An air bubble starts rising from the bottom of a lake its diameter is 3.6 mm at the bottom and moving at the surface the depth of the lake is 250 cm and the temperature at the surface is 40 degree celsius what is the temperature at the bottom of the lake given atmospheric pressure is equal to 76 centimetre of hg and g is equal to 980 centimetre. .
per second square. Specific gravity of mercury is equal to 13.6.

Your question is incomplete, you have not given the diameter or volume or bubble when it reaches the surface of the lake. However here is a hint for you.
Given,
Diameter of bubble=3.6 mm
or radius, r=3.6/2=1.8 mm=0.18 cm
Volume of spherical bubble
43πr3=43×227×0.183=0.024 cm3=2.4×10-8m3Similarly you can find the volume of bubble V2 at the surface of lake.Bubble rises to height, d=250 cm=2.5 mTemperature at bottom, T1=?Temperature at surface,T2=400C=40+273=313 KThe pressure on the surface of lake=P2=1 atm=1.013×105 PaThe pressure at bottom=1 atm+dρg=1.013×105 +2.5×103×9.8=125800 PaWe have,P1V1T1=P2V2T2or T1=P1V1T2P2V2Now putting the values you can get the answer i.e. T1.

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