Law of Boyle

Tutorial 3

        

A gas bubble has a volume of $ 0.550\;mL $ at the bottom of a lake where the pressure is $ 4.50 \cdot 10^3 \; hPa $. What will be its volume at the surface where the pressure is no more than $ 1.00 \; atm $, the temperature remaining constant? Solution (Click): Conversion to $Pa$: ..........$4.5\cdot 10^3\;hPa =4.5\cdot 10^5\;Pa$
$1.00\;atm=1.0133\cdot 10^5\;Pa$ Law of Boyle : ..........$P_1\cdot V_1=P_2\cdot V_2$ Calculation: ..........$V_1=\frac{P_2\cdot V_2}{P_1}$ = $\frac{0.550\cdot 4.5\cdot 10^5}{1.0133\cdot 10^5} = 2.44\;mL$