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

The depth of the lake is 67.164 meters.

Step-by-step explanation:

The combined gas equation is,

`(P_1V_1)/(T_1)=(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:

`(P_1* V)/(277.7 K)=(0.980 atm* 8.00* 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* 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* g`

`6.496 atm=h\rho* g`

`6.496 * 101325 Pa=h\1000 g/m^3* 9.8 m/s^2`

h = 67.164 m

The depth of the lake is 67.164 meters.