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Virty [35]
3 years ago
7

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 wh

ere 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?
Chemistry
1 answer:
Masja [62]3 years ago
4 0

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.

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