

All Questions in this Chapter

Q12.1 Q12.2 Q12.3 Q12.4 Q12.5 Q12.6 Q12.7 Q12.8 Q12.9 Q12.10

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