An air bubble rises from the bottom of a well where the temperature is 25°C to the surface where the temperature is 27°C. Find the percent increase in the volume of the bubble if the depth of the well is 5 m. Atmospheric pressure is 101,528 Pascals.
"T_1 = 25 +273 = 298 \\;K \\\\\n\nT_2 = 27 + 273 = 300 \\;K \\\\\n\np_2 = p_{atm}=101528 \\;Pa = 101528 \\;N\/m^2"
Pressure at bottom due to depth of fluid
"p = \u03c1gh \\\\\n\np = 10^3 \\times 9.81 \\times 5 = 49050 \\;N\/m^2"
Total pressure at bottom
"p_1 = p_{atm} + p \\\\\n\n= 101528 +49050 \\\\\n\n= 150578 \\;N\/m^2 \\\\\n\n\\frac{p_1V_1}{T_1} = \\frac{p_2V_2}{T_2} \\\\\n\n\\frac{150578 \\times V_1}{298} = \\frac{101528 \\times V_2}{300} \\\\\n\nV_2 = \\frac{150578}{101528} \\times \\frac{300}{298} \\times V_1 \\\\\n\nV_2 = 1.4929V_1 \\\\\n\n% \\; increase \\; in \\; volume = \\frac{V_2-V_1}{V_1} \\times 100 \\\\\n\n= \\frac{1.4929V_1-V_1}{V_1} \\times 100 \\\\\n\n= 49.295 \\; \\%"
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