#### 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 $\dpi{100} \small 40^{\circ}C$. What is the temperature (in K) at the bottom of the lake?Option: 1 273.54Option: 2 283.17Option: 3 293.43Option: 4 333.45

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, $T_1=283.17K\: \: or\: \: 10.17^{\circ}C$