Question

The volume of a bubble that starts at the bottom of a lake at 4.55°C increases by a factor of 8.00 as it rises to the surface where the temperature is 18.05°C and the air pressure is 0.980 atm. Assuming that the density of the lake water is 1.00 g/cm3, determine the depth of the lake?

Answer 1

Answer:

The depth of the lake is 67.164 meters.

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas in bubble= ?

$P_2$ = final pressure of gas = 0.980 atm

$V_1$ = initial volume of gas = $V$

$V_2$ = final volume of gas = 8.00 × V

$T_1$ = initial temperature of gas = $4.55^oC=273.15+4.55=277.7 K$

$T_2$ = final temperature of gas = $18.05^oC=273.15+18.05=291.2 K$

Now put all the given values in the above equation, we get:

$\frac{P_1\times V}{277.7 K}=\frac{0.980 atm\times 8.00\times V}{291.2 K}$

$P_1=7.476 atm$

pressure of the gas in bubble initially is equal to the sum of final pressure and pressure exerted by water at depth h.

$P_1=P_2+h\rho\times g$

Where :

$\rho =$ density of water = $1.00 g /cm^3=1000 g/m^3$

g = acceleration due gravity = $9.8 m/s^2$

$7.476 atm=0.980 atm +h\rho\times g$

$6.496 atm=h\rho\times g$

$6.496 \times 101325 Pa=h\1000 g/m^3\times 9.8 m/s^2$

h = 67.164 m

The depth of the lake is 67.164 meters.