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  • An air bubble starts rising from the bottom of a lake, its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and temperature at the surface is

An air bubble starts rising from the bottom of a lake, its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 250 cm and temperature at the surface is \small 40^{\circ}C. What is the temperature (in K) at the bottom of the lake?

Option: 1

273.54


Option: 2

283.17


Option: 3

293.43


Option: 4

333.45


Answers (1)

best_answer

During this process mole is not changing,

we know :

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Now find the volumes and pressure,

Volume of the bubble at the bottom =\frac{4}{3}\pi r^3
                                                  =\frac{4}{3}\pi (0.18)^3
Pressure on the bubble = Atmospheric pressure +  pressure due to lake.

\textrm{P = 76mm of Hg}+\rho gh
Thus,

P=(76\times 13.6 \times 980+1\times980\times250)

Again, the volume of the bubble at the surface =\frac{4}{3}\pi (0.2)^3
And, pressure on the bubble = atmospheric pressure
Thus, P=76\times 13.6 \times 980

Now, we know:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

\frac{(76\times13.6\times980+250\times1\times980)\times\left ( \frac{4}{3} \right )\pi (0.18)^3}{T_1}=\frac{(76\times 13.6 \times 980)\times \left ( \frac{4}{3} \right )\pi (0.2)^3}{313}
Thus, T_1=283.17K\: \: or\: \: 10.17^{\circ}C
 

Posted by

Rishabh

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