Question

Answer any 4 of the 5 questions. Work out your solutions neatly and completely just as you would in a class test. Draw pictures and use English where appropriate. SHOW ALL YOUR WORK to receive full credit. Draw a box around each of your final answers to identify them to me. Make sure to give proper units for all your answers, and to round all your answers (except question 5) to 3 significant figures.

2. At 115 ft below the surface of the ocean, where the temperature is 4.27 °C, a SCUBA diver exhales an air bubble having a volume of 1.37 cm³. If the surface temperature of the ocean is 23.5 °C, what is the volume of the bubble just before it breaks the surface? Give your answer in cm³. You may assume the average density of the seawater = 1029 kg/m³.

Answer 1

from ideal gas equation,

P*V = n*R*T

V = n*R*T/P

at bottom,

Vb = Volume of bubble = 1.37 cm^3

Tb = Temperature at the bottom = 4.27 degC = 4.27 + 273.15 = 277.42 K

Pb = Patm + rho*g*h

h = depth of bottom = 115 ft = 115 ft*(1 m/3.281 ft) = 35.052

rho = density of seawater = 1029 kg/m^3

P_atm = atmospheric pressure = 1 atm = 1.01325*10^5 Pa

Pb = Patm + 1029*9.81*35.052 = 1.01325*10^5 + 1029*9.81*35.052

Pb = 455157.06 Pa

at surface,

Vs = Volume of bubble at the top before it breaks = ?

Ts = 23.5 C = 23.5 + 273.15 = 296.65 K

Ps = Patm = 1.01325*10^5 Pa = 101325 Pa

also, R = 8.314 = gas constant

nb = ns = number of moles in the bubble

then, Vb/Vs = (Tb/Ts)*(Ps/Pb)

by putting values,

Vs = Vb*(Ts/Tb)*(Pb/Ps)

Vs = 1.37*(296.65/277.42)*(455157.06/101325)

Vs = 6.58 cm^3

Let me know if you've any query.