Answer:
56.4 m
Explanation:
volume increases by factor of 6, i.e [tex]\frac{V2}{V1}[/tex] = 6
Initial temperature T1 at bottom of lake = 5.24°C = 278.24 K
Final temperature T2 at top of lake = 18.73°C = 291.73 K
NB to change temperature from °C to K we add 273
Final pressure P2 at the top of the lake = 0.973 atm
Initial pressure P1 at bottom of lake = ?
Using the equation of an ideal gas
[tex]\frac{P1V1}{T1}[/tex] = [tex]\frac{P2V2}{T2}[/tex]
P1 = [tex]\frac{P2V2T1}{V1T2}[/tex] = [tex]\frac{0.973*6*278.24}{291.73}[/tex]
P1 = 5.57 atm
5.57 atm = 5.57 x 101325 = 564380.25 Pa
Density Ρ of lake = 1.02 g/[tex]cm^{3}[/tex] = 1020 kg/[tex]m^{3}[/tex]
acceleration due to gravity g = 9.81 [tex]m/s^{2}[/tex]
Pressure at lake bottom = pgd
where d is the depth of the lake
564380.25 = 1020 x 9.81 x d
d = [tex]\frac{564380.25}{10006.2}[/tex] = 56.4 m